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Pure Math Grade 11

Workbook

Alan Appleby

Robert Letal

Greg Ranieri

Publisher: Absolute Value Publications

Authors: Alan Appleby, Robert Letal, Greg Ranieri

Graphics: Greg Ranieri Danielle Lindsay-Chung (Front Cover)

Copyright © 1997, 1998, 2001, 2002, 2003, 2004, 2006

All rights reserved. This book is not covered under the Cancopy agreement. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher.Printed in Canada.

ISBN 0-9737459-2-4

For information contact:

Absolute Value Publications Inc.P.O. Box 71096 8060 Silver Springs Blvd. N.W.Calgary, AlbertaT3B 5K2

Bus: (403) 313-1442 Fax: (403) 313-2042

e-mail: [email protected] site: www.absolutevaluepublications.com

About the Pure Math Grade 11 Workbook

• The Pure Math Grade 11 Workbook is a complete resource for the Alberta and British Columbia Curriculum. Each curricular topic is subdivided into individual lessons. Most lessons can be covered in one hour (plus homework time), but some may require more time to complete. Lessons are composed of four parts:

• Warm-Ups - which could be review, preview or investigative work.• Class Examples - which are intended to be teacher led.• Assignments - short response, extended response, multiple choice and

numeric response questions are provided for student practice. • Answer Key - answers to the assignment questions.

The Teacher Solution Manual is a copy of the workbook with detailed solutions to all the Warm-ups, Class Examples, and Assignments.

The Student Solution Manual contains detailed solutions to all the Warm-ups, Class Examples, and Assignments without the questions.

The material has been used in several schools and adjustments have been made based on student and teacher feedback. Every effort has been made to achieve a high standard of accuracy in the writing of the workbook, solution manual and answer key. We accept full responsibility for any errors and welcome feedback.

Acknowledgments

We would like to acknowledge the following people for their contributions in the production of this workbook:

• Bruce O’Neil, Terri Marchand, Lynn Darr, Janell Arbour, Linda Binder, Wes Reib, and their students for piloting the material and providing valuable feedback

• our students for their suggestions, opinions, and encouragement.

• David McDougal for his feedback

• Tony Audia, for his support.

Most of all, we would like to thank our families, especially our respective wives, Susan, Linda, and Rose, for their patience and understanding.

Advantages for Students

• Students write in the workbook so that the math theory, worked examples, and assignments are all in one place for easy review.

• Students can write on the diagrams and graphs.

• Provides class examples and assignments so that students can use their time more efficiently by focusing on solving problems and making their own notes.

• For independent learners the workbook plus solution manual fosters self-paced learning.

• Encourages group learning and peer tutoring.

• The design of the workbook ensures that students are fully aware of the course expectations.

• We hope you enjoy using this workbook and that with the help of your teacher you realize the success that thousands of students each year are achieving using the workbook series.

• Written by teachers experienced in preparing students for success in high school and diploma examinations.

• Comprehensively covers the Alberta and British Columbia curriculum.

• Can be used as the main resource, or in conjunction with a textbook, or for extra assignments or review.

• Lessons have been thoroughly piloted in the classroom and modified based on student and teacher feedback.

• Reduces school photocopying costs and time.

• Allows for easy lesson planning in the case of teacher or student absence.

• May be purchased through the Learning Resource Centre (LRC).

Advantages for Teachers

Teacher, student, and parent responses to the workbook series have been very positive. We welcome your feedback. It enables us to produce a high quality resource meeting our goal of success for both teachers and students.

Pure Math Grade 11 Workbook

Table of Contents Linear and Nonlinear Systems

Lesson 1......... Solving Linear Systems of Equations by Graphing.......................................1Lesson 2......... Determining the Number of Solutions to a System of Linear Equations........... 7Lesson 3......... Solving Linear Systems of Equations by Substitution.................................13Lesson 4......... Solving Linear Systems of Equations by Elimination................................. 19Lesson 5......... Applying Systems of Linear Equations - Part One......................................27Lesson 6......... Applying Systems of Linear Equations - Part Two..................................... 35Lesson 7......... Solving Linear Systems of Equations with Three Variables..........................41Lesson 8......... Solving Nonlinear Systems of Equations..................................................51Lesson 9......... Graphing Linear Inequalities in Two Variables...........................................57

Functions

Lesson 1......... Review of Functions, Domain, and Range................................................ 69Lesson 2......... Operations with Functions.....................................................................77Lesson 3......... Composition of Functions.....................................................................85Lesson 4......... The Inverse of Function - Part One..........................................................95Lesson 5......... The Inverse of Function - Part Two........................................................103Lesson 6......... Zeros of a Function.............................................................................111

Quadratic Functions, Equations, and Inequalities

Lesson 1......... Analyzing Quadratic Functions - Part One...............................................119Lesson 2......... Analyzing Quadratic Functions - Part Two.............................................. 127Lesson 3......... Equations and Intercepts from the Vertex and the Point.............................. 135Lesson 4......... Converting from General Form to Standard Form by Completing the Square. 143Lesson 5......... Roots of Quadratic Equations - The Quadratic Formula.............................. 151Lesson 6......... Roots of Quadratic Equations - The Discriminant......................................157Lesson 7......... Applications of Quadratic Functions - A Graphical Approach......................165Lesson 8......... Applications of Quadratic Functions - An Algebraic Approach.................... 173Lesson 9......... Quadratic Inequalities...........................................................................181

Polynomial Functions, Equations, and Inequalities

Lesson 1......... Polynomial Functions.........................................................................189Lesson 2......... The Division Algorithm and Synthetic Division.......................................197Lesson 3......... The Remainder Theorem and the Factor Theorem......................................203Lesson 4......... Factoring Polynomial Expressions.........................................................211Lesson 5......... Solving Polynomial Equations..............................................................217Lesson 6......... Graphing Polynomial Functions - Part One.............................................223Lesson 7......... Graphing Polynomial Functions - Part Two............................................ 229Lesson 8......... Graphing Polynomial Functions - Part Three...........................................237Lesson 9......... Polynomial Functions with a Leading Coefficient other than ±1..................243Lesson 10........Polynomial Inequalities....................................................................... 251

Absolute Value, Radical, and Rational EquationsLesson 1......... Absolute Value Equations - Part One......................................................259Lesson 2......... Absolute Value Equations - Part Two.....................................................267Lesson 3......... Radical Equations............................................................................... 273Lesson 4......... Rational Functions............................................................................. 285Lesson 5......... Rational Equations..............................................................................295Lesson 6......... Inverse of a Rational Function.............................................................. 301

Mathematical ReasoningLesson 1......... Inductive Reasoning, Conjectures, and Counterexamples ...........................307Lesson 2......... Deductive Reasoning...........................................................................315Lesson 3......... Connecting Words - “And”, “Or”, & “Not”..............................................321Lesson 4......... Venn Diagrams - Part One....................................................................329Lesson 5......... Venn Diagrams - Part Two...................................................................335Lesson 6......... If/Then Statements..............................................................................343Lesson 7......... Direct Proof.......................................................................................351

Circle GeometryLesson 1......... Circles and Chords..............................................................................355Lesson 2......... Circles and Angles..............................................................................365Lesson 3......... Cyclic Quadrilaterals........................................................................... 377Lesson 4......... Circles and Tangents........................................................................... 385Lesson 5......... Polygons.......................................................................................... 393Lesson 6......... Relations Between Arcs, Sectors, and Angles...........................................401

Coordinate Geometry and TrigonometryLesson 1......... Coordinate Geometry Review................................................................407Lesson 2......... Distances Between Points and Lines.......................................................415Lesson 3......... Distances Between Parallel Lines ..........................................................421Lesson 4......... Coordinate Geometry and the Circle ...................................................... 425Lesson 5......... Equation of a Circle ........................................................................... 431Lesson 6......... Trigonometry Review..........................................................................435Lesson 7......... Law of Sines and Law of Cosines..........................................................441

Personal FinanceLesson 1......... Income............................................................................................. 447Lesson 2......... Payroll Deductions Part One - CPP, EI, etc.............................................451Lesson 3......... Payroll Deductions Part Two - Income Tax..............................................457Lesson 4......... Bank Statements.................................................................................463Lesson 5......... Investing Money - Simple Interest and Compound Interest......................... 467Lesson 6......... Using TVM (on a Calculator) to Solve Investment Problems......................475Lesson 7......... Annuities..........................................................................................483Lesson 8......... Loans and Consumer Credit..................................................................487Lesson 9......... Mortgages and Property Tax................................................................. 495Lesson 10........Foreign Exchange...............................................................................503Lesson 11........Budgeting..........................................................................................507

Pure Math Grade 11 Workbook ii Table of Contents

Linear and Nonlinear Systems Lesson #1:Solving Linear Systems of Equations by Graphing

Linear and Nonlinear Systems of Equations

• A system of equations consists of two or more equations that are considered together. • A linear system of equations is a system in which each equation represents a straight line. • A nonlinear system of equations is a system in which at least one equation does not

represent a straight line.• The solution to a system of equations must satisfy each equation in the system.

Warm-Up

-5 5x

y

2

–2

A system of equations has been represented on the grid. The system has an integral solution.

a) State the solution x = _____ , y = _____ .

b) Write the solution as an ordered pair.

Class Ex. #1 Consider the system of equations 2x + y = 2, x - 3y = 15.

5 10-5-10

5

10

-5

-10

x

y

a) Solve the system of equations graphically by:

• writing both equations in slope y-intercept form

• making a table of values and plotting the points.

b) Verify the solution by replacing the values in the original equations.

Class Ex. #2 The following system of equations is given: x - y = 7, x + 5y = –5a) Use the x and y-intercepts to graph each equation

5 10-5-10

5

10

-5

-10

x

y

and hence solve the system.

b) Verify the solution.

Solving a System of Equations using a TI-83 Plus Graphing Calculator

Check that the calculator is in “Function” mode.Use the following procedure to find the solution to a system of equations:

1. Write each equation in terms of y.2. Access the “Y= editor” by pressing the Y= key.

3. Enter one equation in Y1 .

4. Enter the other equation in Y2 .

5. Press the GRAPH key to display the graphs.

6. Access the intersect command by pressing 2nd then TRACE and scroll down to “intersect”.

The calculator will return to the display window with the graphs.

7. The calculator will display “First curve?”. Use the cursor key, if necessary, to select the first graph and then press ENTER .

8. The calculator will display “Second curve?”. Use the cursor key, if necessary, to select the second graph and then press ENTER .

9. The calculator will display “Guess?”. Press ENTER .

2 Linear and Nonlinear Systems Lesson #1: Solving Linear Systems by Graphing

Note If a decimal value appears for the x and/or y coordinates, then the x and/or y value can be converted to an exact value (as long as it is not irrational and within the limitations of the calculator) by using the following procedure:

For the x-coordinate

1. Exit the graphing screen by pressing CLEAR twice.

2. Press X,T,q,n key, then press ENTER to import the x-coordinate.3. To display the exact value,

Press MATH , select “Frac”, then press ENTER .

For the y-coordinate

Except for step 2, the instructions to import the y-coordinate are the same as above.For step 2, press ALPHA 1 ENTER to import the y-coordinate value. Then proceed to step 3 above.

Class Ex. #3 Verify the solution to the system of equations from Class Example #1 using a graphing calculator: Class Ex. #1 2x + y = 2

x - 3y = 15

Class Ex. #4 Solve the following system of equations using a graphing calculator.

x

y

List the answers as exact values using the technique above.

2a + 3b = 4–2b + 10a - 7 = 0

Complete Assignment Questions #1 - #7

Linear and Nonlinear Systems Lesson #1: Solving Linear Systems by Graphing 3

Assignment

1. Consider the system of equations x - 2y = 3, x + y = 0.

a) Write each equation in slope y-intercept form.

b) Complete the table of values c) Draw the lines on the grid andfor each equation. state the solution to the system.

x y

–4–2 0 2 4

x - 2y = 3x y

–4–2024

x + y = 0

x

y

5-5

5

-5

2. In each case, solve the systems of equations by using a graphing calculator. Verify the solution by replacing the values in the original equations.

a) x + 3y = 3 b) 3a - 2b = 8x + y = 0 4a + b = 7

c) x = 2y + 9 d) 3x + 2y = 6x + 5y + 5 = 0 x - y = 1


e) y = x - 1 f) 0.4p - 0.5q = 2.2

y = 34 x - 3 7p + 3q = 15

3. Solve the following system of equations using a graphing calculator. List the answers as exact values.

a) 6x + 7y = 5, 3x = 14y b) 7a - 9b + 4 = 0, 14a + 9b - 16 = 0

c) 2x + y + 2 = 0 d) y = 0.46x - 3.14 e) 12x - 3y = 4y = x + 2 y = 2.13x + 4.28 6x + 3y = 1

MultipleChoice 4. The ordered pair (x, y) which satisfies the system of equation x - 3y = 8, x + 4y = –13 is

A. (–1, 3)

B. (–1, –3)

C. (3, –1)

D. (3, 1)

Linear and Nonlinear Systems Lesson #1: Solving Linear Systems by Graphing 5

5. If 4x - 3y = 9 and 2x + 5y = 7, then the value of x, to the nearest tenth, is _____ .Numerical Response

(Record your answer in the numerical response box from left to right)

6. Jamie graphs the equations x - y = – 4 and x + 2y = 4. The y-coordinate, to the nearest hundredth, of the point of intersection is _____ .


7. Two students worked together to solve a system of equations which had integer solutions. Tara made a table of values for the first equation and Jorge made a table of values for the second equation.

x y

–2 0 2 4

x y

–3–1 1 3 5

x

y

5-5

5

-5

Tara Jorge

–6–2 2 6

–2 0 2 4 6

Using the students’ results to determine the solution to the system, the value of x + y is _____ .(Record your answer in the numerical response box from left to right)

Answer Key

1 . a) y = 12 x - 3

2 , y = –x b) x y–4–2024

x y–4–2024

–3.5–2.5–1.5–0.5 0.5

420–2–4

c ) x = 1, y = –1

2 . a) x = –1.5, y = 1.5 b) a = 2, b = –1 c ) x = 5, y = –2d) x = 1.6, y = 0.6 e ) x = –8, y = –9 f ) p = 3, q = –2

3 . a) x = 23 , y = 1

7 b) a = 47 , b = 8

9 c ) x = – 43 , y = 2

3d) x = – 742

167 , y = – 86571670 e ) x = 5

18 , y = – 29

4 . B 5 . 2 . 5 6 . 2 . 6 7 7 . 7


Linear and Nonlinear Systems Lesson #2:Determining the Number of Solutions

to a System of Linear Equations

Warm-Up Determining the Number of Solutions to a System of Linear Equations

Graph each system of equations on the grid provided. State the number of solutions for each system.a) y = x + 1 b) y = 3x - 1 c) y = 3x - 1

y = –x + 3 y = 3x + 4 6x - 2y - 2 = 0

Number of Solutions to a System of Equations

Complete the following chart.

Number of

Solutions

GraphicalExample

Slopesand

Intercepts

infinitely many

Lines intersect at one point

Lines are parallel

Lines are coincident

Slopes are equal and intercepts are different

Class Ex. #1 Without graphing, analyze each system to determine whether the system has one solution, no solution, or infinitely many solutions.

a) 3x + y = 10, x - 2y = 1 b) x - 2y + 6 = 0, y = 12 x + 3 c) 3x + 4y = 12, y = –

34 x


Solving a System Graphically by Changing the Calculator Window

Often the solution to a system of equations will not be visible using the default window of the graphing calculator. When this occurs the window requires to be changed and we use the following graphing calculator window format:

x:[xmin, xmax, xscl] y:[ymin, ymax, yscl]

Class Ex. #5 At a local High School the Students’ Council

-10 10 20 30 40 50

1000

900

800

700

600

500

400

300

200

100

-100

d

n

decided to sell sweaters to students. The cost of designing the sweaters included a fixed cost of $600 plus $5 per sweater.

The Students’ Council planned to sell the sweaters at $25 a piece. The cost and revenue can be represented by the following systems of equations where d represents the dollar value and n represents the number of sweaters sold.

Cost of sweaters in dollars d = 600 + 5nRevenue of sweaters in dollars d = 25n

a) Use a graphing calculator with a window x:[–10, 50, 10] y:[–100, 1000, 100] to determine how many sweaters must be sold to break even. Sketch the graph representing the equations on the grid provided.

b) If all 850 students in the school purchased a sweater, how much profit would the Students’ Council make?


8 Linear and Nonlinear Systems Lesson #2: Determining the Number of Solutions

Assignment

1. How can you tell by graphing a system of linear equations whether the system has no solution, one solution, or infinitely many solutions?

2. Graph each system and determine whether the system has no solution, one solution, or infinitely many solutions.

a) x = 3y - 2 b) 2x - y = 4 c) 2x - 3y = 8

y = 13 (x + 2) y = 2x + 1 3x - 2y = 4

3. How can you tell by writing a system of linear equations in the form y = mx + b whether the system has no solution, one solution, or infinitely many solutions?

4. Rearrange each equation into the form y = mx + b and state whether the system has no solution, one solution, or infinitely many solutions.

a) 5x - y = 4 b) 2x - y = 3 c) 4y + x - 8 = 0

y = 5x + 4 x - 2y = 3 y = –14 x + 2

Linear and Nonlinear Systems Lesson #2: Determining the Number of Solutions 9

5. Write an equation which forms a system with the equation 2x - y = 6 so that the system has:a) no solution b) one solution c) an infinite number of solutions

6. All 480 tickets for a school concert were sold. Seats in the front part of the hall cost $6 each, and seats in the back part of the hall cost $4 each. The total receipts were $2 530.

This information can be represented by the system f + b = 480, 6f + 4b = 2 530.

a) Graph the system to determine the number of tickets sold for each part of the hall.

b) State the graphing window used.

c) Verify the solution.

7. Solve the following systems of equations using a graphing calculator.

a) y = 23 x + 1 b) 4x + 5y = 18

y = 12 x - 2 2x + 3y = 1

8. The solution to the system ÓÏÔÔÔÌ

3x - 4y = 26x - 8y = 1

hasMultipleChoice

A. no solution

B. one solution

C. two solutions

D. infinitely many solutions


9. If 3x + 2y = 48 and 2x + 3y = 12, then the value of x - 2y, to the nearest tenth, is _____ .Numerical Response


10. The value of k, k Œ N, for which the system of equations10x + ky = –8 and –15x - 6y = 12, has an infinite number of solutions is _____ .


11. The value of a, a Œ N, for which the system of equationsax + 5y = 10 and 6x + 2y = 7, has no solution is _____ .(Record your answer in the numerical response box from left to right)

Answer Key

1 . If the lines are parallel there is no solutionIf the lines intersect there is one solutionIf the lines are coincident there are infinitely many solutions

2 . a) infinitely many solutions b) no solution c ) one solution3 . If the values of m are identical but the values of b are different there is no solution

If the values of m are different there is one solutionIf the values of m are identical and the values of b are identical there are infinitely many solutions

4 . a) no solution b) one solution c ) infinitely many solutions

5 . a) eg. 2x - y = 4 b) eg. 2x - 3y = 6 c ) eg. 4x - 2y = 12

6 . a) Front Æ 305 tickets, Back Æ 175 tickets b) x:[–100, 500, 100] y:[–100, 700, 100]

7 . a) x = –18, y = –11 b) x = 24.5, y = –16

8 . A 9 . 4 8 . 0 1 0 . 4 1 1 . 1 5

Linear and Nonlinear Systems Lesson #2: Determining the Number of Solutions 11

Linear and Nonlinear Systems Lesson #3:Solving Linear Systems of Equations by Substitution

Warm-Up

Using graphing to solve a system of equations is an excellent visual tool. An alternative approach is to solve the system algebraically. There are two algebraic methods - substitution and elimination.

Method of Substitution

In using the method of substitution, there are four general steps which are shown in the flowchart below.

Step 1 Step 2 Step 3 Step 4

Choose the simpler equation and express one variable in terms of the other.

Substitute the expression from step 1 into the other equation.

Solve the single variable equation.

Substitute the solution from step 3 into the equation in step 1 to find the value of the other variable.

Class Ex. #1 Consider the following system of equations x + 3y = 112x - 3y = 4

a) Solve the system using the method of substitution.

b) Verify that the solution satisfies both equations.

c) Check the solution using a graphing calculator.

Class Ex. #2 Consider the following system of equations 3x - 2y = 0, 9x + 4y = 5 .

a) Solve and verify the system.

b) Check the solution using a graphing calculator.

Class Ex. #3 Consider the following system of equations 3(x - 2) + y = 7, 4x - 3(y - 1) = 16 .

a) Solve the system using the method of substitution.

b) Verify algebraically that the solution satisfies both equations.

Class Ex. #4 Solve the following system using substitution. 5x - 2y - z = –9y = 2x3x - z + 3 = 0


14 Linear and Nonlinear Systems Lesson #3: Solving Linear Systems by Substitution

Assignment

1. In each of the following systems:• solve the system using the method of substitution

• verify the solution satisfies both equations• check the solution by graphing

a) y = x + 3, 2x + 3y = 4 b) x - 2y = 9, x + 5y + 5 = 0

c) 2p + q = 0, 7p + 5q = 1 d) 6u - v - 1 = 0, 4u = 3v - 4

Linear and Nonlinear Systems Lesson #3: Solving Linear Systems by Substitution 15

2. Solve each of the following systems by substitution. Check each solution.

a) 2x - 3y = 5 b) 2(x + 4) + y = 812 x - y = 1 5x - 2(y - 1) + 16 = 0

c) 4(x - 1) + 2(y - 1) = 5 d) 3(2x + 3) - (y - 8) = –13x - 4(y + 1) = 7 5(1 - 3x) - 2(4 - y) = 42


3. Solve each of the following systems using substitution.

a) y = 3x b) x = 2y + 12x - y = –4 y + 3z + 5 = 0x - 5z = 4 x - 3z = 0

4. The straight line px + qy = 1 passes through the points (–1, 1) and (–5, 4).

a) Substitute the x and y-coordinates into the equation of the line to form two equations in p and q.

b) Solve this system of equations by substitution to determine the values of p and q.

Linear and Nonlinear Systems Lesson #3: Solving Linear Systems by Substitution 17

5. Solve the following system by substitution. Explain the results.

a) y = 2x - 3 b) x = 3y + 1 4x - 2y = 6 2x - 6y = 4

6. If x + y = 12 and x - y = 2, then x + 2y is equal toMultipleChoice

A. 10B. 17C. 19D. 34

7. When solving a system of equations, one of which is x2 -

y3 = 1, a substitution which

can be made is

A. x = 13 (2y + 1)

B. y = 12 (3x - 1)

C. x = 12 (3y + 6)

D. y = 12 (3x - 6)

8. If s - 8t + 20 = 5s - 7t + 1 = 0, then the value of s + t, to the nearest tenth, is _____ .Numerical Response


Answer Key

1 . a) x = –1, y = 2 b) x = 5, y = –2 c ) p = –13

, q = 23

d) u = 12

, v = 2

2 . a) x = 4, y = 1 b) x = –2, y = 4 c ) x = 3, y = –12

d) x = –3, y = 0

3 . a) x = 4, y = 12, z = 0 b) x = –3, y = –2, z = –1

4 . a) –p + q = 1, –5p + 4q = 1 b) p = 3, q = 4

5 . a) There are an infinite number of solutions of the form x = a, y = 2a - 3, a Œ R because the equations are identical, (the resulting equation reduces to 0 = 0).

b) There are no solutions since the graphs of the equations are parallel lines, (the resulting equation reduces to 2 = 4).

6 . B 7 . D 8 . 7 . 0


Linear and Nonlinear Systems Lesson #4:Solving Linear Systems of Equations by Elimination

Warm-Up

So far we have used two methods to solve systems of equations - graphing and substitution. In this lesson we will learn another algebraic technique - the method of elimination. This method is particularly useful when the equations involve fractions.

Method of Elimination

In using the method of elimination, there are four general steps which are shown in the flowchart below.


If necessary, multiply each equation by a constant to obtain coefficients for x (or y) that are identical (except perhaps for the sign).

Add or subtract the two equations to eliminate one of the variables.

Solve the resulting equation to determine the value of one of the variables.

Substitute the solution into either of the original equations to determine the value of the other variable.

Class Ex. #1 Consder the system of equations: 2x + 5y = 113x - 5y = 4

a) Solve the system using elimination.

2x + 5y = 113x - 5y = 4

b) Verify the solution satisfies both equations

c) Check using a graphing calculator

Class Ex. #2 Solve the following system using elimination (addition) and elimination (subtraction)

addition subtraction

3a - 2b = 14 3a - 2b = 142a + b = 7 2a + b = 7

Would using the method of substitution have made this question easier to solve?


Class Ex. #3 Solve the following system using elimination. Check the solution

2p3 -

3q4 =

112

5p9 +

q6 = 3

20 Linear and Nonlinear Systems Lesson #4: Solving Linear Systems by Elimination

Class Ex. #4 Solve the following system using elimination.

x - 23 -

y + 25 = 2

35 (x + 1) -

45 (y - 3) =

212


Assignment 1. In each of the following systems:

• solve the system using the method of elimination

• verify the solution satisfies both equations• check the solution by graphing

a) x + 2y = 3 b) 2a + 5b = 16

–x + 3y = 2 a - b = 1

Linear and Nonlinear Systems Lesson #4: Solving Linear Systems by Elimination 21

c) 4x - 3y = 9 d) 2x + 4y = 72x - 5y = 1 4x - 3y = 3

2. Solve each of the following systems by elimination. Check each solution.

a) 7e + 4f - 1 = 0, 5e + 2f + 1 = 0 b) 5x = 8y, 4x - 3y + 17 = 0

c) 5x - 2y = 0.6, 2x + y = 1.5 d) 3x + 2y - 6 = 0, x = y + 1


3. Solve each of the following systems by elimination.

a) 3x - 12 y = 5 b) m

2 - n + 3

4 = 213 x +

14 y = 3

3m4 -

n5 = 5

c) 12 (2x - y) +

34 x = 6 d) 2x + y

3 - 5 = 012 x -

13 y =

23

3x - y5 = 1


4. Consider the system of equations x - 2y + 1 = 0, 2x + 3y = 12. Solve the system by:

a) elimination b) substitution.

Which method do you prefer?

5. Consider the system of equations 11x + 3y + 7 = 0, 2x + 5y - 21 = 0. Solve the system by:

a) elimination b) substitution.

Which method do you prefer?


6. An arithmetic sequence has an nth term of the form tn = a + (n - 1)d, where a and d are constants.a) If the fourth term, t4, is 23, and the seventh term, t7, is 41, form a system of equations

in a and d.

b) Solve the system.

c) Determine the 250th term of the sequence.

7. Solve the system of equations 2x +

3y = 2,

8x -

9y = 1 by first substituting a for

1x

and b for 1y .

8. Solve each of the following systems by elimination. Explain the results.

a) –2x + 6y - 1 = 0, 5x - 15y + 2.5 = 0 b) 2x - 4y = 7, –7x + 14y = –21


9. When b is eliminated from the equations 2x + b = 8 and 5x + 2b = 2, we obtainMultipleChoice

A. 7x = 10B. 9x = 18

C. x = –14

D. 3x = –6

10. The solution to the systems of equations x + y = 0,12 x +

13 y = 1 is

A. x = 6, y = –6B. x = 1, y = –1

C. x = 0, y = –0

D. x = –6, y = 6

11. If 13 x + 5 =

23 y and

12 x +

13 y =

13 , then the value of y -

12 x, to the nearest tenth, Numerical

Responseis _____ .


Answer Key

1 . a) x = 1, y = 1 b) a = 3, b = 2 c ) x = 3, y = 1 d) x = 32

, y = 1

2 . a) e = –1, f = 2 b) x = –8, y = –5 c ) x = 0.4, y = 0.7 d) x = 85

, y = 35

3 . a) x = 3, y = 8 b) m = 8, n = 5 c ) x = 5, y = 112 d) x = 4, y = 7

4 . x = 3, y = 2 5 . x = –2, y = 5 6 . a) a + 3d = 23 b) a = 5, d = 6 c ) 1499a + 6d = 41

7 . x = 2, y = 3

8 . a) There are an infinite number of solutions of the form x = a, y = 16 (2a + 1), a Œ R because the

equations are identical, (the resulting equation reduces to 0 = 0).b) There are no solutions since the graphs of the equations are parallel lines, (the resulting equation

reduces to eg. 0 = 7).9 . C 1 0 . A 1 1 . 7 . 5


Linear and Nonlinear Systems Lesson #5:Applying Systems of Linear Equations - Part One

Warm-Up Review

Methods for Solving Systems of Equations

Graphing Substitution Elimination

In this lesson we apply these methods in problem solving.

Problem Solving

We can solve a variety of types of problems using a system of equations. There are four general steps to problem solving which are shown in the flowchart below.


Introduce variables to represent the unknown values.

Form a system of equations involvingthe variables.

Answer the problem and check the solution.

Solve the system.

Number Applications

Class Ex. #1 The difference of two numbers is 9. The larger number is 3 more than twice the smaller number. Find the numbers.

Class Ex. #2 The perimeter of a rectangle is 40 metres. The width is 4 metres less than the length. Find the dimensions of the rectangle.

Money Applications

Class Ex. #3 Gary had a total of $260.00 in five-dollar bills and ten-dollar bills. If he has 33 bills in total, how many of each denomination does he have?

Class Ex. #4 Lora invested her inheritance of $48 000 in two different mutual funds. At the end of one year one fund had earned 10.5% interest and the other fund had earned 12% interest. If she received a total of $5520 in interest, how much did she invest in each mutual fund?

Complete Assignment questions #1 - #14

28 Linear and Nonlinear Systems Lesson #5: Applying Systems of Linear Equations Part One

Assignment In problems #1 - #10 use the following procedure:

a) Introduce variables to represent the unknown values.b) Form a system of equations involving the variablesc) Solve the system.d) Answer the problem and check the solution.

1. A rectangle is to be drawn with perimeter 64 cm. If the length is to be 14 cm more than the width, determine the area of the rectangle.

2. The sum of two numbers is 3 and twice the larger number is 36 more than three times the smaller number. Find the numbers.

3. Six pencils and four crayons cost $3.40. Three similar pencils and ten similar crayons cost $4.90. How much would you expect to pay for a set of eight pencils and twelve crayons?

Linear and Nonlinear Systems Lesson #5: Applying Systems of Linear Equations Part One 29

4. The perimeter of a rectangle is 40 cm. If the length were doubled and the width halved, the perimeter would be increased by 16 cm. Find the dimensions of the original rectangle.

5. A small engineering company has an old machine which produces 30 components per hour and has recently installed a new machine which produces 40 components per hour. Yesterday, both machines were in operation for different periods of time. If 545 components were produced when the total number of hours of operation was 15 hours, determine for how many hours each machine was operating.

6. In a hockey arena, a seat at rink level costs three times as much as a seat in the upper level. If five seats at rink level cost $112 more than eight seats in the upper level, find the cost of a seat at rink level.


7. Rachel had been saving quarters and dimes to buy a new toy. She had 103 coins and had saved $21.40. How many coins of each type had she saved?

8. One year a man saved $5000. Next year his income increased by 10% and his expenditure decreased by 16%. He was able to save $14 600. Calculate his income in the second year.

9. Chad invested 34 of his $56 000 lottery winnings in two different mutual funds. At the end

of the year the Balanced Fund had earned 6.5% interest, but the Emerging Markets Fund had lost 3%. If the value of Chad’s funds had increased by $1 590, determine the amount invested in each fund.


10. Shoji invested $7 000, part at 9% interest and part at 6% interest. The interest obtained from the 6% investment was half of the interest obtained from the 9% investment. How much was invested at each rate?

11. The heights, in metres, of the vertical rods of a suspension bridge, as you move out from the centre of the bridge, form the sequence,

1.1, 1.4, 1.9, 2.6, . . . Centre

1.1 1.4 1.9

n = 1 n = 2a) Without a calculator determine the next two terms in the sequence.

b) The height, h metres, of the n th rod is given by the formula h = a + bn2. Using the terms of the sequence given to form a system of equations, determine the values of a and b and state the formula.

c) Use this formula to verify the answers in a).


12. Erika plans to set up an internet connection with Y2K Internet Company. There are three plans to choose from.

• Plan 1 costs $20 per month and includes a user fee of 40¢ per hour.• Plan 2 costs $15 per month and includes a user fee of 80¢ per hour.• Plan 3 costs $60 per month for unlimited use.

a) What would determine which plan is most economical?

b) Use a graphical method to determine when plans 1 and 2 are equally economical to use. State the graphing window used.

c) Verify the solution to b) algebraically.

d) For each of plans 1 and 2 determine the number of hours of use which could be obtained for $60.

e) Devise a simple rule which would determine which plan is most economical depending on the expected number of hours of internet use per month.


13. The diagram shows two parallel lines and a transversal. Numerical Response

12°

(4x + 7y)°(x - 2y)°

The value of x + y, to the nearest whole number, is _____ .


14. A number consists of two digits whose sum is 11. If the digits were reversed, the original number is increased by 27. The original number is _____ .


Answer Key

1 . 207 cm2 2 . 9, –6 3 . $7.20 4 . 12 cm by 8 cm

5 . 5 12 hours old and 9 1

2 hours new 6 . $48 7 . 74 quarters, 29 dimes

8 . $44 000 9 . $30 000, $12 000 1 0 . $3000 at 6% and $4000 at 9%

1 1 . a) 3.5, 4.6, b) a = 1, b = 0.1, h = 1 + 0.1n 2

1 2 . a) The expected number of hours of internet use per monthb) 12.5 hours, eg. x:[0, 50, 10], y:[0, 50, 10] c ) 12.5 hours d) 100 hours, 56 1

4 hourse) Plan 2 for up to 12.5 hours, Plan 1 for between 12.5 and 100 hours, Plan 3 for more than 100 hours.

1 3 . 3 6 1 4 . 4 7


Linear and Nonlinear Systems Lesson #6:Applying Systems of Linear Equations - Part Two

Mixture Applications

Class Ex. #1 Cashew nuts costing $22/kg are mixed with Brazil nuts costing $16/kg. The mixture weighs 50 kg and sells for $18/kg. How many kilograms of each type of nut were used in the mixture?

Class Ex. #2 Earl the chemist has to make 180 ml of 60% hydrochloric acid (HCl) solution . He has available a one litre bottle of 45% HCl solution and a one litre bottle of 70% HCl solution by volume. How many ml of each solution are mixed to make the 60% HCl solution?

ds t

Distance, Speed, and Time Problems

Class Ex. #3 A student drove the 1245 km from Edmonton to Vancouver in 161 2 hours. This included a one hour stop in Golden and a 30 minute stop in Kamloops. She averaged 100 km/h on the divided highways and 75 km/h on the non-divided mountainous roads. How much time did she spend on the divided highways?

Class Ex. #4 A small cruise boat took 3 hours to travel 36 km down a river with the current. On the return trip it took 4 hours against the current. Find the speed of the current and the speed of the small cruise boat in still water.

Now complete Assignment questions #1 - #10

36 Linear and Nonlinear Systems Lesson #6: Applying Systems of Linear Equations Part Two

Assignment

In problems #1 - #8 use the following procedure:

a) Introduce variables to represent the unknown values.b) Form a system of equations involving the variablesc) Solve the system.d) Answer the problem and check the solution.

1. Candy costing $6 per kg is mixed with candy costing $4.50 per kg to produce 112 kg of candy worth $612. How many kg of each type of candy were used?

2. 300 grams of Type A Raisin Bran is mixed with 500 grams of Type B Raisin Bran to produce a mixture which is 11% raisins. Type A Raisin Bran has twice as many raisins per kilogram as Type B. What percentage of raisins are in each type of Raisin Bran?

Linear and Nonlinear Systems Lesson #6: Applying Systems of Linear Equations Part Two 37

3. A scientist has to make 800 ml of 61% sulfuric acid solution . He has available a one litre bottle of 40% sulfuric acid solution and a one litre bottle of 75% sulfuric acid solution by volume.

a) How many ml of each solution are mixed to make the 61% sulfuric acid solution?

b) What is the maximum volume, rounded down to the nearest ml, of 61% sulfuric acid solution which the scientist could make with the original bottles of sulfuric acid?

4. Pure gold is often mixed with other metals to produce jewellery. Pure gold is 24 carat. 12 carat gold is 12 24 or 50% gold, 6 carat gold is 6 24 or 25% gold, etc. A jeweller has some 12 carat gold and some 21 carat gold and wants to produce 90 grams of 75% gold.

a) What percentage of gold is 21 carat?

b) How many grams of 12 carat gold and of 21 carat gold are needed to produce the mixture?


5. A train travels 315 km in the same time that a car travels 265 km. If the train travels on average 20 km/h faster than the car, find the average speed of the car and the time taken to travel 265 km.

6. A small plane flying into a wind takes 3h to travel the 780 km journey from Lethbridge to Fort McMurray. At the same time, a similar plane leaves Fort McMurray and reaches Lethbridge in 2 1

2 h. If the planes have the same cruising speed in windless conditions, determine the speed of the wind.

7. A cyclist leaves home at 7.30 am to cycle to school 7 km away. He cycles at 10 km/h until he has a puncture, then he has to push his bicycle the rest of the way at 3 km/h. He arrives at school at 8.40 am. How far did he have to push his bicycle?

Linear and Nonlinear Systems Lesson #6: Applying Systems of Linear Equations Part Two 39

8. Chris walks at 8 km/h and runs at 12 km/h. One day he walks and runs on the way from his house to the library. It takes him 20 minutes. On his way back from the library he runs twice as far and the journey home takes 17 1

2 minutes. How far is his house from the library?

9. A shopkeeper wishes to mix two types of tea together. One type sells at $8 per kg and MultipleChoice the second type sells at $12 per kg. He wishes to make 100 kg of the mixture to sell

at $11 per kg. The number of kg of the first type of tea in this mixture should be

A. 25B. 33 1

3C. 50D. 75

10. Raj left home at 1pm to travel 675 km to visit his sister. He averaged 110km/h for the Numerical Response first part of the trip during which he had a 1 hour rest, and 90 km/h for the second part of

the trip during which he had a 30 minute rest. He reached his destination at 9pm. The number of minutes taken for the first part of the trip, to the nearest minute, was _____ .(Record your answer in the numerical response box from left to right)

Answer Key 1 . 72 kg of $6/kg candy, 40 kg of $4.50/kg candy 2 . 16% in type A, 8% in type B3 . a) 320 ml of 40% solution, 480 ml of 75% solution b) 1666 ml4 . a) 87.5% b) 30 g of 12 carat gold, 60 g of 21 carat gold 5 . 106 km/h, 2 1

2 h

6 . 26 km/h 7 . 2 km 8 . 3 km 9 . A 1 0 . 2 7 0


Linear and Nonlinear Systems Lesson #7:Solving Linear Systems of Equations

with Three Variables

Warm-Up

We can extend the method of elimination to solve a system of equations involving three variables. To be able to produce a unique solution to a system involving three variables, we require three independent equations.

Starting with three equations in three variables, reduce by eliminating to two equations in two variables and hence to one equation in one variable. Solve and back substitute to determine the value of the remaining variables.

Class Ex. #1 Solve the following system.

a) (i) x + y + z = 1 b) (i) 2x + 3y + 4z = 10(ii) x + y - z = –1 (ii) 4x + 3y + 2z = 8(iii) x - y + z = 3 (iii) 3x - y - 3z = –4

Complete Assignment questions #1 - #3

MatrixA matrix is a rectangular array of numbers arranged in rows and columns. The coefficients of the variables from a linear system of equations can be entered as elements in a matrix. For example, the following linear systems of equations

2x + y = 2 2x + 3y + 4z = 10 x + 3y = 5x - 3y = 15 4x + 3y + 2z = 8 y = 2

3x - y - 3z = –4can be represented by the following matrices

matrix size:2 x 3

matrix size:3 x 4

matrix size:2 x 3 Î

ÈÍÍÍ

˘˙˙

2 1 21 –3 15 Î

ÈÍÍÍ

˘˙˙

1 3 50 1 2

2 rows by 3 columns Î

È

ÍÍÍÍ

˘

˙˙˙

2 3 4 104 3 2 83 –1 –3 –4

• The number of equations in the system defines the number of rows in the matrix. The Note number of columns is one more than the number of rows.

• If there is a missing variable or constant, substitute a zero.

Solving Linear Systems Using the Matrix Features of a Calculator

The graphing calculator can be used to solve linear systems of equations by using the matrix features. The following two part procedure can be used on a TI-83 Plus graphing calculator to solve Class Ex. #1b) (i) 2x + 3y + 4z = 10

(ii) 4x + 3y + 2z = 8(iii) 3x - y - 3z = –4

PART 1: Entering the Equations as Elements of a Matrix

1. Access the matrix menu MATRIX by pressing 2nd then x–1 keys.2. Select “EDIT” in the menu by scrolling.3. Press ENTER to select matrix 1 [A]. To select any other matrix press the number

by the matrix letter (or scroll down to it) then ENTER .

4. Press 3 ENTER 4 ENTER to display a 3 x 4 matrix. 5. Enter the coefficients of each equation in the matrix. To enter the values for the example

above;• Press 2 ENTER 3 ENTER 4 ENTER 10 ENTER

to enter the first row for 2x + 3y + 4z = 10. The cursor then prompts the second row.• Press 4 ENTER 3 ENTER 2 ENTER 8 ENTER

to enter the second row for 4x + 3y + 2z = 8. The cursor then prompts the third row.• Press 3 ENTER –1 ENTER –3 ENTER –4 ENTER

to enter the third row for 3x - y - 3z = –4.

Note The coefficients of each variable in the system must be aligned vertically in the matrix.i.e. the coefficients of x in all the equations must be in the same vertical column

the coefficients of y in all the equations must be in the same vertical column, etc.

42 Linear and Nonlinear Systems Lesson #7: Solving Linear Systems with Three Variables

PART 2: Solving a System of Linear Equations Using the Matrix Features

1. Quit the screen in part 1.

2. Access the MATRIX menu by pressing the keys 2nd x–1 .

3. Access the command “rref( ” to solve the matrix by scrolling right to MATH,

scrolling down to B:rref( and then pressing the ENTER key. Note: “rref” stands for “reduced row-echelon form”.

4. Select the matrix by accessing the MATRIX menu, then scrolling down to the matrix from PART 1, and press ENTER ENTER . This displays the solution to the system of linear equations.

Note • To convert the answers to exact values, access the fraction command, “Frac”, by pressing MATH ENTER ENTER after step 4 above.

• The mathematical theory behind solving a system of equations using matrices will be covered in Linear Algebra courses at post secondary level.

Class Ex. #2 Solve the following systems using your Matrix key on your graphing calculator. Answer as exact values.

a) a + 2b + 3c = –1 b) 2b + 13 = c c) 7x + 2y = 2 d) x2 -

y3 = 2 -

z6

2a + b - c = 5 a - 2c = 78 4x + 5y = 12y2 +

x4 +

z3 = –8

3a - b + 2c = 4 3a + b = 13x2 -

y3 +

z4 = –1

Linear and Nonlinear Systems Lesson #7: Solving Linear Systems with Three Variables 43

Class Ex. #3 Johnny has two gold engraved pens, three hi-polymer pencils, and four erasers that cost $50.00. Three gold engraved pens, two hi-polymer pencils, and one eraser cost $45 . Three gold engraved pens and one hi-polymer pencil cost the same as a dozen erasers. Find the cost of each item.

Now complete Assignment questions #4 - #12

Assignment

1. Algebraically, solve the following systems of equations.

a) 2x + y + z = 9 b) a + b + c = 3x + 2y - z = 6 3a - b + 2c = 13x - y + z = 8 a + b - c = 1


c) x + y + z = 4 d) s + t + v = -12x + 2y - z = 5 s + 2t + 3v = -2x - y = 1 3s - 2t - v = 2

e) 4x - 3y - 2z = -1 f) 0.5p + 0.3q + 0.2r = 462x + 5y + 8z = -1 0.2p - 0.5q + 0.4r = 06x - 2y = -1 0.1p + 0.8q - 0.6r = 26


2. The parabola y = ax2 + bx + c passes through the points (1, 2), (2, 4) and (3, 8).Determine the values of a, b and c and state the equation of the parabola.

3. The circle x2 + y2 + ax + by + c = 0 passes through the points (5, 5), (2, 6) and (7, 1).Determine the values of a, b and c and state the equation of the circle.


4. Use the matrix features of a calculator to solve the following systems of equations.

a) 2x + y - z = 0 b) 2x + y + z = 5 c) a + b + c = 35x + 3y + 7z = - 8 x + 2y - z = 7 3a - b + 2c = 123x - 2y + 4z = -16 4x - 3y + 3z = 1 a + b - c = -7

d) 0.3p - 0.2q - 0.4r = 10 e) x - y = z f) 6x + 5y = 50.6q + 0.2r = - 1 x + z = 11 3x - 2y = 02p - r = 8q z - 2y = 13

5. Use the matrix features of a calculator to solve the following systems of equations.

a) x3 +

y5 -

z2 = 1 b)

r2 -

3s4 -

t7 = 8

x2 -

3y10 -

5z2 = -1

r2 +

s4 -

t5 = 0

5x6 +

y2 + z = -2

r4 = t -

s8

6. Travis has a savings jar full of nickels, dimes and quarters. He has 96 coins with a value of $12.40. The number of dimes is twice as many as the total of the number of nickels and quarters. Form a system of three equations and use the matrix features of a calculator to determine the number of each type of coin that Travis has.


7. John, standing on a bridge above a river, throws a stone up into the air. The height h (in metres) of the stone above the river at time t (in seconds) is given by the equation h = 1

2 at2 + vt + c, where;

a is the acceleration of the stone in m/s2,v is the initial velocity in m/s, and,c is the height of the bridge above the river in m.

The table shows measurements of the height above the ground after 1, 2 and 3 seconds.

time (t), seconds 1 2 3

height(h), metres 45.1 50.4 45.9

a) Explain why the value of h increases from t = 1 to t = 2 but decreases from t = 2 to t = 3?

b) Form a system of three equations in a, v and c and solve using the matrix features of a calculator.

c) State the acceleration, the initial velocity, and the height of the bridge above the river. Use appropriate units.

d) State the formula for h in terms of t and use the formula to calculate h at t = 5.2 and t = 5.3. Explain the result.


8. Consider the following systems of equations.

i ) 2x + 3y = 8 i i ) 2x + 3y = 84x + 6y = 16 4x + 6y = 10

a) Solve each system using an algebraic method.

b) Use the matrix features of a calculator to determine the solution to each system.

Questions #9, 10, and 11 are based on the following information

A company makes three sizes of teddy bears. The large size sells for $18, the medium size for $14, and the small size for $10. During the weekbefore Christmas, the company sold L large, M medium, and S small teddybears. The company’s financial activities for the week can be described by the following system of equations:

L + M + S = 7218L + 14M + 10S = 972

L + M = S + 8

9. The number 972 represents MultipleChoice

A. the number of teddy bears sold in the weekB. the total cost of 42 teddy bearsC. the total sales revenue for the weekD. some other information

10. The total number of teddy bears sold during the week was _____ .Numerical Response (Record your answer in the numerical response box from left to right)

11. The number of small teddy bears sold during the week was _____ .(Record your answer in the numerical response box from left to right)


12. Extension Question: Algebraically, solve the following system and verify using the matrix features of a calculator.

a + b + 2c + 6d = 3a - b + c + d = 3a + 2b + 3c + 7d = 6a + 3b - c + 2d = 0

Answer Key 1 . a) x = 3, y = 2, z = 1 b) a = 1

4 , b = 74 , c = 1 c ) x = 2, y = 1, z = 1

d) s = 0, t = –1, v = 0 e ) x = 0.5, y = 2, z = –1.5 f ) p = 60, q = 40, r = 20 2 . a = 1, b = -1, c = 2, y = x 2 - x + 2 3 . a = -4 , b = - 2, c = - 20, x 2 + y 2 - 4x - 2y - 20 = 0 4 . a) x = –2, y = 3, z = –1 b) x = 2.5, y = 1.5, z = –1.5 c ) a = 0, b = –2, c = 5

d) p = 10, q = 5, r = –20 e ) x = 4, y = –3, z = 7 f ) x = 1027 , y = 5

95 . a) x = –6, y = 10, z = –2 b) r = 4, s = –8, t = 0 6 . 10 nickels, 64 dimes, 22 quarters

7 . a) at t = 1 the stone is moving in an upward direction, and before t = 3 the stone has reached its highest point and is moving down.

b) 12 a + v + c = 45.1, 2a + 2v + c = 50.4, 9

2 a + 3v + c = 45.9, a = –9.8, v = 20, c = 30c ) acceleration = –9.8 m/s2, initial velocity = 20 m/s, height = 30 md) At t = 5.2, h = 1.504, and at t = 5.3, h = –1.641 m.

This means that between 5. 2 and 5.3 seconds the stone hits the water. Note: The formula will not be valid after the stone hits the water, so the height of the stone

at 5.3 seconds will not be –1.641 m.8 . a) i) infinitely many solutions i i ) no solution

b) i) infinitely many solutions i i ) no solution9 . C 1 0 . 7 2 1 1 . 3 2

1 2 . a = 2, b = 1, c = 3, d = –1


Linear and Nonlinear Systems Lesson #8:Solving Nonlinear Systems of Equations

Nonlinear System

A nonlinear system of equations is a system in which at least one equation does not represent a straight line.

Class Ex. #1 Consider the following system of equations: y = x2

x

y

y = x + 6a) Solve the system using a graphing calculator.

b) Verify the solution by replacing the values in the original equations.

Class Ex. #2 Consider the following system of equations: y = x3 - x2 - 4x + 24y = x3 - 2x2 - 4x + 40

x

ya) This system has two solutions. State an appropriate graphing

window and sketch the graphs of the equations.

b) State the solution to the system.

c) Verify the solution..

d) Use the method of substitution to solve the system algebraically.

Class Ex. #3 John invests $750 in a plan which pays 8% interest per annum. The amount of his investment at the end of n years is A = 750(1.08)n.

Sara invests $1000 in a plan which pays 5% interest per annum. The amount of her investment at the end of n years is A = 1000(1.05)n.

a) Using a graphing window x:[0, 20, 2] y:[0, 2000, 100] sketch a graph of investment amount against time for each investment.

10 20

500

2000

1000

1500

0

Time (years)

InvestmentAmount(dollars)

b) Solve the system of equations A = 750(1.08)n by finding the point of intersection.A = 1000(1.05)n

c) If interest is paid at the end of each year, after how many complete years will John’s investment be worth more than Sara’s?

Note More detailed work on investment income will be covered in the Personal Finance unit.


52 Linear and Nonlinear Systems Lesson #8: Solving Nonlinear Systems of Equations

Assignment

1. In each case, solve the systems of equations by graphing. Verify the solution by replacing the values in the original equations.

a) y = x2 - 2 b) y = 8x - x2 c) y = x - 1y = x y = 2x y = x2 - 6x + 5

d) y = 8x e) x + 2y = 3 f) 2x + y = x2 + 5

–x + y = 7 xy = –2 y = 4x

2. Solve the following systems to one decimal place.

a) y = –x2 + 3 b) y = x3 - 3x + 4 c) y = x2

y = x2 - 7 y = 7 y = 2x

Linear and Nonlinear Systems Lesson #8: Solving Nonlinear Systems 53

3. Consider the systems of equations y = 10x

y = mx + b, where m, b Œ ¬.

In each case the graph of y = 10x is given.

Investigate the number of solutions for the given values of m and b and sketch an appropriate line.

a) m = 2, b = –2 b) m = –1, b = 0

x

y

5–5

5

–5 y = 10x

x

y

5–5

5

–5 y = 10x

c) m = 0, b = 0 d) m = 0, b π 0

x

y

5–5

5

–5 y = 10x

x

y

5–5

5

–5 y = 10x

e) m > 0 f) m is undefined (2 possible answers)

x

y

5–5

5

–5 y = 10x

x

y

5–5

5

–5 y = 10x


4. Solve the following systems of equationsi ) algebraically by the method of substitution, and, i i ) graphically

a) y = x3 + 5x + 16 b) y = x4 - 3x2 - 4 c) y = x4 + x2 - 6 y = 5x - 11 y = –3x2 + 12 y = x4 - x2 + 12

5. A farmer has 12 m of fencing with which to erect two sides of a rectangular pen, the other two sides being formed by the corner of a walled garden.

a) If the area of the pen is 24 m2 and the length of one side is x metres, show that x2 + 24 = 12x.

b) Write a system of equations which could be graphed in order to determine the dimensions of the pen. State the dimensions of the pen to the nearest hundredth of a metre.

c) The ancient Babylonians knew that equations of the form x2 + c = bx could be solved by solving the system of equations x + y = b, xy = c. Use this method to write a system of equations which could be graphed to determine the dimensions of the pen. Compare this answer with the answer in b).

Linear and Nonlinear Systems Lesson #8: Solving Nonlinear Systems 55

6. The number of solutions to the system of equations y = x3 - 1, isMultipleChoice y = 2x

A. 0B. 1C. 2D. 3

7. In January 2000, the population of Starville was 25 000 and the population in Numerical Response Moontown was 35 000.

The population of Starville was increasing at the rate of 3% per year which could be modelled by the equation P = 25 000(1.03)n whereas the population of Moontown was decreasing at the rate of 3% per year modelled by the equation P = 35 000(0.97)n. The year in which the population of Starville overtakes the population of Moontown is _____ .


Answer Key

1 . a) x = –1, y = –1 b) x = 0, y = 0 c ) x = 1, y = 0x = 2, y = 2 x = 6, y = 12 x = 6, y = 5

d) x = –8, y = –1 e ) x = –1, y = 2 f ) x = 1, y = 4x = 1, y = 8 x = 4, y = – 1

2 x = 5, y = 20

2 . a) x = –2.2, y = –2.0 b) x = 2.1, y = 7 c ) x = –0.8, y = 0.6x = 2.2, y = –2.0 x = 2.0, y = 4.0

x = 4.0, y = 16.0

3 . a) 2 b) 0 c ) 0 d) 1 e ) 2 f ) 0 if the line is the y-axis, 1 otherwise

4 . a) x = –3, y = –26 b) x = 2, y = 0 c ) x = 3, y = 84 x = –2, y = 0 x = –3, y = 84

5 . b) y = x2 + 24, y = 24 dimensions Æ 2.54 m by 9.46 mc ) y = 12 - x, y = 24

x , dimensions Æ 2.54 m by 9.46 m

6 . C Æ the two solutions are approximately x = 1.59, y = 3.01 and x = 9.94, y = 980.20

7 . 2 0 0 5


Linear and Nonlinear Systems Lesson #9:Graphing Linear Inequalities in Two Variables

Warm-Up

The speed limit on a highway is 100 km/h. A car is breaking the speed limit. This means that the speed, s, at which the car is travelling is given by the inequality s > 100.

Before studying linear inequalities in two variables we will review linear inequalities in one variable.

Linear Inequalities in One Variable

A mathematical inequality must contain one of the following symbols:

< £ > ≥ π

The following are examples of linear inequalities in a single variable :

4x - 1 > 7 1 - 2a £ 5 etc.

The solution to a single variable inequality can be shown on a number line.

In this unit, unless otherwise stated, we assume that the variables are defined on the set of real numbers.

Class Ex. #1 Consider the inequality 4 - 2(3 + x) > 12.

a) Solve the inequality algebraically .

b) Check the solution using a test case.

c) Graph the solution on a number line.

Linear Inequalities in Two Variables

The following are examples of linear inequalities in two variables :2x - 3y ≥ 6 4p + 3q < 10 etc.

The solution region to a linear inequality in two variables can be represented on a coordinate plane using a boundary line.

The boundary line will be solid or broken according to the following rule.- a solid line is used to represent ≥≥≥≥ or ££££- a broken or dashed line is used to represent > or <

Graphing a Linear Inequality Without Using a Graphing Calculator

Use the following procedure to graph the solution region of a two variable linear inequality without using a TI-83 Plus graphing calculator.

1. On a coordinate plane, graph the corresponding linear equation using a table of values or intercepts. Draw the line solid or broken according to the rule above.

2. The line divides the coordinate plane into two regions, called half planes. The solution region will be on one side of the line. To determine which side, choose the coordinates of a point not on the line, called a test point, and determine if the coordinates of the point satisfy the inequality. If the inequality is satisfied, then the solution is the region from which the point was chosen. If not, then the solution region is the other region.

3. Shade the appropriate region.

Class Ex. #2 Graph the following inequality without using a graphing calculator.

x

y

5

5

–5

–5

y < 34 x + 3

Note 1. If the boundary line is given in the form y = mx + b then

• The region above the line represents y > mx + b

• The region below the line represents y < mx + b

2. If the boundary line does not pass through the origin, the simplest test point to choose is ( 0, 0).

58 Linear and Nonlinear Systems Lesson #9: Graphing Linear Inequalities in Two Variables

Class Ex. #3 Graph the inequality 2x - 3y ≥ 12 without using

x

y

5

5

–5

–5

a graphing calculator.

Class Ex. #4 The diagram shows the solution region to an inequality.

x

y

5

5

–5

–5

The boundary line has intercepts at (–2, 0) and (0, 4).

a) Determine the equation of the boundary line.

b) Determine the inequality.


Linear and Nonlinear Systems Lesson #9: Graphing Linear Inequalities in Two Variables 59

Graphing a Linear Inequality Using a Graphing Calculator

Use the following procedure to graph the solution region to a two variable linear inequality using a TI-83 Plus graphing calculator.

1. If necessary, rearrange the inequality by isolating y to the left side so that the equation of the boundary line is in the form y = mx + b.

2. Input the boundary line equation into Y1.

3. To the left of Y1 select the shading which corresponds to the inequality symbol by

pressing the Enter key continuously until the desired symbol appears. Use the following inequality symbols:

For y £ 2x + 1 or y < 2x + 1 use Y1 = 2X +1

For y ≥ 2x + 1 or y > 2x + 1 use Y1 = 2X +1•

•

4. Press the Graph key.

Note The graphing calculator does not distinguish between < or £ and > or ≥ , i.e. broken or solid lines. When sketching a graph from the graphing calculator window, use the appropriate type of line.

Class Ex. #5 Graph the following inequalities using a graphing calculator. Sketch the result on the grid provided.

a) y + x £ 3 b) –2y - x > 5

x

y

5

5

–5

–5x

y

5

5

–5

–5



Assignment

1. Consider the inequality 5x - 3 ≥ 33 - x .

a) Solve algebraically



2. Consider the inequality p + 4

4 - 3p - 9

7 < 12

a) Solve algebraically




3. In each case, graph the inequality without using a graphing calculator.

a) y ≥ 3x + 2 b) y < 5 - x c) y > x2

x

y

5-5

5

-5

x

y

5-5

5

-5

x

y

5-5

5

-5

4. Graph the following inequalities manually.

a) 4x + 3y £ 12 b) 3p - 5q ≥ 30 c) x < 23 y

x

y

5-5

5

-5

p

q

10-10

10

-10

x

y

5-5

5

-5


5. The graph shows the solution to the inequality

x

y

5

5

–5

–5

5x - 2y < 10.

a) Explain why the boundary line is a broken line.

b) Explain why the solution region is above the line and not below the line.

6. Sketch the half-plane represented by the following inequalities:

a) y > –2 b) x < 2 c) x ≥ 0 d) y + 3 £ 0

x

y

5

5

–5

–5x

y

5

5

–5

–5x

y

5

5

–5

–5x

y

5

5

–5

–5

7. In each case, the equation of the boundary line is given. Determine the inequality which the graph represents.

a) b) c)

x + y = 4

x

y

5

5

–5

–5x

y

5

5

–5

–5

3x – 4y + 12 = 0

x

y

5

5

–5

–5

y = 3


8. Determine the following inequalities using the information given.

x

y

5

5

–5

–5 3–2

a)

The boundary line has an x-intercept of 3 and a y-intercept of –2.

x

y

5

5

–5

–5

(–2, 6)b)

The boundary line passes through theorigin and the point (–2, 6).

9. Sketch the region represented by the following inequalities:

a) 1 £ y £ 4 b) -3 < x < 2

x

y

5

5

–5

–5x

y

5

5

–5

–5


10. Graph the following inequalities using a graphing calculator. Sketch the result on the grid provided.

a) y ≥ 12 x + 1 b) 3x - y > 6

x

y

5

5

–5

–5x

y

5

5

–5

–5

c) 2x + 5y £ 10 d) 4x - y + 6 < 0

x

y

5

5

–5

–5x

y

5

5

–5

–5

11. The point which is not in the solution region of the inequality 4x - 3y £ 6 isMultipleChoice

A. (0, 0)B. (–1, 2)

C. (1, –2)

D. (3, 2)

12. The graph shows the solution region to the inequality

x

y

5

5

–5

–5

(– 4, 0)

(0, –2)

A. x + 2y ≥ – 4

B. x + 2y £ – 4

C. 2x + y ≥ –2

D. 2x + y £ –2


The following questions are extension questions

13. Show the solution region to the following system of

x

y

5

5

–5

–5

linear inequalities. x + y ≥ 62x - y < 4

14. Janine is making two kinds of clothing. Sweaters need 500g of wool and take 6 hours to make. Vests need 400g of wool and take 9 hours to make. She has 2 kg of wool and 36 hours of time available.

a) Suppose she makes x sweaters and y vests. Write down a system of four inequalities which represent the information. Give the inequalities in simplest form.

b) Graph the system of inequalities.

5

5

10

10x

yc) State all possible combinations of

the number of sweaters and the number of vests which she could make.


d) If sweaters sell for $36 and vests for $30, how many of each should she make to maximize the value of the items she sells?

Answer Key

1098 11 12 13

654 7 8 91 . a) x ≥ 6 c )

2 . a) p > 10 c )

3 . a) b) c )

4 . a) b ) c )

5 . a) The inequality does not contain “ equal to “, so the line is broken not solidb) Testing the point (0, 0) shows that this point is in the solution region, so the solution region

is above the line.

6 . a) b) c ) d)


7 . a) x + y ≥ 4 b) 3x - 4y + 12 > 0 c ) y £ 3 8 . a) y < 23 x - 2 b) y > –3x

9 . a) b)

1 0 . a) b) c ) d)

1 1 . C 1 2 . B

1 3 .

1 4 . a) x ≥ 0, y ≥ 0, 5x + 4y £ 20, 2x + 3y £ 12b)

c )# sweaters # vests

0000111122233

0123012301201 d) Three sweaters and one vest sell for $138.00


Functions Lesson #1:Review of Functions, Domain, and Range

Warm-Up Review

Complete the following statements:

a) Relation - A connection between two quantities which can be represented graphically by a set of __________ __________ .

b) Domain - The set of all the _________ components of the ordered pairs of a relation.

c) Range - The set of all the _________ components of the ordered pairs of a relation.

Class Ex. #1 State the domain and range of each of the following.

a) (3, 0), (–1, 2), (5, 4) b) y = 3x + 4, x Œ ¬ c) A telephone book

d) e)

x

y

5

5

–5

–5x

y

5

5

–5

–5

f) g)

x

y

5

5

–5

–5x

y

5

5

–5

–5

Function

A function, or mapping, from a set A to a set B is a rule that relates each element in set A to one and only one element in set B. The set of elements in set A is called the domain. The set of images in set B is called the range.

A function can be represented in different ways:• in words• by an equation• by a graph• by an arrow or mapping diagram• by a table of values, or a set of ordered pairs in which no two ordered pairs have the

same first component.

Class Ex. #2 State whether or not the following statement is true or false

“All functions are relations, but not all relations are functions”

Class Ex. #3 What visual test is commonly used to find out if a graph represents a function or not?

Class Ex. #4 Determine whether or not the following are functions. Justify your answer.

a) (3, 0), (-1, 4), (5, 4) b) (3, 0), (5, 4), (3, 2)

c) d) e)

x

y

5

5

–5

–5x

y

5

5

–5

–5x

y

5

5

–5

–5

235

4

15

“is a factor of”

4

9

2–23

–3

“is the square of”f) g)

70 Functions Lesson #1: Review of Functions, Domain, and Range

Function Notation

Consider the equation y = 3x + 2.

There is a free choice for the value of x, but the value of y depends on the value of x.

Because of this free choice; x is called the independent variable, and, y is called the dependent variable.

Note When graphing a function, the horizontal axis corresponds to the independent variable and the vertical axis corresponds to the dependent variable.

In function notation the equation y = 3x + 2 takes the form f(x) = 3x + 2.

The symbol f(x) is read as “f at x” or “f of x” and represents the value of the function for a value of x. The set of all possible values for f(x) represents the range of the function. The set of all possible number(s) which could be replaced or substituted for x represents the domain of the function.

Note In function notation;• f(x) does not mean f times x,• the “name” of the function is f .

Class Ex. #5 Evaluate;a) y in the equation y = 3x + 2 when x = 5. b) f(5) where f(x) = 3x + 2

Class Ex. #6 Consider the function f(x) = 3x2 - 5. Determine:

a) f(2) b) f ËÊ 5

c) f(a) d) f(x - 2)

Class Ex. #7 The graph of a function is shown. The points shown

x

y

5

5

–5

–5

have integer coordinates. a) Complete:

i ) f(1) = i i ) f(0) = iii) f(3) =

b) Write i), ii), iii) as ordered pairs.


Functions Lesson #1: Review of Functions, Domain, and Range 71

Assignment

1. Fill in the following blanks. Each statement refers to the function defined by y = f(x) and its graph.

a) f is the ___________ of the function.

b) x is the _____________ variable. c) y is the _____________ variable.

c) The vertical axis is used to represent the _____________ variable.

d) The horizontal axis is used to represent the _____________ variable.

2. If f(x) = 3x - 2 determine:

a) f(3) b) f ËÊÁÁ–

12

ˆ˜ c) f(3x - 2)

3. If g(x) = 5 - 2x2 determine:

a) g(–3) b) g ËÊ 6 c) g(2x + 7)

4. If f(x) = x - 5 and g(x) = x2 - 3 determine:

a) g(–2) - f(3) b) f( x2 - 2) - g(x - 5) c) f(g(x))


5. For each of the following relations:

i ) state the domain and range (all the answers contain integer values)i i ) determine whether the relation is a function or not.

x

y

5

5

–5

–5x

y

20

20

–20

–20x

y

5

5

–5

–5

a) b) c)

x

y

5

5

–5

–5x

y

10

10

–10

–10

d) e)

x

y

5

5

–5

–5

f)

x

y

5

5

–5

–5

h) i)g)

x

y

10

10

–10

–1020

20

–20

–20x

y


6. Complete the table for each of the following graphs. Answer to the nearest whole number, if necessary.

10

10

–10

–10x

f(x)a)

20

10

–10

–20x

g(x)b)

Ordered Pairf(x)

f(–12) =

f(0) =

f(6) =

f(12) =

Ordered Pairg(x)

g(–24) =

g(0) =

g(8) =

g(20) =

7. The graph that represents a function isMultipleChoice

A. B. C. D.

8. The domain and range, respectively, for the function, y = Ω2xΩ - 5 are A. xΩx Œ ¬, yΩy £ –5, y Œ ¬ B. xΩx £ –5, x Œ ¬, yΩy Œ ¬ C. xΩx £ 5, x Œ ¬, yΩy Œ ¬

D. xΩx Œ ¬, yΩy ≥ –5, y Œ ¬


Questions 9 and 10 are based on the following information.

f(x) = 5 - 3x and h(x) = x2 - 2

9. The value of h(–2) + f(–3) is

A. 8

B. 12

C. 16

D. 18

10. The value of f(x2 + 1) - h(x + 4) is

A. –4x2 + 8x + 16

B. 4x2 + 8x + 12

C. –4x2 - 8x - 6

D. –4x2 - 8x - 12

11. If g(x) = 4x2 + 2, then g ËÊÁÁ–

32

ˆ˜ , to the nearest whole number, is _____ .

Numerical Response



Answer Key

1 . a) name b) independent c ) dependent d) dependent e ) independent

2 . a) 7 b) –72 c ) 9x - 8 3 . a) –13 b) –7 c ) –8x2 - 56x - 93

4 . a) 3 b) 10x - 29 c ) x2 - 8

5 . a) D = –2, –1, 0, 2 b) D = xΩx Œ ¬ c ) D = xΩx ≥ –8, x Œ ¬ R = 0, 1, 2 R = yΩy Œ ¬ R = yΩy ≥ 4, y Œ ¬ function function function

d) D = xΩx ≥ –10, x Œ ¬ e ) D = xΩx Œ ¬ f ) D = xΩ–5 £ x £ 5, x Œ ¬ R = yΩy Œ ¬ R = yΩy £ 4, y Œ ¬ R = yΩ–4 £ y £ 4, y Œ ¬ not a function function not a function

g ) D = xΩ–6 < x < 6, x Œ ¬ h) D = xΩx £ 5, x Œ ¬ i ) D = xΩ–20 £ x £ 16, x Œ ¬R = yΩ0 < y £ 4, y Œ ¬ R = yΩy £ 4, y Œ ¬ R = yΩ–12 £ y £ 24, y Œ ¬ function not a function function

6 . a) b )Ordered

Pairf(x)

f(–12) = –8

f(0) = 4

f(6) = 0

f(12) = –6

Ordered Pairg(x)

g(–24) = 2

g(0) = 0

g(8) = 8

(–12, –8)

(0, 4)

(6, 0)

(12, –6)

(–24, 2)

(0, 0)

(8, 8)

g(20) = 4 (20, 4)

7 . D 8 . D 9 . C 1 0 . D 1 1 . 1 1


Functions Lesson #2:Operations with Functions

Operations with Functions

The following properties apply to functions f and g, provided that x is in the domain of f and g.

The sum of f and g Æ (f + g)(x) = f(x) + g(x)

The difference of f and g Æ (f - g)(x) = f(x) - g(x)

The product of f and g Æ (fg)(x) = f(x)g(x)

The quotient of f and g Æ ËÊÁ f

g¯ˆ (x) =

f(x)g(x) , g(x) π 0

Warm-Up #1

Consider the functions f(x) = 2x - 6 and g(x) = x - 3, x Œ ¬.

a) Find f(x) + g(x)

b) Complete the following statement.

f(x) + g(x) is a __________ of two functions f and g,

and can be rewritten as (f + g)(x) = __________

Warm-Up #2


a) Find f(x) - g(x)


f(x) - g(x) is a __________ of two functions f and g,

and can be rewritten as __________ = __________

Warm-Up #3


a) Find f(x).g(x).

b) Complete the following statement. f(x).g(x) is a __________ of two functions f and g, and can be rewritten as _________ = __________

Warm-Up #4

Consider the functions f(x) = 2x - 6 and g(x) = x - 3.

a) Find f(x)g(x) and state any restrictions

b) Complete the following statements.

•f(x)g(x) is a __________ of two functions f and g,

and can be rewritten as __________ = __________

• The restriction(s) of f(x)g(x) is where __________

Class Ex. #1 Consider the functions f(x) = 3 x - 2 and g(x) = x - 5. Write an expression in simplest form for each of the following functions.

a) (f - g)(x) b) (fg)(x)

78 Functions Lesson #2: Operations with Functions

Class Ex. #2 Consider the functions f(x) =2x 2 - 7x - 15 and g(x) = x - 5.a) State the domain of f and g.

b) Write an expression in simplest form for ËÊÁÁ

fg

ˆ˜ (x). State the domain.

c) Explain two different ways to evaluate ËÊÁÁ

fg

ˆ˜ (4). Calculate Ë

ÊÁÁ

fg

ˆ˜ (4)

Class Ex. #3 Consider the functions f(x) =

2xx - 1 and g(x) =

xx - 3 .

a) State the domain of f and g.

b) Evaluate 3(fg)(2).

c) Write an expression in simplest form for (f + g)(x). State any restrictions on x.

d) Write an expression in simplest form for ËÊÁÁ

fg

ˆ˜ (x). State any restrictions on x.

Functions Lesson #2: Operations with Functions 79

Class Ex. #4 A multi-media production company produces compact discs which costs $3.00 per unit. The fixed costs, which includes the graphics, are $10 000 irrespective of the number of compact discs produced. Each compact disc retails for $15.00.

If x is the number of units produced, answer the following in terms of x.

a) Write the total cost as a function of the number of units produced.

b) Write the revenue as a function of the number of units produced.

c) Write the company’s profit as a function of the number of units produced.


Assignment

1. Given f(x) = 5x - 10 and g(x) = x - 2, determine the following functions in simplest form and state any restrictions on x.

a) (f + g)(x) b) (f - g)(x)

c) (fg)(x) d) ËÊÁÁ

fg

ˆ˜ (x)


2. Given f(x) = x 2 - 9 and g(x) = x + 3, determine the following functions in simplest form and state the restrictions on the variable.

a) (f + g)(x) b) (f - g)(x)


fg

ˆ˜ (x)

3. Given f(x) = x

x - 3 and g(x) = 2x

x + 1 , determine the following functions in simplest form and state any restrictions on x.

a) (f + g)(x) b) (f - g)(x)


fg

ˆ˜ (x)


4. Given f(x) = 2 x - 5 and g(x) = x - 5, determine the following functions in simplest form and state any restrictions on x.

a) (f + g)(x) b) (f - g)(x)


fg

ˆ˜ (x)

5. Given f(x) = 3x

x + 7 and g(x) = x - 1

x , determine the following functions in simplest form. State the domain in each case.

a) (f + g)(x) b) (f - g)(x)


fg

ˆ˜ (x)


6. Given f(x) = x + 1 and g(x) = x 2 - 1, determine the following functions in simplest form and state any restrictions on x.

a) 3(f - g)(x)

b) (ff)(x) - g(x)

c) ËÊÁÁ

gf

ˆ˜ (x)

7. The figure shown has an area A(x) = 2x 2 + 8x - 3 cm2.

2x

x – 1a) Write an expression for the area, B(x) of the bottom rectangular

part of the figure.

b) Find an expression in simplest form for the area, T(x), of the top part of the figure.

c) If the area of the top is 9 cm2, determine the value of x.

8. Consider the functions f(x) = x - 4x + 2 and g(x) =

x - 3x - 1 . Which of the following are Multiple

Choice

restrictions for ËÊÁÁ

fg

ˆ˜ (x)?

A. –2 and 1 onlyB. –2, 1, and 3 only

C. –2, 1, and 4 only

D. –2, 1, 3, and 4


9. If f(x) = 2x 2 and g(x) = x - 42x determine the values of:Numerical

Response

1. (fg)(3) 2. (f - g)(4) 3. ËÊÁÁ

fg

ˆ˜ (2) 4. (f + g)(1)

Rearrange the four answers in increasing order. Write the question number corresponding to the smallest answer in the first box, the question number corresponding to the second smallest answer in the second box, etc.


10. If f(x) = 3x - 2 and g(x) = 5 3x , then 2(g - f)(3), to the nearest tenth, is _____ .


Answer Key 1 . a) 6x - 12 or 6(x - 2) b) 4x - 8 or 4(x - 2)

c ) 5x 2 - 20x + 20 or 5(x - 2)2 d) 5 with restriction x π 2

2 . a) x 2 + x - 6 or (x + 3)(x - 2) b) x 2 - x - 12 or (x + 3)(x - 4)c ) x 3 + 3x 2 - 9x - 27 or (x - 3)(x + 3)2 d) x - 3 with restriction x π –3

3 . a)3x 2 - 5x

(x - 3)(x + 1) or

x(3x - 5)(x - 3)(x + 1)

, x π –1, 3 b)–x 2 + 7x

(x - 3)(x + 1) or

x(7 - x)(x - 3)(x + 1)

, x π –1, 3

c )2x 2

(x - 3)(x + 1), x π –1, 3 d)

x + 12(x - 3)

, x π –1, 0, 3

4 . a) 3 x - 10, x ≥ 0 b ) x , x ≥ 0 c ) 2x - 15 x + 25, x ≥ 0 d)2 x - 5

x - 5, x ≥ 0, x π 25

5 . a)4x 2 + 6x - 7

x(x + 7)b)

2x 2 - 6x + 7x(x + 7)

c )3(x - 1)

x + 7d)

3x 2

(x + 7)(x - 1)

Domain for a), b), c) xΩx π - 7, 0, x Œ ¬ Domain for d) xΩx π - 7, 0, 1, x Œ ¬

6 . a) 6 + 3x - 3x 2 or 3(2 - x)(1 + x) b) 2x + 2 or (2(x + 1) c ) x - 1 with restriction x π –1

7 . a) 2x 2 - 2x cm2 b) 10x - 3 cm2 c ) 65

8 . B 9 . 3 1 4 2 1 0 . 2 8 . 0


Functions Lesson #3:Composition of Functions

Warm-Up #1

When a pebble is dropped in a pool of water, ripples in the shape of circles form on the surface of the water. The radius of the outer ripple is given by the formula r = 0.3t, where r is the radius in metres and t is the time in seconds after the pebble hits the water. If the area of the circle is A = pr2, combine this formula with the radius formula for the outer ripple to write a formula for the area of the circular ripple after t seconds.

Warm-Up #2 The Composition of Two Functions

The 2000 Dodge Durango 5.9l sports utility vehicle (SUV) has a city driving gasoline mileage rating of approximately 0.2 litres per km. The cost of gasoline is $0.75 per litre.a) The volume, l, litres, of fuel used can be written as a function of the distance, d km,

travelled. Complete the following for l in terms of d.l = f(d) = ______________

b) The cost, C dollars, of gasoline used can be written as a function of l. Complete the following for C in terms of l.

C = g(l) = ______________

c) We can find the cost of gasoline in terms of the distance travelled by combining these two functions. If we substitute the formula for the first function into the formula for the second function we can write C as a function of d. Complete:

C = h(d) =

When two functions are combined in this manner, we say that the new function is a composition of the other two functions.

Warm-Up #3 The Composition of Two Functions

Consider the function given by h(x) = 2x + 3. This function can be thought of as being composed of two functions - the “multiply by 2” function denoted by f and the “add 3” function denoted by g.The composite function h, says first multiply by 2 and then add 3. An arrow diagram can be used to help explain this.

g

add 3 2x + 3x 2x

f

multiply by 2

0

1

2

3

5

7

0

2

4

h

The function f applied to x maps to 2x.

The function g applied to 2x maps to 2x + 3.

Complete the following:

f(x) = _________

g(2x) = _________

so g(x) = _________

The function h(x) = 2x + 3 is a composition of two functions f(x) = 2x and g(x) = x + 3.

The composite function h(x) can be written in the form:

h(x) ==== g(f(x)) read as “g of f of x”or

h(x) ==== (g o f)(x).

Note • When h(x) is written as g(f(x)), note that function f is applied first.• h is often referred to as “a function of a function”.

• Class Ex. #1 and Warm-Up #4 are beyond the scope of the grade 11 curriculum, but will benefit students who plan to study calculus (the chain rule) in future years.

• Students who are not covering this work, should proceed to page 87 and omit assignment questions #1 - #3

Class Ex. #1 A composite function h(x) is given. Complete the diagram and write h(x) as a composition of two functions f and g where h(x) = g(f(x)).

g

x

fa) h(x) = x2 - 2

f(x) = __________ g( ___ ) = __________ g(x) = __________

g

x

fb) h(x) = (x + 4)3

f(x) = __________ g( ___ ) = __________ g(x) = __________

Developing a Method for the Composition of Two FunctionsWarm-Up #4

Consider two functions f(x) = 3x and g(x) = x - 5.

a) Complete the diagram to determine a formula for the composite function h(x) = g(f(x)).g

x

f

multiply by 3 subtract 5

h(x) = g(f(x)) = g( ________ ) = _____________

b) Use a similar technique to determine a formula for the composite function k(x) = f(g(x))


86 Functions Lesson #3: Composition of Functions

Composition of Functions

Consider the composite function g(f(x)) = (g o f)(x) where f(x) and g(x) are given.

Use the following procedure to determine g(f(x)).

Start with g(f(x)). Apply the formula for g.

Write the answer in simplest form.

Replace f(x) with the formula for f.

Class Ex. #2 Given f(x) = 10x + 1 and g(x) = 2x - 5, complete the work below to determine (g o f)(x).

STEPS WORK

Step 1: Start with g(f(x)) Step 1: g(f(x))

Step 2: Replace f(x) with the formula for f. Step 2:

Step 3: Apply the formula for g. Step 3:

Step 4: Write the answer in simplest form. Step 4:

Class Ex. #3 If f(x) = 2x2 - 1 and g(x) = 3x - 4, find

a) (g o f)(x) b) (g o g)(x)

Functions Lesson #3: Composition of Functions 87

Domain and Range of a Composite Function

Consider the functions f(x) = x2 - 3 and g(x) = x - 1 .Class Ex. #4

a) Without finding a formula for (f o g)(x) or (g o f)(x) evaluate:

i ) (f o g)(2) i i ) (f o g)(0) iii) (g o f)(2) iv) (g o f)(0)

b) State the domains of f and g. c) State the ranges of f and g.

In order to determine the domain and range of f o g and g o f Aaron found expressions for (f o g)(x) and (g o f)(x) and used his graphing calculator to sketch the graphs of thecomposite functions. He obtained the following graphs.

-5 5

5

-5

x

y(f o g)(x)

-5 5

5

-5

x

y

(g o f)(x)

He concluded that domain of f o g = xΩx Œ ¬ domain of g o f = xΩx £ - 2 or x ≥ 2, x Œ ¬range of f o g = yΩy Œ ¬ range of g o f =yΩy ≥ 1, y Œ ¬

d) There are errors in Aaron’s thinking. Find expressions for (f o g)(x) and (g o f)(x) and using the results from a) and b) complete the following:i ) graph of (f o g)(x) i i ) graph of(g o f)(x)

-5 5

5

-5

x

y

-5 5

5

-5

x

y

domain of f o g = domain of g o f =

range of f o g = range of g o f =

• When we calculate an expression for a composite function there may be restrictions on Note

domain and range.• In most cases the domain and range of a composite function will be different from the

domain and range of the original functions. • The domain of a composite function f o g cannot be more than the domain of g.

The range of a composite function f o g cannot be more than the range of f.



Assignment Questions #1 - #3 are optional, depending on earlier work in this lesson.

1. A composite function h(x) is given. Use the methods of Class Ex. #1 to complete the diagram and write h(x) as a composition of two functions f and g where h(x) = g(f(x)).

g

x

fa) h(x) = (x - 5)2

f(x) = __________ g( ___ ) = __________ g(x) = __________

g

x

fb) h(x) = x3 + 6

f(x) = __________ g( ___ ) = __________ g(x) = __________

g

x

fc) h(x) = (x - 4)

f(x) = __________ g( ___ ) = __________ g(x) = __________

g

x

fd) h(x) = x - 4

f(x) = __________ g( ___ ) = __________ g(x) = __________

g

x

f

e) h(x) = 1

x + 3

f(x) = __________ g( ___ ) = __________ g(x) = __________

2. Consider two functions f(x) = x + 2 and g(x) = x2.


x

f

h(x) = g(f(x)) = g( ________ ) = _____________

b) Use a similar technique to determine a formula for the composite function k(x) = f(g(x)).


3. Consider two functions f(x) = x and g(x) = 4x.


x

f

h(x) = g(f(x)) = g( ________ ) = _____________

b) Use a similar technique to determine a formula for the composite function k(x) = f(g(x)).

4. For each pair of functions, write a formula for (f o g)(x).

a) f(x) = 2x + 1, g(x) = 5x b) f(x) = 5x - 2, g(x) = x3 c) f(x) = 2x, g(x) = x + 4

5. For each pair of functions, write a formula for (g o f)(x).

a) f(x) = 2 - x, g(x) = Ωx + 2Ω b) f(x) = 2x + 1, g(x) = x4 c) f(x) = 3x, g(x) = x - 1


6. Consider the functions f(x) = 3

x + 2 and g(x) = x - 2.

a) Show that (g o f)(x) = –2x - 1x + 2 .

b) For each of the functions f, g, and g o f, state;i ) the domain i i ) range

7. Consider the functions f(x) = x - 3 and g(x) = x2 + 2

a) Find expressions for (f o g)(x) and (g o f)(x).

b) Determine the domains of f, g, f o g, and g o f

c) Determine the ranges of f, g, f o g, and g o f


8. Find (f o g)(x), (g o f)(x), and (f o f)(x) for the following. State any domain restrictions.

a) f(x) = –2x, g(x) = x2 - 3 b) f(x) = 1

3 - x , g(x) = x2 c) f(x) = 3x, g(x) = x - 2

9. If f(x) = 2x + 3 and g(x) = 5 - 2x, find the value of:

a) f(g(5)) b) g(f(–3)) c) (f o g)(0) d) –2(g o f)(0)

10. If f(x) = 2 x and g(x) = 2 + 2x, find the value of:

a) f(g(7)) b) g ËÊÁÁf ËÊÁ

14

ˆ˜ c) (f o g)(5) d) 3(g o f)(5)


11. If f(x) = x + 4 and g(x) = x - 1, find the value(s) of x for which;

a) f(g(x)) = 50 b) (fg)(x) = 50

12. Given f(x) = 4 - x and g(x) = 3 5x then (f o g)(5) is equal toMultipleChoice

A. –71B. –11

C. –1

D. 35

13. Given f(x) = 1

x + 5 and g(x) = 6x - 1 then (g o f)(–2) is equal to

A. 1

B. 16

3

C. –1

8D. –3

14. Given that p(x) = 2x + 1 and q(x) = x2 - 1, then p(q(x)) equals

A. 2x2 + 1B. 2x2 - 1C. 4x2

D. 4x2 + 4x

15. The functions f, g, and h are given by f(x) = x2 - 1, g(x) = 3x + 2, and h(x) = Ωx + 2Ω.Numerical Response The value of (f o g o h)(–8), to the nearest whole number, is _____ .



16. The functions f and g are given by f(x) = 1x , and g(x) =

1x + 1 .

If (f o f)(x) = (g o g) ËÊÁÁ

12

ˆ˜ , then the value of x, to the nearest tenth, is _____ .


Answer Key

1 . a) f(x) = x - 5, g(x) = x2 b) f(x) = x3, g(x) = x + 6 c ) f(x) = x - 4, g(x) = x

d) f(x) = x , g(x) = x - 4 e ) f(x) = x + 3, g(x) = 1x

2 . a) h(x) = (x + 2)2 b) k(x) = x2 + 2

3 . a) h(x) = 4 x b) k(x) = 2 x

4 . a) (f o g)(x) = 10x + 1 b) (f o g)(x) = 5x3 - 2 c ) (f o g)(x) = 2x + 4

5 . a) (g o f)(x) = Ω4 - xΩ b) (g o f)(x) = (2x + 1)4 c ) (g o f)(x) = 3x - 1

6 . b) Domain: f: xΩx π –2, x Œ ¬ g: xΩx Œ ¬ g o f: xΩx π –2, x Œ ¬ Range: f: yΩy π 0, y Œ ¬ g: yΩy Œ ¬ g o f: yΩy π –2, y Œ ¬

7 . a) (f o g)(x) = x2 - 1 , x £ –1 or x ≥ 1, (g o f)(x) = x - 1, x ≥ 3 b) Domain: f: xΩx ≥ 3, x Œ ¬ g: xΩx Œ ¬

f o g: xΩx £ –1 or x ≥ 1, x Œ ¬ g o f: xΩx ≥ 3, x Œ ¬

c ) Range: f: yΩy ≥ 0, y Œ ¬ g: yΩy ≥ 2, y Œ ¬ f o g: yΩy ≥ 0, y Œ ¬ g o f: yΩy ≥ 2, y Œ ¬

8 . a) (f o g)(x) = –2x2 + 6 (g o f)(x) = 4x2 - 3 (f o f)(x) = 4xb) (f o g)(x) = 1

3 - x 2 , x π ± 3 (g o f)(x) = 1(3 - x)2 , x π 3 (f o f)(x) = 3 - x

8 - 3x , x π 83 , 3

c ) (f o g)(x) = 3 x - 2 , x ≥ 2 (g o f)(x) = 3x - 2 , x ≥ 23 (f o f)(x) = 9x

9 . a) –7 b) 11 c ) 13 d) 2

1 0 . a) 8 b) 4 c ) 4 3 d) 6 + 12 5

1 1 . a) x = 47 b) x = –9, 6

1 2 . B 1 3 . A 1 4 . B 1 5 . 3 9 9 1 6 . 0 . 6


Functions Lesson #4:The Inverse of a Function - Part One

Inverse of a Function

A function is a relation in which each element of a set A ( the domain) is mapped to one and only one element of a set B (the range).

4

9

–22

–33

A B“is the square root of”

1234

A B“is half of”

2468

i) ii)

or

eg.

The inverse of a function is a relation which “undoes” what the function does. In other words, the elements in set B are mapped back to elements in set A.

4

9

“is the square of”

–22

–33

B A

1234

B A“is double of”

2468

i) ii)

or

Referring to the cases above, complete the following by choosing the correct answer.

In case (i) the inverse ( is / is not ) a function.

In case (ii) the inverse ( is / is not ) a function.

Note • The domain of the inverse is the range of the original function.• The range of the inverse is the domain of the original function.

• The inverse of a function may or may not be a function.

Class Ex. #1 Consider the “operation” of putting on your socks and then putting on your shoes. What would be the “inverse operation”?

Finding the Inverse of a Function Defined in Words

Class Ex. #2 Complete the table to describe the inverse of the function:

FUNCTION INVERSE Is the inverse a function?multiply by 2 ____________________ _____

square ____________________ _____

take the reciprocal ____________________ _____

divide by 3, then add 1 ____________________ _____

Finding the Inverse of a Function Defined by Ordered Pairs

1234

A B“is half of”

2468

Consider the arrow diagram on the previous page.

The function which maps from A to B can be described by the following set of ordered pairs:

(1, 2), (

1234

B A“is double of”

2468

The inverse function which maps from B to A can be described by the following set of ordered pairs

( 2, 1),

Notice that the ordered pairs for the inverse can be obtained by interchanging the first and second coordinates of the ordered pairs of the original function. This reinforces the rule that the domain of the inverse function is the range of the original function and the range of the inverse function is the domain of the original function.

Class Ex. #3 Consider the function defined by the following set of ordered pairs.

(–4,–2), (–2,–1), (–1, 0), (0,1), (2,4), (3,8)

a) Describe the inverse of the function by a set of ordered pairs.

b) Graph the original function and the

x

y

–5

–5

5

5

inverse.

c) Draw a line which acts as a “mirror” between the original function and its inverse.

d) State the equation of the mirror line.

96 Functions Lesson #4: The Inverse of a Function - Part One

Finding the Inverse of a Function Defined by a Graph

To determine the inverse of a function defined by a graph, reflect the graph of the function in the line y = x.

Alternatively, select the coordinates of some key points, interchange the coordinates and plot the new points.

Class Ex. #4 Sketch the graph of the inverse of the functions defined by the following graphs. Is the inverse a function?

10

10

–10

–10x

y

10

10

–10

–10x

ya) b)

Finding the Inverse of a Function Defined by an Equation - Algebraically

When finding the inverse of a function defined by an equation

interchange x and y in the equation and then solve for y

Class Ex. #5 Consider the function defined by the equation y = 3x + 2.

x

y

5

5

–5

–5

a) Find an equation in the form “y = mx + b” for the inverse of the function.

b) Graph the original and its inverse on the grid.

c) Is the inverse of the function defined by the equation y = 3x + 2 also a function?

Functions Lesson #4: The Inverse of a Function - Part One 97

Graphing the Inverse of a Function Defined by an Equation - by Calculator

A graphing calculator may be used to find the inverse of a function defined by an equation. The following instructions are an example of finding an inverse using the TI-83 Plus.

1. Input the equation into Y1 found in the Y = key and press GRAPH .

2. Access the “draw inverse” command by pressing 2nd then PRGM .

Scroll down to “DrawInv”, and press Enter .

3. To draw the inverse of the function in Y1 press VARS , scroll to “Y-Yars”, then

to “Function”, and press Enter Enter .

4. Press Enter again. The graph of the inverse of the function is shown along with the original function.

Note

• If you are interested in finding only the graph of the inverse of the function and not the graph of the original, use the following procedure:

- clear the “Y= editor”, use “Draw Inv”followed by the equation of the function,

and press Enter .

• If you are doing a series of graphs of inverses by this method, use the following sequence to clear the graph screen between each graph:

- press 2nd then PRGM to “ClrDraw”, and press Enter Enter .

Class Ex. #6 Use the procedure to confirm the graph of the inverse in Class Ex#4a). The original graph has equation y = ΩxΩ - 3.



Assignment

1. The arrow diagram shows a function from set A to set B.

59

2527

A B

3

5

a) Draw an arrow diagram (above right) which represents the inverse of this function.

b) Is the inverse of the function also a function?

2. Complete the table to describe the inverse of the function:

FUNCTION INVERSE Is the inverse a function?

divide by 3 ____________________ _____

cube ____________________ _____

add 10, then multiply by 2 ____________________ _____

square, then subtract 5 ____________________ _____

3. Consider the function defined by the following set of ordered pairs.

(–2, 9), (–1, 7), (0, 5), (1, 3), (2, 1)


b) Is the inverse of the function also a function? Why?

4. Consider the function defined by the following set of ordered pairs.

(–2, 1), (–1, –2), (0, –3), (1, –2), (2, 1),


b) Is the inverse of the function also a function? Why?


5. Sketch the graph of the inverse of the function defined by the following graphs. Is the inverse a function?

5

5

–5

–5x

ya) b)

5

5

–5

–5x

y c)

5

5

–5

–5x

y

6. Find the inverse of the functions defined by the following equations.

a) y = 13 x - 2 b) y = 6 - 8x c) y =

x - 2

5

d) 3y = x - 7 e) 6x - 12 y + 4 = 0 f) y = x2


7. Graph the inverse of the following functions using a graphing calculator.

a) y = 4x - 8 b) y = Ωx + 2Ω c) y = x2 + 3

x

y

x

y

x

y

8. In each case graph the function defined by the equation and the inverse of the function on the grid provided.a) y = 9 - x2 b) y = x - 4

10

10

–10

–10x

y

10

10

–10

–10x

y

9. When a function and its inverse are graphed on the same grid, which of the following lines MultipleChoice must be a line of symmetry for the graph?

A. the x-axisB. the y-axis

C. the line y = x D. the line y = –x

10. A function is defined by the equation y = 2x2 - 3. The inverse of the function has equation

A. y = x + 3

2

B. y = 3 - 2x2

C. y = ±x + 3

2

D. y = ±2

x + 3


11. The point (a, 2) lies on the graph of a function and on the graph of the inverse of the function. The value of a is

A. 2B. 0

C. –2 D. impossible to determine without further information.

Answer Key 1 . a) b) no

59

2527

B A

3

5

2 . FUNCTION INVERSE Is the inverse a function?divide by 3 muyltiply by 3 yescube cube root yesadd 10, then multiply by 2 divide by 2, then subtract10 yessquare, then subtract 5 add 5, then square root no

3 . a) (9, –2), (7, –1), (5, 0), (3, 1), (1, 2)b) yes because each element of the first set (x-coordinates) is mapped to one and only one element of the

second set (y-coordinates).

4 . a) (1, –2), (–2, –1), (–3, 0), (–2, 1), (1, 2)b) no because the elements –2 and 1 in the first set both map to more than one element of the second set.

5 . a) b ) c )

6 . a) y = 3x + 6 b) y = – 1

8 x + 34 c ) y = 5x + 2

d) y = 3x + 7 e ) y = 112 x - 2

3 f ) y = ± x , x ≥ 0

7 . a) b) c )

8 . a) b )

9 . C 1 0 . C 1 1 . A


Functions Lesson #5:The Inverse of a Function - Part Two

Finding the Inverse of a Function in the Form “f(x) =”

The steps are listed below.


Replace f(x) by y

Interchange x and y to obtain the inverse.

Solve for yReplace y by f –1(x)(see note below)

If the inverse of f(x) is a function, then the inverse function is denoted by f –1(x) (read as f inverse of x)

If the inverse of f(x) is not a function, then the notation f –1(x) should not be used.

Note that some textbooks do not make this distinction and use f –1(x) even when the inverse is not a function.

Note • f –1(x) is the notation used for the inverse of f(x) when the inverse is a function.

• Although 10–1 = 110 , note that f –1(x) π

1f(x) .

f –1 represents the inverse function, not the reciprocal function.

• The inverse of a rational function will be studied in a later unit.

Class Ex. #1 Find f–1(x) for the following.

a) f(x) = 2x - 3 b) f(x) = x3 + 4

Class Ex. #2 The following questions deal with function f(x) = x 2 - 4.

10

10

–10

–10x

ya) Graph f(x) on the grid provided and state the domain and range.

b) Find the inverse of f(x).

c) Graph the inverse of f(x) on the grid provided and state the domain and range.

d) Is the inverse of f a function? If not, how could the domain or range of f be restricted so that the inverse of f is also a function?

Class Ex. #3 Don incorrectly determined the inverse of the function

x

y

5

5

–5

–5

defined by y = x - 3 to be the equation y = x2 + 3.

He graphed the inverse of the function and obtained a parabola.

Explain why Don’s equation of the inverse is not complete and why the graph of the correct inverse is not a parabola.

Note When determining the inverse of a function, domain restrictions, if any, must be included in the equation of the inverse. There may also be restrictions on the range of the inverse function.

104 Functions Lesson #5: The Inverse of a Function - Part Two

Verifying that Functions are Inverses of Each Other

Find (f o f –1)(x) and (f –1o f)(x) for Class Ex. #1, where f(x) = 2x - 3. What do you notice?Class Ex. #4

Two Functions f and g are Inverses of each other if (f o g)(x) = x and (g o f)(x) = x


Assignment1. Find the inverse of the following functions.

a) f(x) = 4x + 5 b) g(x) = 3x - 1

7 c) f(x) = x3 - 1

2. For each of the following functions;

• find the inverse function using the notation f–1(x), where appropriate,• state the domain and range of the inverse function.

a) f(x) = x + 2 b) f(x) = (x - 2)2

Functions Lesson #5: The Inverse of a Function - Part Two105

c) f(x) = x2 - 25 d) f(x) = 16 - x2

3. Find f–1(x) for each of the following functions with restricted domains.

a) f(x) = x2, x ≥ 0 b) f(x) = (x - 3)2, x ≥ 3 c) f(x) = x2 + 1, x £ 0

4. Kaleb incorrectly determined the inverse of y = 4 - –x to be y = –(x - 4)2 and used the graphing calculator to obtain a parabola. Explain why the graph of the correct inverse is not a complete parabola.


5. Functions f and g are defined as f(x) = 2x + 6 and g(x) = 3x.

a) Determine f –1(x) and g–1(x).

b) Find expressions for:

i ) (f –1 o g –1)(x) i i ) (g–1 o f–1)(x)

i i i) (f o g)–1(x) iv) (g o f)–1(x)

c) Compare the answers in b). What do you notice?


6. Given that f(x) = 1 - 2x, x Œ ¬ then f –1(x) isMultipleChoice

A. –x

2 - 1

B. x

2 - 1

C. 1 - x

2

D. x - 1

2

7. If g(x) = x3 - 1

2 , x Œ ¬ then g-1(x) is

A. 2

x3 - 1

B. 2x + 13

C. 2 x3

+ 1

D. 2 x + 13

8. Given that f(x) = (x - 2)2, x Œ ¬ then f –1(x) is

A. x + 2B. x + 2 C. – x + 2

D. not defined

9. If f(x) = 3x - 2 and g(x) = 13 x + 1, then (f o g)-1(x) equals

A. x - 1

B. 1 - x

C. x + 1

D. 1

3 (3x - 1)

10. Given that f(x) = 2x and g(x) = 3 - 5x, then (g o f)-1(x) equals

A. 3

11

B. 6

11

C. 1

10 (3 - x)

D. 1

10 (6 - x)


Use the following graphs to answer question #11.

x

y

x

y

x

y1. 2. 3.

x

y

x

y

x

y4. 5. 6.

11. Consider the following questions: Numerical Response

a) Which graph represents a function whose inverse is also a function?

b) Which graph does not represent a function, but could be made to represent a function if the range were restricted to y ≥ 0?

c) Which graph represents a function whose inverse is not a function, but could be made to represent a function whose inverse is also a function if the domain were restricted to x £ 0?

d) Which graph represents a function whose inverse is not a function but could not be made to represent a function whose inverse is also a function if the domain were restricted to x £ 0?

Write the graph number corresponding to answer a) in the first box, the graph number corresponding to answer b) in the second box, etc.(Record your answer in the numerical response box from left to right)


Answer Key

1 . a) f –1(x) = 14

x - 54

b) g–1(x) = 7x + 13 c ) f –1(x) = x + 1

3

2 . a) f –1(x) = x2 - 2,x ≥ 0 Domain: xΩx ≥ 0, x Œ ¬ Range: yΩy ≥ –2, y Œ ¬ b) y = ± x + 2 Domain: xΩx ≥ 0, x Œ ¬ Range: yΩy ≥ 2, y Œ ¬ c ) y = ± x + 25 Domain: xΩx ≥ –25, x Œ ¬ Range: yΩ y Œ ¬

d) y = ± 16 - x2 , 0 £ x £ 4 Domain: xΩ 0 £ x £ 4, x Œ ¬ Range: yΩ –4 £ y £ 4, y Œ ¬

3 . a) f –1(x) = x , x ≥ 0 b) f –1(x) = x + 3, x ≥ 0 c ) f –1(x) = – x - 1 , x ≥ 1

4 . The correct inverse is y = –(x - 4)2, x £ 4, the graph of which is only half of a parabola.

5 . a) f –1(x) = 12 x - 3, g–1(x) = 1

3 x b) i ) 16 x - 3 i i ) 1

6 x - 1 i i i ) 16 x - 1 i v ) 1

6 x - 3

c ) (f o g)–1(x) = (g–1 o f–1)(x) (g o f)–1(x) = (f –1 o g–1)(x)

6 . C 7 . B 8 . D 9 . A 1 0 . C 1 1 . 6 3 4 2


Functions Lesson #6:Zeros of a Function

x-intercept - is the x-coordinate of the point(s) where a graph touches or crosses the x-axis.

y-intercept - is the y-coordinate of the point(s) where a graph touches or crosses y-axis.

Root of an Equation - is the value of the variable which satisfies the equation.

Warm-Up #1 Roots and Factors

-5 5

5

-5

x

y

The graph of y = x2 - x - 6 is shown. a) Write the equation y = x2 - x - 6 in factored form.

b) Find the roots of the equation x2 - x - 6 = 0.

c) Explain the connection between the factors of y = x2 - x - 6 and the roots of the equation x2 - x - 6 = 0.

d) State the x-intercepts of the graph.

e) Explain the connection between the x-intercepts of the graph of y = x2 - x - 6 and the roots of the equation x2 - x - 6 = 0.

Zero(s) of a Function

A zero of a function is a value of the independent variable which makes the value of the function equal to zero. Zero(s) of a function can be found by solving the equation f(x) = 0.

Class Ex. #1 Find the zero of the function f where f(x) = 3x - 12.

Warm-Up #2

x5

5

y

The graph of f(x) = 2x2 - 7x + 3 is shown to the right.

a) Write the function f(x) = 2x2 - 7x + 3 in factored form.

b) Find the zeros of the function f(x) = 2x2 - 7x + 3.

c) State the x-intercepts of the graph.

d) Explain the connection between the zeros of the function f(x) = 2x2 - 7x + 3 and the x-intercepts of the graph of the function.

Class Ex. #2 a) Fill in the blanks in the following statement.

“The _________ of a function, the ___________ of the graph of the function, and

the __________ of the corresponding equation y = 0, are the ________ numbers.”

b) The graph of f(x) = x2 - x - 6 is shown. Fill in the blanks.

-5 5

5

-5

x

y The graph of

f x x x( )= - -2 6

has x-intercepts

x = _____ and

x = _____ with y-intercept

y = _____

The equation x x2 6 0- - = has the roots

x = _____ and

x = _____

The function f x x x( )= - -2 6

= + -( )( )x x2 3has zeros _____ and _____

112 Functions Lesson #6: Zeros of a Function

Finding Zeros of a Function

To find the zeros of a function, f(x), either;

• substitute zero for f(x) and find the roots of the resulting equationor

• graph the function and determine the x-intercepts of the graph.

Finding the Roots of an Equation Algebraically

Finding the roots of a single variable equation may involve factoring. Except in the case of a linear equation, set the equation to zero before factoring.

Recall the following techniques for factoring:• common factors, difference of two squares, trinomials of the form x2 + bx + c = 0,

and trinomials of the form ax2 + bx + c = 0.

Class Ex. #3 Find the roots of the following equations

a) 2x + 1 = 2 b) x2 + 8x = 33 c) 6(4x + 5)(x - 3) = 0 d) 2x2 - 8 = 0

Class Ex. #4 For the following functions:i ) find the zeros i i ) find the y-intercept of the graph of the function.

a) f(x) = 5x2 + 15x - 20 b) f(x) = 3x2 - 11x + 10 c) g(x) = 2x(2x + 1)(x - 3)(3x - 4)


Functions Lesson #6: Zeros of a Function 113

Using A Graphing Calculator to Find Zeros

A graphing calculator can be used to find the zeros of a function by finding the correspondingx-intercepts of its graph. Use the following procedure on a TI-83 Plus calculator.

1. Enter the equation of the function into Y1 and press GRAPH .

2. Access the CALC feature by entering 2nd then TRACE .

3. Select “zero”.

4. On the bottom left hand side of the screen the calculator will ask for a left bound. Select a value on the left side of the max/min point and press ENTER .

5. On the bottom left hand side of the screen the calculator will ask for a right bound. Select a value on the left side of the max/min point and press ENTER .

6. On the bottom left hand side of the screen the calculator will ask for a guess. Press ENTER . The x value will be the x-intercept.

Class Ex. #5 For each of the following functions, use a graphing calculator to:

• sketch the graph of the function, and,• find the zeros (as exact values) of the function.

a) f(x) = 3x2 + 4x - 7 b) g(x) = 4x3 - 7x2 - 4x + 7



Assignment

-5 5

20

-20

x

y

1. The graph represents a function, f. The x and y-intercepts of the graph are integers. a) State the x and y-intercepts of the graph.

b) State the zeros of the function f.

2. Find the roots of the following equations. a) x2 + 5x + 6 = 0 b) 2x + 5 = 0 c) 2x(x + 3) = 0

d) 2x2 - 10x + 12 = 0 e) x2 - 64 = 0 f) x2 - 16x + 64 = 0

g) x2 - 6x = 16 h) x3 + 8x2 = 20x i ) 4x2 + 4x - 3 = 0

3. Find the zeros of the following functions.

a) f(x) = x3 + 5 b) g(x) = x2 - 20x + 36 c) P(x) = 3(2x - 5)(x + 1)

d) g(x) = 25x2 - 64 e) f(x) = 3x - 7 f) h(x) = x2


g) f(x) = x4 - 16 h) P(x) = x(x - 3)(2x + 1) i ) f(x) = 30x2 + 140x - 50

4. For the following functions:i ) Find the zerosi i ) Find the y-intercept of the graph of the function.a) f(x) = 3(x - 5)(5x - 9) b) f(x) = 5x2 - 35x c) f(x) = 3x(x2 - 49)

d) f(x) = 2x2 - x - 15 e) P(x) = 8x2 + 14x - 15 f) g(x) = 2x2 - 56x - 120

5. In each case the graph of a function is shown where the x and y-intercepts are integers. Determine the:

• zeros of the function,• factors of the equation of the function, and,• y-intercept of the graph of the function.

5

5

-5

x

y

-5 5

5

-5

-10

x

y

-5 5

20

10

-10

x

ya) b) c)


6. Use a graphing calculator to find the zeros (as exact values) of the following functions.

a) g(x) = x 2 - 3x - 4 b) f(x) = 18x 2 - 5x - 7 c) g(x) = 3x 3 - 11x 2 + 6x

7. Use a graphing calculator to find the roots, (to the nearest tenth) of the following equations.

a) x 2 - 3x = 11 b) 21x 3 - 41x 2 = –11x - 9

8. Use a graphing calculator to write the equation in factored form.

a) y = x 2 - 7x + 6 b) y = 2x 2 - 3x - 9 c) y = 5x 3 - 7x 2 - 21x - 9

9. The zeros of the function f(x) = 2(x - 3)(4x + 7) are:MultipleChoice

A. 3, –74

B. –3, 74

C. 0, 3, –74

D. 2, 3, –74

10. The roots of the equation 3x(x + 1) = 6

A. 0, –1B. 2, 5C. 2, –1D. –2, 1

11. The least possible zero of the function f(x) = 2x3 - 7x2 + 3x is

A. 0

B. 12

C. 3D. –3

12. The y-intercept of the graph of the function f(x) = (x + 4)(3 - 2x)(x + 1), Numerical Response to the nearest whole number, is _____ .



Answer Key 1 . a) x-intercepts are –6, 4 and y-intercept is –24. b) -6, 4

2 . a) –3, –2 b) – 52 c ) –3, 0 d) 2, 3 e ) –8, 8

f ) 8 g ) –2, 8 h) –10, 0, 2 i ) – 32 , 1

2

3 . a) –15 b) 2, 18 c ) –1, 52 d) – 8

5 , 85 e ) 7

3f ) 0 g ) –2, 2 h) – 1

2 , 0, 3 i ) –5, 13

4 . a) i) 95 , 5 i i ) 135 b) i) 0, 7 i i ) 0 c) i) –7, 0, 7 i i ) 0

d) i) – 52 , 3 i i ) –15 e) i) – 5

2 , 34 i i ) –15 f) i) –2, 30 i i ) –120

5 . a) zero: 3 b) zeros: –2, 4 c ) zeros: –2, 3, 4factor: x - 3 factors: x + 2, x - 4 factors: x + 2, x - 3, x - 4y-intercept: 6 y-intercept:–8 y-intercept: 24

6 . a) –1, 4 b) –12 , 7

9 c ) 23 , 0, 3

7 . a) –2.1, 5.1 b) –0.3, 1.0, 1.3

8 . a) y = (x - 6)(x - 1) b) y = (2x + 3)(x - 3) c ) y = (5x + 3)(x - 3)(x + 1)

9 . A 1 0 . D 1 1 . A 1 2 . 1 2


Quadratic Functions and Equations Lesson #1:Analyzing Quadratic Functions - Part One

Quadratic Function

A quadratic function is a function which can be written in the form

f(x) = ax2 + bx + c, where a, b, c Œ ¬, and a π 0

or in equation form as

y = ax2 + bx + c, where a, b ,c Œ ¬, and a π 0

Quadratic Equation

A quadratic equation is an equation which can be written in the form

ax2 + bx + c = 0, where a, b, c Œ ¬, and a π 0.

The roots of the quadratic equation ax2 + bx + c = 0 are the zeros of the related quadratic function f(x) = ax2 + bx + c.

General and Standard Forms

Quadratic function can be written in general or standard form.

General Form: y = ax2 + bx + c, where a, b, c Œ ¬, and a π 0.

Standard Form: y = a(x - p)2 + q, where a, p, q Œ ¬, and a π 0

In this unit we will study both the general form and standard form, beginning with the standard form in this lesson.

Warm-Up #1 Analyzing the Graph of the Function with Equation y ==== x2.

x

y

5-5

5

0

10• Graph the function with equation y = x2 by completing the table of values. Join the points with a smooth curve. The graph of this function is called a parabola.

x –3 –2 –1 0 1 2 3

y

• The axis of symmetry is the “mirror” line which splits the parabola in half.State the equation of the axis of symmetry for this parabola.

• The vertex of a parabola is where the axis of symmetry intersects the parabola.The vertex can represent a minimum point or maximum point depending on whether the parabola opens up or down.

Label the vertex (V) on the graph and state its coordinates.

• The maximum or minimum value of a quadratic function occurs at the vertex and is represented by the y-coordinate of the vertex. Complete the following:

The _____________ value of the function with equation y = x2 is ____ .

• State the domain and range of the function with equation y = x2, x Œ ¬.

Domain: _____________________ Range: _____________________

Warm-Up #2 Analyzing the Function with Equation y ==== a(x ---- p)))) 2 + q, a = 1.

The next three explorations help us explore some general transformations on the graph of y = x2 and the relationship they have to the standard form y = a(x - p)2 + q where a = 1.

A transformation is an operation which moves (or maps) a figure from an original position to a new position.

In each exploration use a graphing calculator to sketch the equations.

The following explorations can be completed as a class lesson or as an individual assignment. The process used in these explorations will be further developed in grade 12 mathematics.

120 Quadratic Functions and Equations Lesson #1: Analyzing Quadratic Functions - Part One

Exploration 1 Analyzing the Graph of y ==== x 2 ++++ q.

5-5

5

-5

x

yy = x2

The graph of y = f(x) = x2 is shown.

a) Write an equation which represents each of the following:

• y = f(x) + 3 • y = f(x) - 3

b) Use a graphing calculator to sketch y = f(x) + 3 and y = f(x) - 3 on the grid.

c) Complete the following chart.

Function Equation Representing

Function

Vertex Max/Min Value

Equation of Axis of Symmetry

Description of Transformation

y = f(x)

y = f(x) + 3

y = f(x) – 3

y = f(x) + q

y = x2(0, 0) min, 0 x = 0 no transformation

__________ translation

_____ units ______

d) What is the effect of the parameter q on the graph of y = x2 + q?

e) Compared to the graph of y = x2, the graph of y = x2 + q results in

a _________________ translation (or shift) of q units.

If q > 0, the parabola moves _______. If q < 0, the parabola moves _______.

Quadratic Functions and Equations Lesson #1: Analyzing Quadratic Functions - Part One121

Exploration 2 Analyzing the Graph of y ==== (x ---- p)2.

x

y

5-5

5

-5

y = x2The graph of y = f(x) = x2 is shown.

a) Write an equation which represents each of the following:• y = f(x + 3) • y = f(x - 3)

b) Use a graphing calculator to sketch y = f(x + 3) and y = f(x - 3) on the grid.

c) Complete the following chart. Function Equation

Representing Function




y = f(x)

y = f(x + 3)

y = f(x – 3)

y = f(x – p)


__________ translation

_____ units ______

d) What is the effect of the parameter p on the graph of y = (x - p)2?

e) Compared to the graph of y = x2, the graph of y = (x - p)2 results in a _________________ translation (shift) of p units. If p > 0, the parabola moves _______. If p < 0, the parabola moves _______.

Exploration 3 Analyzing the Graph of y ==== (x ---- p )2 ++++ q .

Consider the function f(x) = x2.a) Write an equation which represents f(x + 2) - 4.

b) Predict the transformations on y = x2 in a). Use a graphing calculator to verify the results.

c) Complete the following chart. Function Equation

Representing Function




y = f(x)

y = f(x + 2) – 4


y = f(x – p) + q


Class Ex. #1 Describe how the graphs of the following functions relate to the graph of y = x2.

a) y = (x + 10)2 b) y = x2 + 4 c) y + 8 = (x - 5)2

Class Ex. #2 The following transformation(s) are applied to the graph of y = x2. Write the equation of the image function for each.

a) A horizontal translation of 5 units right.

b) A translation of 6 units down and 4 units left.

Class Ex. #3 Write the coordinates of the image of the point (3, 9) on the graph y = x2 when a translation of two units up and seven units right is applied.


Assignment 1. Describe how the graphs of the following functions relate to the graph of y = x2.

a) y = (x + 5)2 b) y = x2 - 7 c) y - 8 = x2

d) y = 5 +(x - 2)2 e) y + 7 = (x + 1)2 - 10 f) y = (x - a)2 - b


2. Consider the graph of the function f(x) = (x - 2)2 + 3.

x

y

5

5

–5

–5

a) Without using a graphing calculator, sketch the graph on the grid.

b) State the coordinate of the vertex.

c) State the maximum or minimum value of the function.

d) State the domain and range of the function.

3. The following transformation(s) are applied to the graph of y = x2. Write the equation of the image function for each.

a) A horizontal translation of 7 units right.

b) A vertical translation of 2 units down.

c) A translation 3 units left and 8 units up.

d) A translation c units down and d units right.

4. Complete the following table.

Function

Coordinates of Vertex

Max/Min Value

Eqn. of Axis of Symmetry

Domain

Range

y = x2 + 5 y = (x + 3)2 - 4 y + 9 = (x - 6)2 + 1 y - w = (x + r)2

5. Write the coordinates of the image of the point (–2, 4) on the graph y = x2 when each of the following transformations are applied:

a) a horizontal translation of 2 units to the left

b) a translation of 3 units up and 11 units right.


6. After a combination of a horizontal and a vertical translation, the graph of y = x2 has an image graph with a vertex at (2, –6). Describe the translations.

7. Which of the following transformations shifts the graph of y = x2 to the MultipleChoice graph of y + a = (x - b)2?

A. a units right and b units down.B. b units right and a units down .C. b units up and a units right.

D. a units down and b units left.

8. The function defined by the equation y = x2 is transformed to y = (x + 2)2 + 4. If the point (2, 4) lies on the graph of y = x2, which of the following points must lie on the graph of y = (x + 2)2 + 4?

A. (0, 0)B. (4, 0)

C. (4, 8)

D. (0, 8)

Use the following information to answer questions #9 and #10.

5 10

10

5

x

y

• The graph of a quadratic function is shown.

• The four points marked have integer coordinates.

9. The domain and range, respectively, of the function are

A. x Œ ¬ and y Œ ¬B. x ≥ –2 and y Œ ¬

C. x Œ ¬ and y ≥ –2

D. 2 £ x £ 6 and y ≥ –2

10. The sum of the x and y-intercepts is _____ .Numerical Response



Answer Key

1 . a) horizontal translation 5 units left b) vertical translation 7 units downc ) vertical translation 8 units up d) translation 2 units right and 5 units upe ) translation 1 unit left and 17 units down f ) translation a units right and b units down

2 . a)

b) (2, 3) c ) minimum value of 3 d) Domain: xΩx Œ ¬ Range:yΩy ≥ 3, y Œ ¬

3 . a) y = (x - 7)2 b) y = x2 - 2 c ) y = (x + 3)2 + 8 d) y = (x - d)2 - c

4 .

Function

Coordinates of Vertex

Max/Min Value

Eqn. of Axis of Symmetry

Domain

Range

y = x2 + 5 y = (x + 3)2 - 4 y + 9 = (x - 6)2 + 1 y - w = (x + r)2

(0, 5)

min, 5

x = 0

xΩx Œ ¬

yΩy ≥ 5, y Œ ¬

(–3, –4)

min, –4

x = –3

yΩy ≥ –4, y Œ ¬

xΩx Œ ¬

yΩy ≥ –8, y Œ ¬

x = 6

min, –8

(6, –8)

xΩx Œ ¬

(–r, w)

min, w

x = –r

yΩy ≥ w, y Œ ¬

xΩx Œ ¬

5 . a) (–4, 4) b) (9, 7)

6 . horizontal translation 2 units right, vertical translation 6 units down.

7 . B 8 . D 9 . C 1 0 . 1 4


Quadratic Functions and Equations Lesson #2:Analyzing Quadratic Functions - Part Two

In the last lesson we analyzed the graph of y = (x - p)2 + q and discovered transformations associated with the parameters p and q. In this lesson we investigate the effect of the parameter a on the graph of y = a(x - p)2 + q. The following explorations can be completed as a class lesson or as an individual assignment.

Exploration 1 Analyzing the Graph of y ==== a (x ---- p )2, a > 0.

x5-5

5

y

10The graph of y = f(x) = (x - 2)2 is shown.

a) Write an equation which represents each of the following:

• y = 2f(x) • y = 12 f(x)

b) Use a graphing calculator to sketch

y = 2f(x) and y = 12 f(x) on the grid.

c) Complete the following by circling the correct choice and filling in the blank.

• Compared to the graph of y = f(x), the number 2 in the graph of y = 2f(x) results in a vertical expansion / compression by a factor of _____ .

• the y intercept of the graph of y = 2f(x) is ____________ the y-intercept of the graph of y = f(x).

d) Complete the following by circling the correct choice and filling in the blank.

• Compared to the graph of y = f(x), the number 12 in the graph of y =

12 f(x) results in a

vertical expansion / compression by a factor of _____ .

• the y intercept of the graph of y = 12 f(x) is ____________ the y-intercept of the

graph of y = f(x).

Note • In mathematics the general name given to an expansion or a compression is a stretch.• A vertical stretch is “anchored” by the x-axis, i.e. the x-coordinate of every point on the

original graph will not change and the y-coordinate of every point is multiplied by a factor of a.

• In some texts a compression is called a contraction.

e) Describe the effect of the parameter a on the graph of y = a(x - p)2 where a > 0.

f) Compared to the graph of y = x2, the graph of y = ax2 results in a vertical stretch of factor _____ .If a > 1, the parabola undergoes a vertical ____________ by a factor of _____ . If 0 < a < 1, the parabola undergoes a vertical ____________ by a factor of _____ .

Exploration 2 Analyzing the Graph of y ==== ax2, a < 0.

5-5

5

-5

x

yy = x2The graph of y = f(x) = x2 is shown.

a) Write an equation which represents:

• y = –f(x) • y = –2f(x)

b) Use a graphing calculator to sketchy = –f(x) and y = –2f(x).

c) Complete the following chart. The first row is done.

Function Equation Representing

Function




y = f(x)

y = –f(x)


y = –2f(x)

y = af(x), where a < 0

d) How does the graph of y = –x2 compare to the graph of y = x2?

e) Compared to the graph of y = x2, the graph of y = ax2, a < 0 results in a

____________ in the ___________ and a _____________ stretch by a factor of _____ .

128 Quadratic Functions and Equations Lesson #2: Analyzing Quadratic Functions - Part Two

Transformations Associated with the Parameters of y ==== a(x - p)2 ++++ q

Compared to the graph of y = x2, the following transformations are associated with the parameters of y = a(x - p)2 + q:

a Æ a vertical stretch • if a > 1 there is an expansion • if 0 < a < 1 there is a compression • if a < 0, there is also a reflection in the x-axis

p Æ a horizontal translation where ÓÏÔÔÔÌ

p > 0 the parabola shifts p units rightp < 0 the parabola shifts p units left

q Æ a vertical translation where ÓÏÔÔÔÌ

q > 0 is q units up q < 0 is q units down

(p, q) Æ the coordinates of the vertex

x = p Æ the equation of the axis of symmetry

Class Ex. #1 Consider the function f(x) = 2(x + 4)2 - 3.a) State the transformations applied to the graph of y = x2 which would result in the

graph of y = 2(x + 4)2 - 3.

b) Marika and Curtis were discussing how to graph this function without using a graphing calculator. Marika suggested doing the stretch followed by the translation. Curtis suggested doing the translation followed by the stretch.

• Complete the grids below to show the graphs obtained by each student.• Use a graphing calculator to determine which student is correct.

5 10-5-10

5

10

-5

-10

x

y

y = x2Marika’s Graph

5 10-5-10

5

10

-5

-10

x

y

y = x2Curtis’s Graph

Quadratic Functions and Equations Lesson #2: Analyzing Quadratic Functions - Part Two 129

Note Unless otherwise indicated, use the following order to describe how to transform from one graph to another.

1. Expansions and/or compressions2. Reflections3. Translations

Class Ex. #2 Describe how the graphs of the following functions relate to the graph of y = x2.

a) y = –1

4 x2 b) 13 y = (x + 6)2

Class Ex. #3 The following three transformations are applied, in order, to the graph of y = x2: a reflection in

the x-axis, a vertical compression by a factor of 13 , and a translation 7 units right. At the end

of the three transformations, the point (1, t) is on the resulting graph.

a) Find the equation of the image function after each transformation.

b) State the coordinates of the vertex of the final graph.

c) Find the value of t.

Class Ex. #4 Complete the following table.

Function Vertex Max/Min Value


Domain Range

y = –(x + 3)2 - 4

y = 3(x - 9)2



Assignment

1. Describe how the graphs of the following functions relate to the graph of y = x2.

a) y = –3x2 b) y = x2 - 15

c) y = –23 (x + 4)2 - 1 d) 2y = (x - 8)2 + 12

2. The following transformations are applied to the graph of y = x2 in the order given. Write the equation of the image function for each.

a) A reflection in the x-axis and a vertical expansion by a factor of 4.

b) A vertical compression by a factor of 35 , and a translation of 5 units down.

c) A vertical expansion by a factor of 8, a reflection in the x-axis, a vertical translation of 3 units up, and a horizontal translation 9 units left.

d) A vertical stretch by a factor of c, a reflection in the x-axis, and a translation of e units right and f units down.

3. Complete the following table.

Function Vertex Max/Min Value


Domain Range

y = 3x2

y = 2x2 + 1

y = –(x + 7)2

y - 10 = (x + 5)2

y + 3 = –3(x - 1)2 + 2


4. The following transformations are applied, in order to, the graph of y = x2

• a reflection in the x-axis• a vertical stretch of factor 3• a translation of 5 units right and 2 units down.

a) Find the equation of the image function after each transformation.

b) At the end of all the transformations, the point (4, y) is on the final graph of the parabola. Find the y-coordinate for the final graph when x = 4.

5. The graph of f(x) = x2 undergoes a series of transformations.

x

y

5

5

–5

–5

a) State the transformations applied to the graph of f(x) = x2

which would result in the graph of f(x) = –12 (x - 2)2 + 1.

b) Without using a graphing calculator, sketch the graph of f(x) = –12 (x - 2)2 + 1.

c) Verify using a graphing calculator

6. Write the coordinates of the image of the point (–3, 9) on the graph y = x2 when each of the following transformations are applied:

a) A reflection in the x-axis, followed by a vertical translation of 4 units up.

b) A vertical compression by a factor of 13 .

7. Write the equation of a quadratic function which is the image of y = x2 after a vertical stretch by the given factor of a and after a translation which results in the given vertex.

a) a = 3, vertex (4, –1) b) a = 12 , vertex (–3, 2)

c) a = –4, vertex (0, 5) d) a = –13 , vertex (–6, –3)


8. The quadratic function f(x) = x2 is transformed to f(x) = –12 (x + 3)2 + 1. Multiple

ChoiceThe point (1, 1) on the graph of y = x2 is transformed to which point on the graph of

y = –12 (x + 3)2 + 1?

A. ËÊÁÁ–2,

12

ˆ˜

B. ËÊÁÁ–2,

32

ˆ˜

C. (–2, –1)

D. ËÊÁÁ

52 , 2

ˆ˜

-5 5 10

10

5

-5

-10

x

y

1

2

3

4

9. The diagram shows the graphs of four Numerical Response quadratic functions.

In the first box, write the numbercorresponding to the graph of

y = 12 (x - 5)2 - 3.

In the second box, write the numbercorresponding to the graph of y = –3(x + 4)2 + 2.

In the third box, write the numbercorresponding to the graph of y = (x - 3)2 + 4.

In the last box, write the numbercorresponding to the graph of

y + 3 = –14 (x - 5)2.


10. The following transformations are applied, in order to, the graph of y = x2

• a vertical stretch of factor 2

• a reflection in the x-axis• a vertical translation of 12 units up.

At the end of all the transformations, the point (2, y) is on the final graph of the parabola. The value of y, to the nearest tenth, is _____ .



Answer Key 1 . a) vertical expansion by a factor of 3 and a reflection in the x-axis

b) vertical translation 15 units down

c ) vertical compression by a factor of 23

, a reflection in the x-axis, a translation 4 units left, 1 unit down.

d) vertical compression by a factor of 12

, and a translation 8 units right, 6 units up.

2 . a) y = –4x2 b) y = 35 x2 - 5 c ) y = –8(x + 9)2 + 3 d) y = –c(x - e)2 - f

3 .Function Vertex Max/Min

ValueEquation

of Axis of Symmetry

Domain Range

(0, 0) min, 0

(0, 1) min, 1

(–7, 0) max, 0

(–5, 10) min, 10

(1, –1) max, –1

y = 3x2

y = 2x2 + 1

y = –(x + 7)2

y - 10 = (x + 5)2

y + 3 = –3(x - 1)2 + 2

x = 0

x = 0

x = –7

x = –5

x = 1

xΩx Œ ¬

xΩx Œ ¬

xΩx Œ ¬

xΩx Œ ¬

xΩx Œ ¬

yΩy ≥ 0, y Œ ¬

yΩy ≥ 1, y Œ ¬

yΩy £ 0, y Œ ¬

yΩy ≥ 10, y Œ ¬

yΩy £ –1, y Œ ¬

4 . a) y = –x2, y = –3x2, y = –3(x - 5)2 - 2 b) –5

5 . a) vertical compression by a factor of 12

, a reflection in the x-axis, a translation 2 units right, 1 unit up.

6 . a) (–3, –5) b) (–3, 3)

7 . a) y = 3(x - 4)2 - 1 b) y = 12 (x + 3)2 + 2 c ) y = –4x2 + 5 d) y = – 1

3 (x + 6)2 - 3

8 . A 9 . 1 4 3 2 1 0 . 4 . 0


Quadratic Functions and Equations Lesson #3: Equations and Intercepts from the Vertex and a Point

Warm-Up

In the last lesson we analyzed the graphs of quadratic functions with equations in standard form y = a(x - p)2 + q. In this lesson we determine the equation of a quadratic function from the graph. To do this we need the vertex of the parabola and a point on it. We will also learn how to find intercepts from the standard form of the equation.

Determining the Equation from the Vertex and a Point

The following procedure will enable us to write quadratic functions in standard form if we are given the coordinates of the vertex and of another point on the parabola.


Replace p and qwith the coordinates of the vertex.

Replace x and ywith the coordinates of the point.

Solve for a.Replace the valuesof a, p and q in the equationy = a(x - p)2 + q.

Class Ex. #1 The graph of a quadratic function has vertex (–2, 8) and passes

5-5

5

-5

x

y

through the point (–1, 7).

a) Find the equation of the function in standard form y = a(x - p)2 + q.

b) Rewrite the equation in general form y = ax2+ bx + c.

c) Use a graphing calculator to sketch the graph and determine the x and y-intercepts ofthe graph of the function. Answer to the nearest hundredth if necessary.

Finding Intercepts from the Standard Form

We can use the equation of a quadratic function written in standard form to algebraically determine the x- and y-intercepts of the graph of the function.

Class Ex. #2 Determine, as exact values, the x and y-intercepts of the graph of the function f(x) = 3(x - 1)2 - 9.

Class Ex. #3 The graph of a quadratic function is shown.

-5 5

5

-5

x

y

(–1, 2)(–3, 1)

a) Find the equation of the function in standard form.

b) Find, algebraically, the x-intercepts and y-intercepts of the graph. Answer both as exact values and to the nearest hundredth.

c) State the domain, range and equation of the axis of symmetry.


136 Quadratic Functions and Equations Lesson #3: Equations and Intercepts from the Vertex and a Point

Assignment

1. The graph of a quadratic function has vertex (3, –4) and

5-5

5

-5

x

y

passes through the point (4, 1).

a) Find the equation of the function in standard form.

b) Rewrite the equation in general form.

c) Use a graphing calculator to sketch the graph and determine the x and y-intercepts ofthe graph of the function. Answer to the nearest hundredth if necessary.

2. In each case write an equation in standard form for the parabola with the given vertex and passing through the given point.

a) vertex (7, –6), point (9, –4) b) vertex (–2, 5), point (–4, 21)

c) vertex (–1, 0), point (–5, –12) d) vertex (3, –8), y-intercept is 10

Quadratic Functions and Equations Lesson #3: Equations and Intercepts from the Vertex and a Point 137

3. The graph of a quadratic function has a vertex at ËÊÁÁ

53 , 1

ˆ˜ and one x-intercept is

23 .

a) Determine the equation of the function in standard form.

b) Determine the equation of the function in general form.

c) State the other x-intercept.


e) State the equation of the axis of symmetry of the graph.

4. Determine, as exact values, the x and y-intercepts of the graph of the following functions. a) f(x) = (x - 4)2 - 16 b) f(x) = –3(x + 2)2 + 3

c) f(x) = 2(x - 6)2 - 6 d) f(x) = –14 (x + 1)2 + 5


5. The graph of a quadratic function is shown.

-10 -5 5

20

10

x

y

(1, 24)

(–9, 9)

The equation of the axis of symmetry is x = –5.

a) Find the equation of the graph of the function in standard form.

b) Find, algebraically, the x-and y-intercepts of the graph.

c) State the domain, range, and equation of the axis of symmetry.

6. Write an equation of the form y = a(x - p)2 + q for each parabola.

-5 5

10

5

x

y

(0, 5)

(–2, 8)

-5 5

5

-5

x

y

(–4, 1)

(–3, 4)

a) b)


7. A function of the form p(x) = ax2 + q has two x-intercepts, one of which is 9. Determine the other x-intercept and explain how you arrived at your answer.

8. The parabola with equation y = a(x - 2)2 + q passes through the points (–2, 5) and (4, –1).Determine the coordinates of the vertex of the parabola.

9. The graph of the function with equation y = a(x + 5)2 + q passes through the points (–6, 2) and (–3, 20).Determine whether the function has a maximum or minimum value and state the value.

10. The parabola with equation y = a(x - p)2 + q has a maximum value of 8. MultipleChoice The line x = 2 is the axis of symmetry of the parabola. If the graph passes through

the origin, then the value of a is

A. 2B. 1

32

C. –2

D. – 132


11. The graph of the function g(x) = –2(x - 3)2 + q passes through the point (–5, –2). Numerical Response The value of q, to the nearest whole number, is _____ .


12. The graph of a function of the form f(x) = a(x + 2)2 - 7 has two x-intercepts, one of which is –6.5. The other x-intercept, to the nearest tenth , is _____ .(Record your answer in the numerical response box from left to right)

Answer Key 1 . a) y = 5(x - 3)2 - 4 b) y = 5x2 - 30x + 41 c ) x-intercepts are 2.11, 3.89

y-intercept is 412 . a) y = 1

2 (x - 7)2 - 6 b) y = 4(x + 2)2 + 5 c ) y = – 34 (x + 1)2 d) y = 2(x - 3)2 - 8

3 . a) y = - ËÊÁx - 5

3 2 + 1 b) y = –x2 + 10

3 x - 169 c ) 8

3

d) D: xΩx Œ ¬, R: yΩy £ 1, y Œ ¬ e ) axis of symmetry x = 53

4 . a) x - intercepts 0 and 8 b) x - intercepts -3 and -1 y -intercept 0 y -intercept –9

c ) x - intercepts 6 + 3 and 6 - 3 d) x - intercepts -1 + 2 5 and -1 - 2 5y -intercept 66 y -intercept 19

4

5 . a) y = 34 (x + 5)2 - 3 b) x-intercepts are –7 and –3, y-intercept 63

4c ) Domain: xΩx Œ ¬, Range: yΩy ≥ –3, y Œ ¬, axis of symmetry x = –5

6 . a) y = 34 x2 + 5 b) y = –3(x + 3)2 + 4

7 . The vertex of the parabola is on the y-axis so the x-intercepts are an equal distance on either side of x = 0.If one x-intercept is 9, the other must be -9.

8 . (2, –3) 9 . minimum value of –4

1 0 . C 1 1 . 1 2 6 1 2 . 2 . 5


Quadratic Functions and Equations Lesson #4:Converting from General Form to Standard Form

by Completing the Square

Warm-Up #1 Review

The general form of a quadratic function has the equation y = ax2 + bx + c.

The standard form of a quadratic function has the equation y = a(x - p)2 + q.

Writing a function in standard form enables us to analyze the function more easilye.g. we can determine the vertex, axis of symmetry and maximum / minimum value of the function.

Warm-Up #2 Completing the Square

(x + 4)2, (x - 5)2 etc are examples of perfect squares.

a) Expand the following perfect squares.

(x + 4)2 = (x + 4)(x + 4) = ____________ (x + 7)2 = (x + 7)(x + 7) = ____________

(x - 5)2 = (x - 5)(x - 5) = ____________ (x - 1)2 = (x - 1)(x - 1) = ____________

(x + a)2 = ____________ (x - a)2 = ____________

b) Factor the following expressions into perfect squares.

x2 + 6x + 9 = ___________ x2 + 12x + 36 = ___________

x2 - 4x + 4 = ___________ x2 - 16x + 64 = ___________

c) Add an appropriate constant so that the following expressions can be written as perfect squares.

x2 + 2x + __ = _______ x2 + 18x + __ = _______

x2 - 3x + __ = _______ x2 - 14 x + __ = _______

The process of adding a constant term to a quadratic expression to make it a perfect square is called completing the square.

To complete the square of x2 + bx, add ËÊÁÁ

12 coefficient of x

ˆ˜

2 i.e. Ë

ÊÁÁ

12 b

ˆ˜

2 to give Ë

ÊÁÁx + 1

2 b ˆ˜

2

Writing f(x) ==== x2 ++++ bx ++++ c in Standard Form by Completing the Square

Use the following process to convert a function of the form f(x) = x2 + bx + c into standard form.

Add and subtract the squared number in step 1. (This keeps the value of the function the same).

Take half of the numerical coefficient of the x-term and square it.

Form a perfect square, write it infactored form, and simplify.

Step 1 Step 2 Step 3

Class Ex. #1 Express y = x2 + 10x + 16 in completed square form. Use a graphing calculator to verify that both equations are represented by identical graphs.

Class Ex. #2 A function f is defined by f(x) = x2 - 9x - 20. Determine the minimum value of f by writing the function in standard form.


144 Quadratic Functions and Equations Lesson #4: Completing the Square

Writing f(x) ==== ax2 ++++ bx ++++ c in Standard Form by Completing the Square

Add and subtract the squared number in step 2 inside the brackets.

Take half of the numerical coefficient of the x-term and square it.

Use the coefficient of x2 as a common factor for the x2 term and the x term only.

Form a perfect square within the brackets and write it in factored form.

Step 1 Step 2 Step 3

Step 4 Step 5

Multiply the term which has been subtracted in step 3 by the coefficient of x2 and combine with the constant term.

Class Ex. #3 Convert f(x) = 3x2 - 18x + 20 to standard form by completing the square. Determine whether the graph of the function f has a maximum or minimum value and state the value.

Class Ex. #4 Convert y = 7 + 10x – 2x2 to standard form by completing the square. What direction does the parabola open? What are the coordinates of the vertex of the parabola?

Quadratic Functions and Equations Lesson #4: Completing the Square 145

Class Ex. #5 Express f(x) = 3x2 - 12x - 8 in completed square form and use this form to determine the zeros of the function. Answer to the nearest hundredth.


Assignment 1. What number must be added to each to make a perfect square?

a) x2 + 8x b) x2 - 24x c) x2 + 40x d) x2 - x e) x2 + 12 x f) x2 -

23 x

2. Complete the square in each part. a) x2 + 6x + __ = (x + __ )2 b) x2 - 20x + __ = (x __ )2

c) x2 + 5x + __ = (x __ )2 d) x2 - 9x + __ = (x __ )2

e) x2 + 0.6x + __ = (x __ )2 f) x2 - 34 x + __ = (x __ )2

3. Express the following in completed square form. a) y = x2 + 10x + 3 b) y = x2 - 4x - 21 c) y = x2 + 14x - 2

d) y = x2 + 9x + 22 e) y = x2 - x + 1 f) y = x2 + bx + c


4. Express f(x) = x2 - 14x - 40 in completed square form. Hence state the coordinates of the vertex and the equation of the axis of symmetry of the graph of the function.

5. Express the following in completed square form.

a) y = 2x2 + 12x + 5 b) y = 3x2 - 18x - 19 c) y = 2x2 + 14x - 11

d) y = -x2 + 10x + 20 e) y = -4x2 - 8x + 7 f) y = –x2 + bx + c

g) y =11x - x2 h) y = 5x2 - 20x + m i ) y = -3x2 + 12x - 11


6. Write the function f(x) = ax2 + bx + c in completed square form.

7. When y = 2x2 + 5x + 10 is converted to the form y = a(x - p)2 + q, the value of q isMultipleChoice

A. - 2.5B. 3.75

C. 6.875

D. 8.4375

8. The x-coordinate of the vertex of the graph of the function with equation y = bx - 4x2 is

A. b4

B. b8

C. b16

D. b2

16

9. A high school student was asked to arrange the equation y = –3x2 - 6x - 5 in the form y = a(x - p)2 + q by completing the square. The student’s procedure is shown:

Step I: y = –3(x2 + 2x ) - 5Step II: y = –3(x2 + 2x + 1 - 1) - 5Step III: y = –3(x + 1)2 - 5 - 1Step IV: y = –3(x + 1)2 - 6

The student made an error in

A. Step IB. Step IIC. Step IIID. Step IV


10. The x-intercepts of the graph of the function f(x) = x2 - 8x - 4 are

A. 2 ± 4 5B. 4 ± 5 C. 4 ± 2 5

D. –4 ± 2 5

11. The maximum value, to the nearest tenth, of the function f(x) = -5x2 + 10x + 12 is _____Numerical Response


Answer Key

1 . a) 16 b) 144 c ) 400 d)14

e )116

f )19

2 . a) x2 + 6x + 9 = (x + 3)2 b) x2 - 20x + 100 = (x - 10)2 c) x2 + 5x + 254

= ËÊÁÁx +

52

ˆ˜2

d) x2 - 9x + 814

= ËÊÁÁx -

92

ˆ˜2

e ) x2 + 0.6 + 0.09 = (x + 0.3)2 f) x2 - 34

x + 964

= ËÊÁÁx -

38

ˆ˜2

3 . a) y = (x + 5)2 - 22 b) y = (x - 2)2 - 25 c ) y = (x + 7)2 - 51

d) y = ËÊÁx + 9

2 2 + 7

4 e ) y = ËÊÁx - 1

2 2 + 3

4 f ) y = ËÊÁx + b

2 2 + c - b 2

4

4 . (7, –89), x = 7

5 . a) y = 2(x + 3) 2 - 13 b) y = 3(x - 3) 2 - 46 c ) y = 2 ËÊÁx + 7

2 2 - 71

2

d) y = –(x - 5) 2 + 45 e ) y = –4(x + 1) 2 + 11 f ) y = – ËÊÁx - b

2 2 + c + b 2

4

g ) y = – ËÊÁx - 11

2 2 + 121

4 h) y = 5(x - 2) 2 + m - 20 i ) y = –3(x - 2) 2 + 1

6 . f(x) = a ËÊÁx + b

2a 2 + 4ac - b 2

4a

7 . C 8 . B 9 . C 1 0 . C 1 1 . 1 7 . 0


Quadratic Functions and Equations Lesson #5:Roots of Quadratic Equations - The Quadratic Formula

Warm-Up Review

A quadratic equation is an equation of the form ax2 + bx + c = 0.

a) Find the roots of the equation x2 + 7x - 18 = 0 by inspection.

b) Find the roots of the equation 6x2 - x - 12 = 0 by decomposition.

c) Is it possible to solve the quadratic equation 2x2 - 8x + 5 = 0 by either of the above methods? Explain.

Developing the Quadratic Formula

In the warm-up above, we were unable to solve the equation 2x2 - 8x + 5 = 0 by inspection or decomposition. Another method is required.

There are two further algebraic methods which can be used - completing the square and the quadratic formula. We will solve the equation 2x2 - 8x + 5 = 0 by completing the square and use this technique to develop the quadratic formula.

Class Ex. #1 Solve the following equations by completing the square.a) 2x2 - 8x + 5 = 0 b) ax2 + bx + c = 0

The solution to Class Ex. #1b) is a formula which can be used to solve any quadratic equation of the form ax2 + bx + c = 0. The formula is known as the quadratic formula.

Solving a quadratic equation by completing the square is rarely used as the quadratic formula is usually a more efficient method.

152 Quadratic Functions and Equations Lesson #5: The Quadratic Formula

The Quadratic Formula

The quadratic equation ax2 + bx + c = 0, a π 0 has the roots

x = –b ± b2 - 4ac

2a

Class Ex. #2 Find the roots of the following equations using the quadratic formula. Give answers as exact values in simplest form and to the nearest tenth.

a) x2 + 2x - 1 = 0 b) 4x2 - 12x + 3 = 0 c) 4x2 = 3(4x + 5)

Class Ex. #3 Find the zeros of the quadratic function f(x) = –3x2 + 4x + 1. Give answers as exact values in simplest form and to the nearest hundredth.


Quadratic Functions and Equations Lesson #5: The Quadratic Formula 153

Assignment

1. Solve the equation x2 - 3x - 10 = 0 by using; a) inspection b) the quadratic formula

2. Solve the equation 4x2 - 11x - 3 = 0 by using;a) decomposition b) the quadratic formula

3. Find the exact roots of the equation 6x2 + 5x + 1 = 0 by using; a) graphing b) the quadratic formula


4. Find the roots of the following quadratic equations (to the nearest tenth) using the quadratic formula. a) 2x2 + x - 4 = 0 b) 2x2 - 3x - 4 = 0 c) 10t2 = 7t + 1

5. Solve the following quadratic equations (as exact values) using the quadratic formula. a) x2 - 10x - 15 = 0 b) x2 + 6x + 7 = 0 c) 3x2 - 12x + 11 = 0

6. Find the zeros of the following quadratic functions Give answers as exact values in simplest form and to the nearest hundredth. a) f(x) = x2 + 20x + 15 b) f(x) = 5x2 + 12x - 5

Quadratic Functions and Equations Lesson #5: The Quadratic Formula 155

7. The roots of the quadratic equation dx2 + ex + f = 0 areMultipleChoice

A. x = e ± e2 - 4df

2d

B. x = –e ± e2 - 4df

2d

C. x = e ± e2 + 4df

2d

D. x = –e ± e2 + 4df

2d

8. The zeros of the quadratic function f(x) = 6x2 + 2x - 1 are

A. –1 ± 146

B. –1 ± 2 76

C. –1 ± 76

D. –2 ± 76

9. The quadratic equation 2x2 + 15x + p = 0 has a positive root of –12 when p has the whole Numerical

Responsenumber value of _____ .(Record your answer in the numerical response box from left to right)

Answer Key 1 . a) –2, 5 b) –2, 5 2 . a) –

14

, 3 b) –14

, 3

3 . a) –13

, –12

b) –13

, –12

4 . a) –1.7, 1.2 b) –0.9, 2.4 c ) –0.1, 0.8

5 . a) 5 ± 2 10 b) –3 ± 2 c )6 ± 3

3

6 . a) –10 ± 85 –0.78, –19.22 b)–6 ± 61

5 –2.76, 0.36

7 . B 8 . C 9 . 7


Quadratic Functions and Equations Lesson #6:Roots of Quadratic Equations - The Discriminant

Warm-Up #1 Review

Find the roots of the quadratic equation x2 - 8x + 12 = 0 by each of the following methods:

i) by graphing ii) by factoring

iii) by completing the square iv) by the quadratic formula

x

y

Class Ex. #1 Discuss when each of the following methods might be appropriate or not appropriate for solving a quadratic equation. • by factoring using inspection or decomposition

• by quadratic formula

• by completing the square

• by graphing

Class Ex. #2 Form a quadratic equation and solve.

2a2

+ 3a

= –1, a π 0


Warm-Up #2

Insert the missing values.

5

15

10

5

x

y

5

15

10

5

x

y

x5

10

5

15

y

y = x2 - 6x + 5

y = x2 - 6x + 9y = x2 - 6x + 13

Equation #1

x 2 - 6x + 5 = 0

x =–b ± b2 - 4ac

2a

x =6 ±

2

=6 ±

2

=6 +

2 and6 -

2

\ the roots are

x = and x =

Equation #2

x 2 - 6x + 9 = 0

x =–b ± b2 - 4ac

2a

x =6 ±

2

=6 ±

2

=6 +

2 and6 -

2

\ the roots are

x = and x =

Equation #3

x 2 - 6x + 13 = 0

x =–b ± b2 - 4ac

2a

x =6 ±

2

=6 ±

2

\ the roots are

158 Quadratic Functions and Equations Lesson #6: The Discriminant

The Nature of the Roots of a Quadratic Equation

The roots of a quadratic equation are represented by the x-intercepts of the graph of the corresponding quadratic function.

The roots of a quadratic equation can be equal or unequal and real or non-real.

Consider the graphs from the previous page.

• In graph 1 the roots of the equation x2 - 6x + 5 = 0 are real and unequal (distinct).

• In graph 2 the roots of the equation x2 - 6x + 9 = 0 are real and equal.

• In graph 3 the roots of the equation x2 - 6x + 13 = 0 are non-real.

The Discriminant

The nature of the roots of a quadratic equation can be determined without actually solving the equation or drawing its graph.

The number b2 - 4ac, which appears under the radical symbol in the quadratic formula can be used to discriminate between the different types of roots and is called the discriminant.

discriminant = b2 - 4ac

Class Ex. #3 a) Complete the table using the calculations from Warm-Up #2

Equation Roots Nature of Roots b2 ---- 4ac

x 2 - 6x + 5 = 0

x 2 - 6x + 9 = 0

x 2 - 6x + 13 = 0

b) Complete the following:

• If the discriminant b2 - 4ac = 0 then the roots are ____________ and ____________ .

• If the discriminant b2 - 4ac > 0 then the roots are ____________ and ____________ .

• If the discriminant b2 - 4ac < 0 then the roots are ____________ .

Quadratic Functions and Equations Lesson #6: The Discriminant 159

Class Ex. #4 Determine the nature of the roots of the following equations without solving or graphing. a) 6x2 - x - 1 = 0 b) x2 + 16 = 8x c) 5x2 + 2x + 1 = 0.

Class Ex. #5 Determine for what value(s) of m the quadratic equation x2 - 8x + m has: a) real and distinct roots b) real and equal roots c) non-real roots

Class Ex. #6 a) State a condition for b2 - 4ac so that the equation ax2 + bx + c = 0 has real roots.

b) Show that the roots of the equation (m - 2)x2 - (3m - 2)x + 2m = 0 are always real.



Assignment

1. Form a quadratic equation and solve. Answer to the nearest tenth.

a) x + 1

x = 3, x π 0 b) (2x - 1)(3x + 2) = (x + 3)(2x + 1)

2. Form a quadratic equation and solve. Give answers as exact values in simplest form

a) 4

x2 +

2x

= 3 b) 3x(x - 4) = 8 c) 3(x - 1)(x + 2) - (x2 + 3) = 0

3. Find a quadratic equation in simplest form which is equivalent to the given equation, but has integral coefficients. Hence find the roots of the given equation to the nearest tenth.

a) 1.4x2 - 2.8x = 1.8 b) x2

2 - x - 54 = 0


4. Find the value of the discriminant in each of the following equations. a) x2 + x + 9 = 0 b) 3x2 - 18x + 27 = 0

5. Determine the nature of the roots of the following equations without solving or graphing. a) 2x2 + 4x + 8 = 0 b) 9x2 - 24x + 16 = 0 c) –2x2 - x + 3 = 0

d) –2(x + 3)2 + 40 = 0 e) x2 + 10 + 3x = 0 f) 4x2 + 4x + 1 = 0

6. Determine for what value(s) of d the quadratic equation 5x2 - 10x + d = 0 has:a) real and distinct roots b) real and equal roots c) non-real roots


7. For what values of n does each equation have real roots?

a) nx2 - 2x + 1 = 0 b) 2x2 + 20x + n = 0

8. For what values of a does the equation ax2 + (2a - 3)x + a = 0 have non-real roots?

9. Show that the roots of the equation x(x - 3) = k2 - 2, k Œ ¬, are always real.


Use the following information for questions #10 and #11

I. f(x) = x2 - x - 11 II. f(x) = 2x2 - x + 3

III. f(x) = (3x - 1)(x - 2)(x + 3) IV. f(x) = 4x2 - 12x + 9

Rosa was analyzing the following four functions:

10. Which of these functions is a quadratic function with real and equal zeros?MultipleChoice A. I

B. IIC. IIID. IV

11. Which of these functions is a quadratic function with no real zeros?A. IB. IIC. IIID. IV

12. The discriminant for the quadratic equation 3x2 - 8 - 7x = 0 is _____ .Numerical Response


Answer Key

1 . a) 0.4, 2.6 b) –0.6, 2.1

2 . a)1 ± 13

3b)

6 ± 2 153

c ) –3, 32

3 . a) 7x2 - 14x - 9 = 0 , –0.5, 2.5 b) 2x2 - 4x - 5 = 0, –0.9, 2.9

4 . a) –35 b) 0

5 . a) non-real b) real and equal c ) real and unequald) real and unequal e ) non-real f ) real and equal

6 . a) d < 5 b) d = 5 c ) d > 5

7 . a) n £ 1 b) n £ 50 8 . a > 34

9 . b2 - 4ac = 1 + 4k 2 which is always positive. 1 0 . D 1 1 . B1 2 . 1 4 5


Quadratic Functions and Equations Lesson #7:Applications of Quadratic Functions - A Graphical Approach

Using A Graphing Calculator to Find Maximum or Minimum Values

1. Enter the equation of the function into Y1 and press GRAPH .

2. Access the CALC feature by entering 2nd then TRACE .

3. Select “minimum” or “maximum”.

4. On the bottom left hand side of the screen the calculator will ask for a left bound. Select a value on the left side of the max/min point and press ENTER .

5. On the bottom left hand side of the screen the calculator will ask for a right bound. Select a value on the right side of the max/min point and press ENTER .

6. On the bottom left hand side of the screen the calculator will ask for a guess. Press ENTER . The y value will be the max/min answer.

Class Ex. #1 The height, h, in metres above the ground, of a projectile at any time, t, in seconds, after the launch is defined by the function h(t) = –4t2 + 48t + 3.

Use a graphing calculator to answer the following:

a) Sketch the relevant part of the parabola on the grid.

b) Find the height of the projectile 3 seconds after the launch.

c) Find the maximum height reached by the projectile.

d) How many seconds after the launch is the maximum height reached?

e) What was the height of the projectile at the launch?

f) Determine when the projectile hit the ground to the nearest tenth of a second.

Class Ex. #2 Last season, a struggling hockey club had only 7 200 season ticket holders. The owner of the hockey club has decided to raise the price of a package of season tickets for the new season to generate more revenue. The existing cost of a package of season tickets is $1 400. Before raising the price of a package of season tickets, he hired a market research company to gather data on the proposed increase. The research company reported that for every $25 increase in price, approximately 100 season ticket holders would not renew their season tickets.

If the price increase is to be a multiple of $25, use the following procedure to determine what price would maximize the revenue from season tickets.

a) Let x be the number of $25 increases from the current price of a season ticket. Write expressions in x for the cost of a package of season tickets and the potential number of season ticket holders.

b) Use the results of a) to generate an expression which represents the revenue obtained.

c) Determine the price of a package of season tickets which would generate maximum revenue.

d) How many season ticket holders would there be if this plan was implemented?

e) How much more revenue would be generated if the plan in c) was implemented?

166 Quadratic Functions and Equations Lesson #7: Applications of Quadratic Functions - Graphical

Class Ex. #3 Barry, a high school student, found that driving a truck can be a costly venture depending on how fast he drives. He knew from his Mechanics class that if he drives his truck too slowly, the cost per km is high because the engine does not run efficiently. He also knows from Physics class that if he drives his truck too fast the cost per km is also high because of high wind resistance. He accumulated the following data.

Speed (km/hr) 30 50 65 100 120

Cost per kilometre 27.4¢ 19.4¢ 16.0¢ 16.9¢ 22.9¢

x

y

speed (km/hr)

cost per km (¢)

10

20

30

40

20 40 60 80 100 120

a) If x represents the speed in km/h andy represents the cost per km in cents, plot the data on a Cartesian plane and join the points with a smooth curve.

b) Looking at the graph Barry thought that the data could be modelled by a quadratic function with equation y = ax2 + bx + c.

Use the first, third, and fifth data points in the table to form a system of three equations in the variables a, b, and c.

c) Use the matrix features of a graphing calculator to solve this system and write the equation of the quadratic function which models the data. Answer to three decimal places.

d) Which speed, to the nearest km/h, results in the lowest cost per kilometre? What is this cost to the nearest tenth of a cent?

e) Does it make sense to extend the parabola to the left or right of the data points?


Quadratic Functions and Equations Lesson #7: Applications of Quadratic Functions - Graphical 167

Assignment

1. A football punted during a high school football game followed the path of a parabola.The path can be modelled by the function

d(t) = –5t2 + 15t + 1, t ≥ 0

where t is the number of seconds which have elapsed since the football was punted. d(t) is the number of metres above the ground after t seconds

a) Sketch the graph on the grid.

In the following questions, answer to the nearest hundredth of a unit where necessary.

b) What was the height of the football above the ground as the punter makes contact with the football?

c) What was the height of the football above the ground 1 second after contact?

d) What is the maximum height reached by the football? What relation does this have to the vertex of the parabola?

e) How many seconds had elapsed when the football reached its maximum height?What relation does this have to the vertex?

f) The punt is not fielded by the opposition and the football hits the ground. How many seconds did it take for the football to hit the ground?

g) The original domain was given as t ≥ 0. Write a more accurate domain for the function which describes the path of the football.


2. The cross section of a river, from one bank to the other, can be represented by the function

d(w) = 114 w2 -

57 w

whered(w) is the depth, in metres, of the river w metres from the river bank.

a) Sketch the graph of the cross section of the river using a graphing calculator.

b) Determine the depth of the river 3 metres from the edge.

c) What is the maximum depth of the river, to the nearest hundredth of a metre?

d) How far from the edge of the river, to the nearest tenth of a metre, is the deepest part of the river?

e) What is the width of the river to the nearest tenth of a metre?

3. Recall the following information from Class Ex. #2 on page 166. The hockey club had 7 200 season ticket holders who each paid $1 400 for a package of season tickets. The owner had suggested raising the price to generate more revenue, but knew that the number of season ticket holders would be reduced.

The general manager suggested that more revenue might be obtained by decreasing the price and thus attracting more fans to buy a package of season tickets. The research company that the owner hired to explore the general manager’s suggestion, reported that for every $50 decrease in price, approximately 400 new season ticket holders would be generated.

If the price decrease is to be a multiple of $50, determine the following:

a) The price of a package of season tickets which would generate maximum revenue.

b) The number of season ticket holders which would be generated.

c) The revenue which would be generated if the plan in a) was implemented?

d) What advice would you give the owner in regards to the direction he should take to obtain maximum revenue?


4. The cost of car insurance depends on many factors, one of which is the age of the driver. Insurance companies know that younger drivers under the age of 25 and older drivers over the age of 70 are statistically more likely to have accidents than drivers between the ages of 25 and 70. The following data shows the number of accidents of per million kilometres driven by drivers of a particular age.

Age (x) 18 30 45 60 75

Number of Accidents (y) 5.2 3.1 2.2 2.8 4.7

x

y

age

# ofaccidents

2

4

6

8

20 30 40 50 60 70

a) If x represents the age of drivers andy represents the number of accidents per million kilometres driven, plot the

data on a Cartesian plane and join the points with a smooth curve.

b) The data looks like it could be modelled by a quadratic function with equation y = ax2 + bx + c.

Use the first, third, and fifth data points in the table to form a system of three equations in the variables a, b, and c.

c) Use the matrix features of a graphing calculator to solve this system and write the equation of the quadratic function which models the data. Answer to four decimal places.

d) What age, to the nearest year, results in the lowest number of accidents per million kilometres? What is this number of accidents, to the nearest tenth?

e) Based on this model, who is more likely to have an accident - a 17 year old student or a 79 year old senior?


5. Luigi owns a potato farm in southern Alberta. Each year he faces a dilemma as to when to harvest his crop of potatoes. He know that if he harvests early, the price will be high but his yield will be low and if he harvests late, the price will be low but the yield will be high. From past experience, he knows that if he harvests on July 15, he can expect approximately 2000 kg of potatoes which he could sell at $0.60 per kg. For each week he waits after July 15, he can expect an extra 400 kg of potatoes, but the price will reduce by $0.05 per kg.

When should he harvest his crop for maximum revenue?

Use the following information to answer questions #6 and #7

Researchers predict that the world population will peak sometime during the 21stcentury before starting to decline. In the year 2000, the world population was approximately 6 100 000 000 (or 6.1 billion).

The following model has been suggested as an approximate relationship (up to the year 2100), between the number of years, x, since the year 2000 and the world population, y.

The equation of the relationship is y = –595 000x2 + 83 000 000x + 6 100 000 000

6. The world population is expected to peak in the year _____ .


7. The maximum population, to the nearest tenth of a billion, is expected to be _____ .



Answer Key

1 . b) 1 metre c ) 11 metres d) 12.25 metres. It is the y-coordinate of the vertexe ) 1.50 seconds. It is the x-coordinate of the vertex. f ) 3.07 seconds g ) 0 £ t £ 3.07

2 . b) 1.5 metres c ) 1.79 metres d) 5.0 metres e ) 10.0 metres

3 . a) $1,150.00 b) 9 200 c ) $10,580 000d) It would be better to reduce the price to $1,150 than to increase the price to $1,600.00

4 . b) 324a + 18b + c = 5.2, c ) a = 0.0034 b = -0.3260 c = 9.96322025a + 45b + c = 2.2 d) 48 years, 2.1 accidents5625a + 75b + c =4 .7 e ) both are about equally likely

5 . 3 12 weeks after July 15. 6 . 2 0 6 9 7 . 9 . 0


Quadratic Functions and Equations Lesson #8:Applications of Quadratic Functions - An Algebraic Approach

Maximum/Minimum Applications

ReviewThe standard form of quadratic functions is useful to solve, analyze, and interpret problems which involve the shape of a parabola. Complete the following statements of the standard form of parabola y = a(x - p)2 + q as a review for this lesson.

a) The coordinates of the vertex are _______ .

b) When a___0, the maximum value is _____ .

When a___0, the minimum value is _____ .

c) The equation of the axis of symmetry is _______ .

In this lesson all the questions are intended to be completed algebraically.

Class Ex. #1 Consider the following information taken from Lesson 7, page 165, Class Ex. #1.

“ The height, h, in metres above the ground, of a projectile at any time, t, in seconds, after the launch is defined by the function h(t) = –4t2 + 48t + 3.”

a) Complete the square to write h in standard form.

b) Find the height of the projectile 3 seconds after the launch.

c) Find the maximum height reached by the projectile.

d) How many seconds after the launch is the maximum height reached?

e) What was the height of the projectile at the launch?

f) Determine when the projectile hits the ground to the nearest tenth of a second.

g) Compare the answers from b) - f) with those on page 165.

Class Ex. #2 A rancher has 300 m of fencing with which to form a rectangular corral (an enclosure for confining livestock), one of whose sides is an existing wall which does not require fencing.a) If two of the sides of the rectangle are each x metres in length, show that the area of the

corral can be expressed in the form A(x) = 300x - 2x2.

b) Use the method of completing the square to determine the maximum area possible.

c) State the dimensions of the rectangle which gives the maximum area.

Class Ex. #3 Ashley was asked by her Math teacher to find two numbers which differ by 8 and whose product is a minimum.

a) If x represents the smaller number, write a quadratic expression in x for the product of the two numbers.

b) Write the product in completed square form.

c) Determine the numbers and the minimum product.


174 Quadratic Functions and Equations Lesson #8: Applications of Quadratic Functions - Algebraic

Extension: The Vertex Formula

The coordinates of the vertex of the graph of a quadratic function f(x) = ax2 + bx + c can be found by completing the square as follows:

f(x) = ax2 + bx + c

= a ËÊÁÁx2 +

ba x

ˆ˜ + c

= a Ë

ÊÁÁÁx2 +

ba

x + b2

4a2 -

b2

4a2

ˆ˜˜ + c

= a Ë

ÊÁÁÁx2 +

ba

x + b2

4a2

ˆ˜˜ + c -

b2

4a

= a ËÊÁÁx +

b2a

ˆ˜

2

+ 4ac - b2

4a Vertex = Ë

ÊÁÁ–b2a ,

4ac - b2

4a ˆ˜

Maximum / Minimum value = 4ac - b2

4a

Class Ex. #4 Use an appropriate procedure to determine the coordinates of the vertex of the graph of each of the following functions. State the maximum or minimum value of each function. a) f(x) = 2(x + 5)2 + 8 b) P(x) = –2x2 + 12x - 13

Complete Assignment Question #10

Quadratic Functions and Equations Lesson #8: Applications of Quadratic Functions - Algebraic 175

Assignment In this assignment all the questions are intended to be completed algebraically.

1. At a local golf course, on the par 3, eighth hole, Linda used a seven iron to reach the green. Her golf ball followed the path of a parabola, approximated by the function

h(t) = –5t2 + 25t + 0.05 where t is the number of seconds which have elapsed since Linda hit the ball, and,

h(t) is the height, in metres, of the ball above the ground after t seconds.

a) Write the function in standard form.

b) Find the height of the golf ball 2 seconds after the ball is hit.

c) Find the maximum height reached by the golf ball.

d) How many seconds did it take for the golf ball to reach its maximum height?

e) How high, in centimetres, did Linda tee up her golf ball before she hit it?

f) How long, to the nearest tenth of a second, did it take for the golf ball to hit the ground?

2. The sum of a number, x, and its reciprocal, is 2910 . Form an equation and find the

original number.


3. The perimeter of a rectangular plot of land is 84 metres and its area is 320 metres2. If the length of the plot is represented by x metres, form a quadratic equation in x, and solve it to find the length and width of the plot.

4. The paved walkway from the main school building to the Physical Education block at a school is “L” shaped with the total distance being 180 metres. A student, taking a short cut diagonally across the grass, shortens the distance to 130 m.

a) Draw a sketch to illustrate this information.

b) If one of the “L” shaped sides has a length of x metres, state the length of the other “L” shaped side in terms of x.

c) Use the Pythagorean Theorem to write a quadratic equation in x and hence solve to find the length of the two legs of the paved walkway. Answer to the nearest tenth of a metre.


5. A stone is thrown vertically upward at a speed of 22 m/s. Its height, h metres, after t seconds, is given approximately by the function h(t) = 22t - 5t2. Use this formula to find, to the nearest tenth of a second, when the stone is 15 metres up and explain the double answer.

6. Two numbers have a difference of 20. When the squares of the numbers are added MultipleChoice together, the result is a minimum. The larger of the two numbers is

A. 0

B. 10

C. 20

D. 30

7. A springboard diver’s height, in metres, above the water, is given by the formula Numerical Response h(t) = –5t2 + 8t + 4

where t is the number of seconds which have elapsed since the start of the dive, and,h(t) is the height, in metres, of the diver above the water after t seconds.

The time taken, to the nearest tenth of a second, for the diver to enter the water is _____ .



8. One positive integer is 3 greater than 4 times another positive integer. If the product of the two integers is 76, then the sum of the two integers is _____ .


9. A whole number is multiplied by 5 and added to 3 times its reciprocal to give a sum of 16. The number is _____ .


10. Use the vertex formula to determine the coordinates of the vertex of the graph of each of the following functions. State the maximum or minimum value of each function. a) f(x) = 5x2 + 3x - 2 b) f(x) = –3x 2 - 7x - 1 c) f(x) = x 2 + 9x + 4

Answer Key 1 . a) h(t) = –5(t - 2.5)2 + 31.3 b) 30.05 metres c ) 31.3 metres

d) 2.5 seconds e ) 5 cm f ) 5.0 seconds

2 . 25 or 5

2

3 . length = 32 metres, width = 10 metres

4 . b) (180 - x) metres c ) 2x2 - 360x + 15500 = 0, 108.7m and 71.3m

5 . 0.8 seconds and 3.6 seconds. There are two answers as the stone goes up and then comes down.

6 . B 7 . 2 . 0 8 . 2 3 9 . 3

1 0 . a) vertex ËÊÁ– 3

10 , – 4920 minimum value is – 49

20 b) vertex ËÊÁ– 7

6 , 3712 maximum value is 37

12

c ) vertex ËÊÁ– 9

2 , – 654 minimum value is – 65

4


Quadratic Functions and Equations Lesson #9:Quadratic Inequalities

Quadratic Inequality

An inequality is a mathematical statement which includes one of the following symbols.< > £ ≥

A quadratic inequality takes one of the following forms:

• ax2 + bx + c < 0 • ax2 + bx + c £ 0• ax2 + bx + c > 0 • ax2 + bx + c ≥ 0

Solving Quadratic Inequalities by Graphing

5 10-5-10

5

10

-5

-10

x

y

y = x2 + 2x - 8

y > 0

y < 0

x2 + 2x - 8 > 0 and x2 + 2x - 8 < 0 are quadratic inequalities whose solutions can be determined from the graph of the associated function f(x) = x2 + 2x - 8.

The graph of the function whose equation isy = x2 + 2x - 8 is shown. The x-intercepts are –4 and 2.

The x-coordinates of the points on the part of the graph which is above the x-axis give the solution to the inequality x2 + 2x - 8 > 0.

The x-coordinates of the points on the part of the graph which is below the x-axis give the solution to the inequality x2 + 2x - 8 < 0.

Class Ex. #1 Complete the following using the above graph:

a) The solution to the inequality x2 + 2x - 8 > 0 is _____________ or _____________ .

b) The solution to the inequality x2 + 2x - 8 < 0 is _____________ .

Note The solution in a) can be represented on a number line as

2–4

x

The solution in b) can be represented on a number line as

2–4x

Class Ex. #2 a) Solve the inequality 12 + 4x - x2 ≥ 0.

b) Show the solution on a number line.

c) State the solution to the inequality 12 + 4x - x2 £ 0


Solving Quadratic Inequalities Algebraically

Colin wondered if he could solve the inequality x2 + 2x - 8 > 0 from Class Ex. #1 without drawing a graph.

His work is shown below.

x2 + 2x - 8 > 0

(x + 4)(x - 2) > 0

x + 4 > 0 and x - 2 > 0

x > –4 and x > 2

He concluded that the solution was x > 2.

When he checked his answer with Class Ex. #1 he discovered that he only had part of the solution.

Class Ex. #3 Explain why Colin’s process did not produce the complete solution and determine the correct solution algebraically.

182 Quadratic Functions and Equations Lesson #9: Quadratic Inequalities

Class Ex. #4 Solve the inequality x2 - x - 20 £ 0 by an algebraic process similar to Class Ex. #3.

Alternative Method to Solving Quadratic Inequalities Algebraically

Colin’s elder brother who had just completed high school, showed him a different method which was useful in solving more complicated inequalities.

The method for solving x2 + 2x - 8 > 0 is shown below.

Step 1: Factor the quadratic expression. (x + 4)(x - 2) > 0

Step 2: Determine the zeros of the quadratic expression. –4 and 2

Step 3: Use a chart which shows the sign of each factor (+, 0, –) to the left and right of each of the zeros.

x Æ –4 Æ 2 Æ

x + 4 – 0 + + +

x - 2 – – – 0 +

Product + 0 – 0 +

Step 4: Since the original inequality symbol is > 0, look for the solution under the + in the product row. The solution is x < –4 or x > 2.

Note Although this method may appear complex at first, it certainly makes the solution to polynomial inequalities easier to understand and is used in calculus courses in higher level math courses.

Class Ex. #5 Use the above chart to state the solution to the inequality x2 + 2x - 8 < 0.

Quadratic Functions and Equations Lesson #9: Quadratic Inequalities 183

Class Ex. #6 Solve the inequality 2x2 - 13x + 6 £ 0 by an algebraic technique.


Assignment

1. The graph of y = x2 - 9 is shown.

5 10-5-10

5

10

-5

-10

x

y

y = x2 - 9Write the solution to each of the following.

a) x2 - 9 = 0

b) x2 - 9 £ 0

c) x2 - 9 ≥ 0


2. The graph of y = –15 x2 +

25 x + 7 is shown.

-5 5

5

-5

x

y y = –15 x 2 +

25 x + 7

Write the solution to each of the following.

a) –15 x2 +

25 x + 7 = 0

b) –15 x2 +

25 x + 7 < 0

c) –15 x2 +

25 x + 7 > 0

3. Use a graph to solve each of the following inequalities and show the solution on a number line.

a) x2 - 4x + 3 < 0 b) 2 + x - x2 ≥ 0 c) 2x2 + 7x > –5

d) x2 + 4x > 0 e) x2 - 6x + 9 £ 0 f) –4x2 - 8x + 21 < 0

4. Use a graph to solve each of the following inequalities. Answer to the nearest tenth.

a) x2 - 6x + 1 > 0 b) 7 + 2x - x2 ≥ 0 c) 3x2 - 9x < 4


5. Solve each of the following inequalities algebraically.

a) x2 - 7x + 10 > 0 b) x2 + 5x - 14 < 0

c) 2x2 - x - 15 ≥ 0 d) x2 - 9x £ 0

e) 3x2 + 5x - 2 > 0 f) 24 - 2x - x2 < 0


6. In each case write a quadratic inequality which has the solution given.

a) –5 £ x £ –1 b) x < -2 or x > 3

7. The height, h metres, of an object, at time t seconds, is given by the equation h = 60t - 5t2.

a) Find t when h = 0.

b) Explain why the height of the object cannot be greater than 180 metres.

c) Find the interval for t such that h ≥ 100.

8. Consider the equation f(x) = x2 + mx + 4. Determine the range of values of m for which the equation has two unequal roots.


9. Consider the number line shownMultipleChoice

3–1x

The inequality which has the solution shown is

A. (x - 1)(x + 3) £ 0B. (x - 1)(x + 3) ≥ 0C. (x + 1)(x - 3) £ 0D. (x + 1)(x - 3) ≥ 0

Answer Key

1 . a) x = –3 or 3 b) –3 £ x £ 3 c ) x £ –3 or x ≥ 3

2 . a) x = –5 or 7 b) x < –5 or x > 7 c ) –5 < x < 7

3 . a) 1 < x < 3 b) –1 £ x £ 2 c ) x < – 52 or x > –1

d) x < –4 or x > 0 e ) x = 3 f ) x < – 72 or x > 3

2

4 . a) x < 0.2 or x > 5.8 b) –1.8 £ x £ 3.8 c ) –0.4 < x < 3.4

5 . a) x < 2 or x > 5 b) –7 < x < 2 c ) x £ – 52 or x ≥ 3

d) 0 £ x £ 9 e ) x < –2 or x > 13 f ) x < –6 or x > 4

6 . a) x2 + 6x + 5 £ 0 b) x2 - x - 6 > 0

7 . a) t = 0 or t = 12 b) The maximum height is at t = 6, where h = 180 c ) 2 £ t £ 10

8 . m < –4 or m > 4 9 . C


Polynomial Functions and Equations Lesson #1:Polynomial Functions

Polynomial Function

A polynomial function is a type of function in the form

f(x) = anx n + an - 1x n - 1 + an - 2x n - 2 + ... + a2x 2 + a1x + a0,where

• a0, a1, a2, ... an are real numbers, • an π 0, • and n Œ W, (that is the exponents are non-negative integers)

The values, a1, a2, ... an are called coefficients. The coefficient of the highest power of x, an, is called the leading coefficient and a0 is the constant term. The value of n is the degree of the polynomial.

Class Ex. #1 Consider the polynomial f(x) = x4 + 7x3 - 8x2 + 5. State:

a) the degree of the polynomial

b) the leading coefficient

c) the constant term

Recognizing a Polynomial Function

Expressions containing roots of variables, negative or fractional powers of a variable, or any coefficient which is non-real are NOT polynomial functions.

Class Ex. #2 State whether or not the following are polynomial functions. If they are not polynomial functions explain why.

a) f(x) = - 5x 3 + x12 - 4

b) g(x) = 2x 2 - 7x -1 - 3

c) P(x) = x 4 + 9029x 3 - 17 x 2 + 3897

d) P(x) = 5x 3 - 3x 2 + 2x - 4

e) P(x) = 5x 3 - 3x 2 + 2x - 4

f) P(x) = x + 3x + 2

g) P(x) = 3 x 3 - –3 x

Degree, Leading Coefficient, and Constant Term

Sometimes a polynomial function can be in a “disguised” form.

Class Ex. #3 State the degree, the leading coefficient, and the constant term of each polynomial function.

a) f(x) = 34x2 - 25x3 + 2x - 39 b) f(x) = (5x - 1)(2x + 7)degreeÆ degreeÆleading coefficientÆ leading coefficientÆconstantÆ constantÆ

c) P(x) = 2(3x - 1)2(5x 2 - x + 1) d) P(x) = 2(3x - 1)2 + (5x2 - x + 1)degreeÆ degreeÆleading coefficientÆ leading coefficientÆconstantÆ constantÆ

Types of Common Polynomial Functions

Class Ex. #4 Consider the polynomial function P(x) = 7.

a) Explain why the polynomial function P(x) = 7 has a degree of zero.

b) Explain why this type of function is called a constant function.

Class Ex. #5 Complete the chart. Polynomial Function Degree Type

P(x) = c

P(x) = ax + b, a π 0

P(x) = ax2 + bx + c, a π 0

P(x) = ax3 + bx2 + cx + d, a π 0 cubic

P(x) = ax4 + bx3 + cx2 + dx + e, a π 0 quartic

190 Polynomial Functions and Equations Lesson #1: Polynomial Functions

Note Polynomial functions can also be classified according to their coefficients. For example,

• 3x 4 - 5x 2 + x + 7 is an integral polynomial function.

• 3x 4 - 25 x 2 + x + 7 is a rational polynomial function.

• 3 x 4 - 25 x 2 + x + 7 is a real polynomial function.

Class Ex. #6 If P(x) = -3x2 + ax + 8 and P(1) = -9, then find the value of a.

Class Ex. #7 Determine the values of a and b in P(x) = -2x 2 + ax + b if P(2) = -18 and P(-3) = -13


Polynomial Functions and Equations Lesson #1: Polynomial Functions 191

Assignment

1. Which are not polynomial functions of x? Explain.

a) P(x) = 12 x 2 + 5x b) f(x) = x -4 + x 3 c) g(x) =

x - 3x - 1

d) P(x) = x + 3x - 1 e) g(x) = 4( x - 1)( x + 1)

f) P(x) = 10x 4 + 3x g) P(x) = x 2 + 4 h) P(x) = 3x + 1

2. State the degree, leading coefficient and constant term for each of the following polynomial functions.a) P(x) = 5x 3 - 7x 4 + 2 b) P(x) = (x + 1)(x + 2)(x - 3)

degreeÆ degreeÆleading coefficientÆ leading coefficientÆconstantÆ constantÆ

c) P(x) = 4(4x + 1)2(2x 2 - x + 5) d) P(x) = 3(4x + 1)2 - (4x 2 + x - 1)


e) P(x) = 4x 2 - x 4 + 2x 3 + x 4 f) g(x) = -5(2x 3 + 3x - 2)4



3. Which of the following is a polynomial of degree 5?

a) 5x 4 + 5x b) 3x 5

+ 5 c) x 2 - 3x 5 + x - 4

d) x5 - 2x7 e) 2x 5 + 3x - 7x -3 f) (x 2 + 3)(x 3 + 4)

4. For each polynomial, use substitution to find the indicated value.

a) P(x) = 3x 3 - 5x 2 - 2; P(-3) b) P(x) = 4x 4 - x 3 - x + 3; P ËÊÁÁ

12

ˆ˜

5. Which of the following are polynomial functions when simplified? For those which are, answer the following:

i) Write the expression in simplest polynomial form in descending powers of x.ii) Name the type of polynomial (integral, rational, real) according to its coefficients.iii) State the degree, leading coefficient, and constant term.

a) P(x) = 3x -4 + 2x 3 - 3x -4 + 2x 3 b) P(x) = 2x 2 - 3x -1 + 5x 2 + 7x -1

c) P(x) = 7x 3 - 3x 2 + 2x

2 d) P(x) = 3 - 10x 2 + 3x 4

e) P(x) = 8x 5 - 36x + 2 f) P(x) = ËÊ 3 x - 2 ËÊ 3 x + 2


6. Determine the values of a and b in P(x) = ax 3 + bx 2 + 3x - 4 if P(1) = -2 and P(2) = 2.

7. Determine the values of a, b, and c in P(x) = ax 2 + bx + c if P(0) = 1, P(1) = 6, and P(-1) = 2.

8. Determine the values of p, q, and r in P(x) = px 3 + qx + r if P(0) = 2, P(2) = 3, and P(-1) = 5.

9. Which of the following is an integral polynomial of degree 4?MultipleChoice

A. 2x 4 - 13 x 2 + 1

B. x + x 2 + 3x 4

C. x 4 + 2xx

D. 3x 6 + 2x 4 + 2x + 1


Use the following functions to answer questions #10 - #13

A. f(x) = (x - 1)(2x + 3)2(4 - x)

B. g(x) = 2x4 - 3x5 + x3 - 7x2 + 2x - 3

C. h(x) = 3x3 - 2x2 + 4x–1 + 14

D. P(x) = –5x2 +12 x + 3x3 + 36

10. Which one of these functions is a polynomial function with a leading coefficient of 3?

11. Which one of these functions is a polynomial function with a degree of 4?

12. Which one of these functions would be classified as an integral polynomial function?

13. Which one of these functions has a constant term of –36?

14. Consider the polynomial function P(x) = px 3 + qx + r, where P(0) = 1, P(1) = 3, Numerical Response and P(–2) = –33. The value of p, to the nearest tenth, is _____ .


15. The degree of the polynomial 4(x + 1)3 + (x3 - 2)2 - x2(x4 + 12) is _____ .



Answer Key

1 . No to b,c,d,g,h

2 . a) 4, -7, 2 b) 3, 1, -6 c ) 4, 128, 20d) 2, 44, 4 e ) 3, 2, 0 f ) 12, -80, –80

3 . c, f 4 . a) -128 b)218

5 . a) (i) P(x) = 4x 3 (ii) integral (iii) 3, 4, 0b) not a polynomial function

c ) (i) P(x) = 72

x 3 - 32

x2 + x (ii) rational (iii) 3, 72

, 0

d) (i) P(x) = 3x 4 - 10 x + 3 (ii) real (iii) 4, 3, 3e ) not a polynomial functionf ) (i) P(x) = 3x 2 - 2 (ii) integral (iii) 2, 3, -2

6 . a = 1, b = -2 7 . a = 3, b = 2, c = 1 8 . p = 76

, q = -256

, r = 2

9 . B 1 0 . D 1 1 . A

1 2 . A 1 3 . A 1 4 . 5 . 0 1 5 . 1


Polynomial Functions and Equations Lesson #2:The Division Algorithm and Synthetic Division

Warm-Up

In order to help us understand today’s lesson, do the following review examples.

a) Divide 5773 by 25 using long division b) divide 5775 by 25 using long division

c) Label the quotient, the divisor, the dividend, and the remainder in a) and b)

d) Complete the following:

i ) 5773 = 25 ¥ ________ + ________ i i ) 5775 = 25 ¥ ________ + ________

e) In which of the above examples is 25 a factor of the dividend? Explain why.

Class Ex. #1 a) Use long division to divide f(x) = 3x3 + 4x - 5 by x - 1. Label the quotient, the divisor, the dividend and the remainder.

b) Complete the following:

i ) 3x3 + 4x - 5 = (x - 1) ¥ ________________ + ______ .

The polynomial = the divisor ¥ the quotient + the remainder.

Note When dividing polynomials, the degree of the remainder must be less than the degree of the divisor.

The Division Algorithm

When a polynomial P(x) is divided by a polynomial D(x), there exist unique polynomials Q(x) and R(x) such that

P(x) ==== D(x)....Q(x) ++++ R(x).

• If R(x) = 0, then D(x) and Q(x) are both factors of P(x). • If R(x) π 0, then D(x) and Q(x) are not factors of P(x) and the degree of R(x) is less than

the degree of D(x).

Class Ex. #2 Use the division algorithm to determine the polynomial P(x).

x 2 + 4 2x - 3 P(x)

-2

Synthetic Division

Synthetic division is a technique for dividing a polynomial by a linear binomial, which is much quicker than using long division.

Class Ex. #3 Use synthetic division to divide P(x) = 4x3 - 3x2 - 5x - 15 by x - 2 and express the polynomial in the form of the division algorithm.

Class Ex. #4 a) Use synthetic division to find the quotient and remainder when P(x) = x4 + 2x2 - 2x + 1 is divided by x + 1.

b) Find the value of P(-1). c) Comment on the answers to a) and b).

198 Polynomial Functions and Equations Lesson #2: The Division Algorithm and Synthetic Division

Class Ex. #5 If x + 3 is the divisor in the following synthetic division, calculate the values of m and p.

2 2 -m 16 n

2 2m p


Assignment

1. In the synthetic division below, a polynomial P(x) is divided by x - 2.1 -2 6 3

1 0 6 15a) State the polynomial P(x).

b) State the quotient.

c) State the remainder.

d) Write the above synthetic division in the form of the division algorithm.

2. Use synthetic division to divide the polynomial by the binomial and express each in the form P(x) = D(x).Q(x) + R(x). a) x3 + 2x2 + 3x + 6; x - 2 b) 2x3 - 4x2 - 5x + 9; x + 2

c) x 4 - x 2 + 7; x + 1 d) 2y4 - y5 - y3 + 4y; y - 3

Polynomial Functions and Equations Lesson #2: The Division Algorithm and Synthetic Division 199

3. Find p, q, and r in the synthetic division below in which the divisor is x - 1.

2 3 q 1

2 p 7 r

4. Find m and n in the synthetic division below in which the divisor is x + 2.

2 m -3 n

-4 m

5. Find the remainder on dividing x3 - 3x2 + x + 8 by x - 2. Compare this with f(2) where f(x) = x 3 - 3x 2 + x + 8.

6. Find the remainder on dividing 12 - 5x + 3x2 + 2x3 by x + 3. Compare this with P(-3) where P(x) = 12 - 5x + 3x 2 + 2x 3.

7. In each case determine the polynomial P(x). 2x 3x - 2

a) x + 3 P(x) b) x 2 - 3x + 2 P(x)

-1

x + 7


8. Given that the degree of D(x) = 4, state the possible degrees of R(x) in P(x) = D(x).Q(x) + R(x).

9. When the polynomial function P(x) = 6x3 - x2 + 19x + p is divided by 3x - 2, the quotient is 2x 2 + x + 7 and the remainder is 3. Use the division algorithm to find the value of p.

10. When 5x2 - 6x3 + 4x - 5 is divided by x + 1, the remainder is MultipleChoice

A. 2B. –2C. 20D. –20

11. When the polynomial 2x3 - 5x2 + ax - 5 is divided by x - 3, the remainder is 16. Numerical Response The value of a is _____ .


12. The value of b in the following synthetic division where the divisor is x + 1 is _____ .

3 –4 3 –b

3 c a –25


13. The value of a + b + c + d + e + f in the following synthetic division where the divisor is x - 2 is _____ .

3 –5 –2 1 b d f

c e 1

a


Polynomial Functions and Equations Lesson #2: The Division Algorithm and Synthetic Division 201

Answer Key

1 . a) x 3 - 2x 2 + 6x + 3b) x 2 + 6c ) 15d) x 3 - 2x 2 + 6x + 3 = (x - 2)(x 2 + 6) + 15

2 . a) x 3 + 2x 2 + 3x + 6 = (x - 2)(x 2 + 4x + 11) + 28b) 2x 3 - 4x 2 - 5x + 9 = (x + 2)(2x 2 - 8x + 11) - 13c ) x 4 - x 2 + 7 = (x + 1)(x3 - x 2) + 7d) –y 5 + 2y 4 - y 3 + 4y = (y - 3)(–y 4 - y 3 - 4y 2 - 12y - 32) - 96

3 . p = 5, q = 2, r = 8 4 . m = 0, n = 10 5 . 6, 6 6 . 0, 0

7 . a) 2x 2 + 6x - 1b) 3x 3 - 11x 2 + 13x + 3

8 . 0, 1, 2, 3 9 . –11 10 . A

1 1 . 4 1 2 . 1 5 1 3 . 1 1


Polynomial Functions and Equations Lesson #3:The Remainder Theorem and the Factor Theorem

Warm-Up

a) Use synthetic division to divide Calculate P(-1) in P(x) = x3 - 2x2 - 4 by x + 1 P(x) = x3 - 2x2 - 4

b) Use synthetic division to divide Calculate P(2) in P(x) = x 2 - 2x - 5 by x - 2 P(x) = x 2 - 2x - 5

c) Complete the following statements based on your observations in a) and b).

• When P(x) = x3 - 2x2 - 4 is divided by x + 1, the __________ is equal to P( ).

• When P(x) = x 2 - 2x - 5 is divided by x - 2, the __________ is equal to P( ).

The Remainder Theorem

When a polynomial function, P(x), is divided by a binomial, (x ---- a), the remainder obtained is equal to the value of the polynomial when x = a, i.e. the remainder is P(a).

Proof:The division algorithm states P(x) = D(x).Q(x) + R(x).

Using x - a as the divisor, we get P(x) = (x - a).Q(x) + R(x)

To find P(a) we can substitute a for x to get

P(a) = (a - a).Q(a) + R

= 0.Q(a) + R

= 0 + R

\ P(a) = R which is what the remainder theorem states.

Class Ex. #1 Use the remainder theorem to find the remainder when P(x) = 6x3 - 4x2 + 8x + 6 is divided by x + 1.

Class Ex. #2 Find a if the remainder is 131 when P(x) = 2x 4 - x 3 - ax + 8 is divided by x - 3;

a) using synthetic division b) using the remainder theorem

Class Ex. #3 Find the coefficients d and c in P(x) = 2x 4 + dx 3 - cx 2 + 5x - 8 if the remainder is -41 when divided by x + 3 and the remainder is 74 when divided by x - 2.


204 Polynomial Functions and Equations Lesson #3: The Remainder Theorem and the Factor Theorem

The Factor Theorem

The binomial x - a is a factor of the polynomial function P(x) if and only if P(a) = 0.Note that a is then a zero of the polynomial function P(x).

Class Ex. #4 Show that x - 4 is a factor of P(x) = x2 + 2x - 24 by using

a) synthetic division b) the factor theorem

Class Ex. #5 Write a binomial factor with integral coefficients of the polynomial P(x) if;

a) P(3) = 0 b) P ËÊÁÁ–

23

ˆ˜ = 0

Class Ex. #6 If P(5) = P(-2) = 0, determine a second degree factor of the polynomial P(x).

Class Ex. #7 Use the factor theorem to determine which of the following is a factor of 4x3 - 16x2 - x + 4.

a) x + 2 b) 2x - 1

Polynomial Functions and Equations Lesson #3: The Remainder Theorem and the Factor Theorem 205

Class Ex. #8 Show that 1 is a root of the equation x3 - 9x2 + 20x - 12 = 0 and find the other roots.

Note Class Ex. #8 shows that if the sum of the coefficients of a polynomial function is equal to zero, then 1 is a zero of the function.


Assignment

1. Use the remainder theorem to find the remainder when each of the following polynomials is divided by the binomial.

a) P(x) = 3x3 - x2 + 2x + 1 b) P(x) = x4 + x2 - 8x + 5 is divided by x + 5. is divided by x - 4.

2. Find the values of p and q if x 3 + px + q yields remainders of -3 and 2 when divided by x - 2 and x + 1 respectively.


3. When P(x) = x 4 + mx 3 - nx 2 + 28x - 24 is divided by x - 3, the remainder is 6. If P(1) = –4, find the values of m and n.

4. When x4 + ax2 - 16 is divided by x + 1, the remainder is –14. What is the remainder when x4 + ax2 - 16 is divided by x - 2 ?

5. If x - a is a factor of the polynomial P(x), what is the remainder obtained when P(x) is divided by x - a?

6. Which of the following are factors of P(x) = x3 - 4x2 - x + 4 ?

( i ) x - 1 ( i i) x - 2 (iii) x + 2 (iv) x - 4

7. Find the value of a so that x + 1 is a factor of x4 + 4x3 + ax2 + 4x + 1.


8. When P(x) = 2x3 + ax2 + bx + 6 is divided by x + 2, the remainder is –12. If x - 1 is a factor of the polynomial, find the values of a and b.

9. If P(x) = x3 + kx2 - x - 2 and P(-2) = 0, determine the complete factoring of P(x).

10. Show that –4 is a zero of P(x) = 6x3 + 25x2 + 2x - 8 and find the other roots.


11. Given that x2 + 2x - 3 is a factor of f(x) = x4 + 2x3 - 7x2 + ax + b, find a and b and hence factor f(x) completely.

12. Given that f(1) = 4 and f(2) = 3, then find the values of a and b for f(x) = ax4 + bx2 + 1.

13. For P(x) = x2 - x + 1, find a if P(a) = 3.


14. When a polynomial P(x) is divided by x - 2, the remainder is 3. If the polynomialMultipleChoice A(x) = 2P(x) is divided by x - 2, the remainder will be

A. 1.5B. 2C. 3D. 6

15. When a polynomial P(x) is divided by x + 5, the remainder is –2. Which of the following statements is true?

A. P(–2) = –5B. P(–5) = 0C. P(5) = –2D. P(–5) = –2

16. The polynomial P(x) = 2x3 - ax2 - 11x + 2a has a remainder of 126 when divided Numerical Response by x - 5. The value of a, to the nearest tenth, is _____ .


Answer Key

1 . a) –409 b) 245 2 . p = - 143 , q = - 5

3 3 . m = -3, n = 6

4 . 4 5 . 0 6 . (i) and (iv) 7 . a = 6 8 . a = -3, b = -5

9 . (x + 2)(x + 1)(x - 1) 10. - 23 , 1

2 1 1 . a = - 8, b = 12 (x + 3)(x - 1)(x + 2)(x - 2)

1 2 . a = –56

b = 236

1 3 . a = –1 or a = 2 1 4 . D 1 5 . D

1 6 . 3 . 0


Polynomial Functions and Equations Lesson #4:Factoring Polynomial Expressions

Factoring Polynomial Expressions

In the last two lessons we have used synthetic division or the factor theorem to enable us to factor polynomials of degree greater than two. In each case we were given a binomial factor to start with. In this lesson we will factor polynomials algebraically without being given one of the factors.

In order to do this we have to “guess” a potential factor, x - a, and do the calculation to check whether the guess is correct. For this to work, the value of a must be an integer which divides into the constant term of the polynomial.

Warm-Up

Consider the polynomial P(x) = x3 - 7x - 6.

a) List the integers which divide into the constant term.

b) List the potential factors of the polynomial.

c) Divide the polynomial by one of these potential factors (try x - 1) to determine if it is a factor. Continue this process until one factor is found. Complete the factoring of the polynomial by synthetic division or the factor theorem.

Class Ex. #1 Factor the polynomial P(x) = x4 + 2x3 - 7x2 - 8x + 12.

Factoring By Grouping

Sometimes, a polynomial expression can be factored by grouping, as in the next example.

Class Ex. #2 Factor 2x3 + 3x2 - 8x - 12 by grouping.


212 Polynomial Functions and Equations Lesson #4: Factoring Polynomial Expressions

Assignment

1. Factor the following polynomials algebraically.

a) P(x) = x 3 + x 2 - 5x + 3 b) P(x) = x 3 + x 2 - 17x + 15

c) P(x) = 2x 3 - 9x 2 - 20x + 12 d) P(x) = x4 + x 3 - 7x 2 - x + 6

e) P(x) = 3x 3 - 2x 2 - 19x - 6 f) P(x) = 3x4 - 5x 3 - 17x 2 + 13x + 6

Polynomial Functions and Equations Lesson #4: Factoring Polynomial Expressions 213

2. Factor the following by grouping.

a) 2x 3 - 3x 2 - 18x + 27 b) x 3 + x 2 - 4x - 4

c) 3x 3 - x 2 - 27x + 9 d) x 3 + x 2 + 2x + 2

3. Jenny is attempting to algebraically find the factors of the polynomial functionMultipleChoice P(x) = 6x3 - 7x2 - x + 2. Which of the following factors should she NOT consider as a

possible factor?

A. x - 1B. x - 6C. x + 1D. x + 2

4. The zeros of the function f(x) = 2x3 - x2 - 18x + 9 are

A. 12 , –3, 3

B. 12 , 3, 3

C. –12 , 3, 3

D. –12 , –3, 3


5. The polynomial function P(x) = x3 - 7x2 + 16x - 12 can be written in the Numerical Response form P(x) = (x - a)(x - b)2 where a and b are integers. The value of b is _____ .


Answer Key

1 . a) P(x) = (x - 1) 2(x + 3) b) P(x) = (x + 5)(x - 1)(x - 3)c ) P(x) = (x + 2)(x - 6)(2x - 1) d) P(x) = (x + 1)(x - 2)(x - 1)(x + 3)e ) P(x) = (x + 2)(x - 3)(3x + 1) f ) P(x) = (x - 1)(x + 2)(x - 3)(3x + 1)

2 . a) (x + 3)(x - 3)(2x - 3) b) (x - 2)(x + 1)(x + 2)c ) (x - 3)(x + 3)(3x - 1) d) (x + 1)(x2 + 2)

3 . B 4 . A 5 . 2

Polynomial Functions and Equations Lesson #4: Factoring Polynomial Expressions 215

Polynomial Functions and Equations Lesson #5:Solving Polynomial Equations

Solving Polynomial Equations Algebraically

To solve a polynomial equation algebraically, factor the polynomial expression and set each factor to zero.

Class Ex. #1 Solve the equations algebraically.

a) (2x - 5)(x - 7)(3x + 1) = 0

b) x3 - 19x + 30 = 0

c) x3 - 2x2 - 4x + 3 = 0

Complete Assignment Questions #1

Solving Polynomial Equations Graphically

To solve polynomial equations graphically, graph the equation and find the x-intercepts of the graph using the “zero” feature of the calculator.

Class Ex. #2 Solve the equation x4 + 2x3 - 13x2 - 14x + 24 = 0 by using a graphing calculator.

Class Ex. #3 Consider the polynomial function P(x) = x3 - 3x2 - 3x + 1.

a) Find the zeros of P(x), to the nearest hundredth, using a graphing calculator.

b) Use the integral zero in a) and synthetic division to determine the exact value of the zeros.


Assignment

1. Solve the equations algebraically. Answer as exact values.

a) (x - 4)(x - 9)(2x + 5) = 0 b) 3x(x + 2)(3x - 1)(4x + 3) = 0

218 Polynomial Functions and Equations Lesson #5: Solving Polynomial Equations

c) x 3 - 2x 2 - 5x + 6 = 0 d) 6x 3 + 13x 2 + x - 2 = 0

e) 2x4 + 3x 3 - 3x 2 - 2x = 0 f) x 3 + 4x 2 + 3x - 2 = 0

g) x4 - 2x3 - 9x2 + 2x + 8 = 0 h) x4 - 8x3 + 13x2 + 12x - 18 = 0

Polynomial Functions and Equations Lesson #5: Solving Polynomial Equations 219

2. Solve the following equations graphically.

a) x4 + 6x3 + x2 - 24x - 20 = 0 b) x4 - 7x2 - 6x = 0

c) x3 - x + 6 = 0 d) x4 - 9x2 + 4x + 12 = 0

3. Consider the polynomial function P(x) = x3 + 10x2 + 8x - 16.

a) Find the zeros of P(x), to the nearest hundredth, using a graphing calculator.

b) Use the integral zero in a) and synthetic division to determine the exact value of the zeros.

4. Determine the exact roots of the equation x4 - 12x2 + 4x + 15 = 0.


Questions #5 - #7 refer to the graph below.

PNQ

R

S T

M

y x= + 2

y x x= - +2 6 8

5–5

5

x

y

V

5. The complete solution to the equation x2 - 6x + 8 = 0 is found by looking at the point(s)MultipleChoice

A. QB. N and PC. T and VD. M, N, and P

6. The complete solution to the equation x + 2 = 0 is found by looking at the point(s)

A. MB. SC. T and VD. M, S, T, and V

7. The complete solution to the system of equations Ó

ÏÔÔÔÌ

y = x2 - 6x + 8 y = x + 2

is found by

looking at the point(s)

A. QB. R and SC. T and VD. M, N, and P

Polynomial Functions and Equations Lesson #5: Solving Polynomial Equations 221

8. The smallest positive root, to the nearest hundredth, of the equationNumerical Response x5 - 2x3 - 4x2 - 7x + 6 = 0 is _____ .


Answer Key

1 . a) 9, 4, – 52 b) 1

3 , 0, – 34 , –2 c ) –2, 1, 3

d) –2, – 12 , 1

3 e ) –2, – 12 , 0, 1 f ) –2, –1 - 2 , –1 + 2

g ) –2, –1, 1, 4 h) 1, 3, 2 - 10 , 2 + 10

2 . a) –5, –2, –1, 2 b) –2, –1, 0, 3 c ) –2 d) –3, –1, 2

3 . a) –8.90, –2.00, 0.90 b) –2, –4 - 2 6 , –4 + 2 6

4 . –1, 3, –1 - 6 , –1 + 6

5 . B 6 . A 7 . C 8 . 0 . 6 0


Polynomial Functions and Equations Lesson #6:Graphing Polynomial Functions - Part One

Warm-Up Review of Zeros, Roots, and x-intercepts

Fill in the following blanks:

The _________ of a function, the __________ of an equation, and

the ________________ of a graph are the ________ numbers.

Unique Factorization Theorem

This theorem states that every polynomial function of degree n ≥ 1 can be written as the product of a leading coefficient c, and n linear factors to get

P(x) = c(x - a1)(x - a2)(x - a3)...(x - an)

The above theorem implies the following two important points for polynomial functions of degree n ≥ 1:

Point #1Every polynomial function can be written as a product of its factors with a leading coefficient.

Point #2Every polynomial function has the same number of factors as its degree. The factors may be real or complex.

Note In this lesson we will consider only polynomial functions where the leading coefficient, c, is either 1 or –1.

Class Ex. #1 The graph of a cubic polynomial function

5

5

–5

–5x

yP(x) = x3 + bx2 + cx + d is shown.

a) State the zeros of the polynomial.

b) State the factors of the polynomial.

c) Write the polynomial in factored form.

Class Ex. #2 a) Use a graphing calculator to sketch the graph of the

x

ypolynomial function P(x) = –x 3 + 7x 2 – 7x - 15.

b) Use the graph to determine the zeros of the function.

c) Write the polynomial in factored form.

d) Determine the local maximum and local minimum values of the function to one decimal place.


Assignment

1. In each question use a graphing calculator to:

i ) sketch the graph of the polynomial functioni i ) state the zeros of the polynomial functioni i i) write the polynomial function in factored form.iv) state any local maximum and/or local minimum values to one decimal place.

a) P(x) = x - 2 b) P(x) = x2 - 6x + 8 c) P(x) = –x2 + 6x - 8

i i ) i i ) i i )

iii) P(x) = iii) P(x) = iii) P(x) =

iv) iv) iv)

224 Polynomial Functions and Equations Lesson #6: Graphing Polynomial Functions - Part One

d) P(x) = x3 - 7x2 + 7x + 15 e) P(x) = –x3 + 7x2 - 7x - 15

i i ) i i )

iii) iii)

iv) iv)

f) P(x) = x3 - x2 - 12x g) P(x) = –x3 + x2 + 12x

i i ) i i )

iii) iii)

iv) iv)

h) P(x) = x4 - 5x2 + 4 i ) P(x) = –x4 + 5x2 - 4

i i ) i i )

iii) iii)

iv) iv)

Polynomial Functions and Equations Lesson #6: Graphing Polynomial Functions - Part One 225

2. In each question use a graphing calculator to:i ) sketch the graph of the polynomial function

i i ) state the zeros of the polynomial functioni i i) write the polynomial function in factored form.

a) P(x) = x4 - 7x3 + 7x2 + 15x b) P(x) = –x4 + 7x 3 - 7x 2 - 15x

i i ) i i )

iii) iii)

c) P(x) = x5 - 3x 4 - 5x 3 + 15x 2 + 4x - 12 d) P(x) = –x5 + 3x 4 + 5x 3 - 15x 2 - 4x + 12

i i ) i i )

iii) iii)

e) P(x) = x6 - 14x 4 + 49x 2 - 36 f) P(x) = –x6 + 14x 4 - 49x 2 + 36

i i ) i i )

iii) iii)


3. Based on your observations from questions #1 - #2 circle the correct choice in each of the following statements.

a) If the graph of a polynomial has two raising arms, then the

degree of the polynomial is ( even, odd ) and

the leading coefficient is ( positive, negative ).

b) If the graph of a polynomial has two drooping arms, then the



c) If the graph of a polynomial has a right arm raised up and

the left arm drooping down, then the



d) If the graph of a polynomial has a right arm drooping down and

the left arm raised up, then the



e) The leading coefficient is positive if the ( left, right ) arm

is ( raised up, drooping down ).

f) The leading coefficient is negative if the ( left, right ) arm

is ( raised up, drooping down ).

Polynomial Functions and Equations Lesson #6: Graphing Polynomial Functions - Part One 227

Answer Key

1 . a) i i ) 2 i i i ) P(x) = (x - 2) i v ) none

b) i i ) 2, 4 i i i ) P(x) = (x - 2)(x - 4) i v ) local min –1.0

c ) i i ) 2, 4 i i i ) P(x) = –(x - 2)(x - 4) i v ) local max 1.0

d) i i ) –1, 3, 5 i i i ) P(x) = (x + 1)(x - 3)(x - 5) i v ) local max 16.9, local min –5.0

e ) i i ) –1, 3, 5 i i i ) P(x) = –(x + 1)(x - 3)(x - 5) i v ) local max 5.0, local min –16.9

f ) i i ) –3, 0, 4 i i i ) P(x) = x(x + 3)(x - 4) i v ) local max 12.6, local min –20.7

g ) i i ) –3, 0, 4 i i i ) P(x) = –x(x + 3)(x - 4) i v ) local max 20.7, local min –12.6

h) ii)–2, –1, 1, 2 i i i ) P(x) = (x + 2)(x + 1)(x - 1)(x - 2) i v ) local max 4.0, local min –2.3 & –2.3

i ) i i ) –2, –1, 1, 2 i i i ) P(x) = –(x + 2)(x + 1)(x - 1)(x - 2) i v ) local max 2.3 & 2.3, local min –4.0

2 . a) i i ) –1, 0, 3, 5 i i i ) P(x) = x(x + 1)(x - 3)(x - 5)

b) i i ) –1, 0, 3, 5 i i i ) P(x) = –x(x + 1)(x - 3)(x - 5)

c ) i i ) –2, –1, 1, 2, 3 i i i ) P(x) = (x + 2)(x + 1)(x - 1)(x - 2)(x - 3)

d) i i ) –2, –1, 1, 2, 3 i i i ) P(x) = –(x + 2)(x + 1)(x - 1)(x - 2)(x - 3)

e ) i i ) –3, –2, –1, 1, 2, 3 i i i ) P(x) = (x + 3)(x + 2)(x + 1)(x - 1)(x - 2)(x - 3)

f ) i i ) –3, –2, –1, 1, 2, 3 i i i ) P(x) = –(x + 3)(x + 2)(x + 1)(x - 1)(x - 2)(x - 3)

3 . a) even, positive b) even, negative c ) odd, positive

d) odd, negative e ) right, raised up f ) right, drooping down


Polynomial Functions and Equations Lesson #7:Graphing Polynomial Functions - Part Two

Warm-Up

-3 -2 -1 1 2 3 4 5

10

5

-5

x

y

The graph of the polynomial functionP(x) = (x + 1)(x - 3)2 is shown.

The function has two factors, one of which is repeated. This means that the function has two zeros, one of which is a repeated zero.

The factors are (x - 3) which is repeated, and (x + 1).

The zeros of the function are therefore, 3, which is repeated, and –1.

The repeated zero of 3 is said to be a zero of multiplicity 2. The zero of –1 is a zero of multiplicity 1.

Multiplicity

The multiplicity of a zero corresponds to the number of times a factor is repeated in the function.

In this lesson, we will investigate how the multiplicity of a zero affects the shape of the graph of a polynomial function. In order to do this, we have to define the following terms.

Tangent

orx xA polynomial graph is tangent to the x-axis at a point where the graph touches the x-axis and does not cross through it.

Point of Inflection

orx x

concave up

concave down

concave up

concave down

A polynomial graph has a point of inflection on the x-axis if the graph changes concavity at a point on the x-axis.

Note In this lesson we will consider only polynomials where the leading coefficient, c, is either 1 or –1.

Class Ex. #1 Consider the polynomial function P(x) = x6 - x5 - 11x4 + 13x3 + 26x2 - 20x - 24

= (x + 3)(x + 1)2(x - 2)3;

a) Sketch the graph of P(x) using the windowx: [–5, 5, 1] y: [–100, 100, 20]

b) Complete the chart below to state the zeros of P(x), their multiplicities, and whether each zero;

• passes straight through the x-axis,• is tangent to the x-axis, or,• has a point of inflection.

zero multiplicity description

c) Complete the following for P(x)

• Degree of P(x) _____ • Sum of the multiplicities of the zeros of P(x) _____ .

Class Ex. #2 A polynomial function has the equation P(x) = x4 + 2x3 - 15x2 - 32x - 16.

a) Sketch the graph of P(x) using the windowx: [–6, 6, 1] y: [–150, 100, 20]

b) Complete the chart below zero multiplicity description

c) Complete the following for P(x)

• Degree of P(x) _____ • Sum of the multiplicities of the zeros of P(x) _____ .

d) Write the polynomial in the form P(x) = (x - a)(x - b)(x - c)2.


230 Polynomial Functions and Equations Lesson #7: Graphing Polynomial Functions - Part Two

Assignment In this assignment, choose appropriate windows which will enable you to investigate all the characteristics of the functions

1. a) Graph the polynomial function

x

P(x)P(x) = x 3 - 4x 2 - 3x + 18

b) Complete:zero multiplicity

c) Write the polynomial in the formP(x) = (x - a)(x - b) 2, where a, b Œ I.


x

P(x)P(x) = x4 + x 3 - 18x 2 - 52x - 40


c) Write the polynomial function in factored form.


x

P(x)P(x) = –x 3 - 6x 2 + 32


c) Write the polynomial in the formP(x) = – (x - a)(x - b) 2, a, b Œ I.

Polynomial Functions and Equations Lesson #7: Graphing Polynomial Functions - Part Two 231

4. Two polynomial functions have the equations

x

yP(x) = x 4 - 4x 3 - 2x 2 + 12x + 9 and Q(x) = –x 4 + 4x 3 + 2x 2 - 12x - 9

a) Sketch the graphs of P(x) and Q(x) on the same grid.

b) State the zeros, their multiplicities, and y-intercept of each polynomial function.

c) Write the equations of the polynomials in factored form.

P(x) = Q(x) =

5. A cubic polynomial function has the equation

x

P(x)P(x) = ax 3 + bx 2 + cx + d with a leading coefficient of 1. The zeros of the polynomial are –6, 1, and 3.

a) Sketch the graph of P(x).

b) Write the equation of the polynomial in factored form.

c) Determine the values of a, b, c, and d in P(x).

6. A cubic polynomial function has the equation

x

P(x)P(x) = ax 3 + bx 2 + cx + d with a leading coefficient of 1. The function has a zero of 2 with multiplicity one and a zero of –3 with multiplicity two.

a) Sketch the graph of the function.

b) Write the equation of the polynomial in factored form.

c) Determine the values of a, b, c, and d in P(x).

d) A new function is formed by changing the signs of each of the values of a, b, c, and d.Describe how the graph of the new function compares to the graph of P(x).


7. A polynomial function has the equation

x

P(x)P(x) = x 2(x - 2)(x + 3)

a) Make a rough sketch without using a graphing calculator. Verify using a graphing calculator.

b) State the zeros, their multiplicities, and they-intercept of P(x).


x

P(x)P(x) = –(x - 4)2(x + 3) 2




x

P(x)P(x) = (x - 1)3(x + 3)




x

P(x)P(x) = (x + 2)2(x + 5)(3 - x)




11. Complete the following based on your observations from questions #1 - #10.

a) If a polynomial function has a zero of multiplicity 1 at x = a, then the graph of the function at x = a _________________________________________ .

b) If a polynomial function has a zero of multiplicity 2 at x = b, then the graph of the function at x = b _________________________________________ .

c) If a polynomial function has a zero of multiplicity 3 at x = c, then the graph of the function at x = c _________________________________________ .

x

P(x)d) A polynomial function with a leading coefficient of 1 has:

• a zero of multiplicity 1 at x = a, • a zero of multiplicity 2 at x = b, and,• a zero of multiplicity 3 at x = c.

If a < b < c make a rough sketch of a polynomial which satisfies these conditions.

12. The graphs shown each represent a cubic polynomial function with equationP(x) = ax 3 + bx 2 + cx + d, where a is 1 or –1.

-3 -2 -1 1 2 3

1510

5

-5

i) ii) iii)

-4 -2 2 4 6

105

-5-103

-2 -1 1 2

1510

5

-5-10

In each case;a) Write the equation of the polynomial function in factored form.

b) Determine the values of a, b, c, and d.


13. The graphs shown below each represent a quartic polynomial function with equationP(x) = ax 4 + bx 3 + cx 2 + dx + e, where a is 1 or –1.

In each case write the equation of the polynomial function in factored form and determine the value of e.

-5 -4 -3 -2 -1 1 2 3 4 5

100

-100

i) iii)

4-3

100

-100

x

y

1 2 3 4

5

ii)

x

y

x

y

Use the following information to answer questions #14 and #15.

x

y

1 4–3

The partial graph of a fourth degree polynomial function P(x) is shown. The leading coefficient is 1 and the x-intercepts of the graph are integers.

14. If the polynomial function is written in the form P(x) = c(x - a)2(x - b)(x + d), Numerical Response where a, b, c, and d are all positive integers, then the respective numerical values of a, b, c,

d from left to right are _____ .(Record your answer in the numerical response box from left to right)

15. The graph crosses the y-axis at (0, –m). The value of m is _____ .(Record your answer in the numerical response box from left to right)


Answer Key 1 . b) The zero of –2 has a multiplicity of 1. 2 . b) The zero of –2 has a multiplicity of 3.

The zero of 3 has a multiplicity of 2. The zero of 5 has a multiplicity of 1.c ) P(x) = (x + 2)(x - 3) 2 c ) P(x) = (x - 5)(x + 2) 3

3 . b) The zero of –4 has a multiplicity of 2.The zero of 2 has a multiplicity of 1.

c ) P(x) = –(x - 2)(x + 4) 2

4 . b) P(x): The zero of –1 has a multiplicity of 2. Q(x): The zero of –1 has a multiplicity of 2.The zero of 3 has a multiplicity of 2. The zero of 3 has a multiplicity of 2.The y-intercept is 9. The y-intercept is –9.

c ) P(x) = (x +1) 2(x - 3) 2 Q(x) = –(x + 1) 2(x - 3) 2

5 . b) P(x) = (x + 6)(x - 1)(x - 3) c ) a = 1, b = 2, c = –21, d = 18

6 . b) P(x) = (x - 2)(x + 3) 2 c ) a = 1, b = 4, c = –3, d = –18d) The graph of the new function is a reflection in the x-axis of the graph of P(x).

7 . b) The zero of –3 has a multiplicity of 1. 8 . b) The zero of –3 has a multiplicity of 2.The zero of 0 has a multiplicity of 2. The zero of 4 has a multiplicity of 2.The zero of 2 has a multiplicity of 1. The y-intercept is –144.The y-intercept is 0.

9 . b) The zero of –3 has a multiplicity of 1. 1 0 . b) The zero of –5 has a multiplicity of 1.The zero of 1 has a multiplicity of 3. The zero of –2 has a multiplicity of 2.The y-intercept is –3. The zero of 3 has a multiplicity of 1.

The y-intercept is 60.1 1 . a) passes straight through the x-axis

b) is tangent to the x-axis c ) has a point of inflectiond) answers may vary, one possible answer is

a b c

1 2 . a) i ) P(x) = –x(x + 2)(x - 3) i i ) P(x) = (x + 3)(x - 2) 2 i i i ) P(x) = –(x - 4) 3

b) i ) a = –1, b = 1, c = 6, d = 0i i ) a = 1, b = –1, c = –8, d = 12i i i ) a = –1, b = 12, c = –48, d = 64

1 3 . i ) P(x) = –(x + 5)(x + 3)(x - 1)(x - 4); e = –60i i ) P(x) = (x - 4)(x - 2) 3; e = 32i i i ) P(x) = x(x - 4)(x + 3) 2; e = 0

1 4 . 1 4 1 3 1 5 . 1 2


Polynomial Functions and Equations Lesson #8:Graphing Polynomial Functions - Part Three

In this lesson we will graph polynomial functions which have zeros with multiplicities greater than 3.

Warm-Up #1

Graph the following functions on the grid

50

2–1

P(x)

–50

1 3x

showing the x- and y-intercepts.

1. P(x) = (x - 2)3

2. P(x) = (x - 2)5

3. P(x) = (x - 2)7

What happens as the multiplicity of the zero increases through the odd numbers?

Warm-Up #2

Graph the following functions on the grid

50

2–1

P(x)

–50

1 3x

showing the x- and y-intercepts.

1. P(x) = (x - 2)2

2. P(x) = (x - 2)4

3. P(x) = (x - 2)6

What happens as the multiplicity of the zero increases through the even numbers?

Even and Odd Multiplicities of a Zero

A real zero of even multiplicity (i.e. 2 or 4 or 6 or 8 or . . .) occurs where the graph of a polynomial function is tangent to the x-axis.

A real zero of odd multiplicity greater than 1 (i.e. 3 or 5 or 7 or . . . ) occurs where the graph of a polynomial function has a point of inflection on the x-axis.

The sum of the multiplicities of the zeros of a polynomial function is equal to the degree of the polynomial function.

Class Ex. #1 Consider the graphs below.

a) In each case state the number of zeros and the possible multiplicities of each zero.

i) ii)

iii) iv)

b) Which graph could represent a fourth-degree polynomial function?

c) In which of the graphs is the leading coefficient positive?

238 Polynomial Functions and Equations Lesson #8: Graphing Polynomial Functions - Part Three

Class Ex. #2 Without using a graphing calculator make a rough sketch of the graph of

f(x) = -x(x - 1) 4(x - 4).

Verify using a graphing calculator.

Class Ex. #3 The following graphs represent functions of lowest possible degree.State the degree in each case.

a) b)


Assignment

1. How does the concept number of zeros differ from the concept multiplicity of zeros?

Polynomial Functions and Equations Lesson #8: Graphing Polynomial Functions - Part Three 239

2. Consider the graphs below.

a) In each case state the number of zeros and the possible multiplicities of each zero.

x

y

x

y

x

y

x

y

i) ii)

iii) iv

b) Which graph(s) could represent a seventh-degree polynomial function?

c) In which of the graphs is the leading coefficient negative?

3. The following graphs represent functions of lowest possible degree.

State the degree in each case.

a) b)

c) d)


4. The graph represents a polynomial

-3 -2 -1 1

20

–20

x

P(x)function P(x) of degree 5.

Write the equation of P(x) in factored form if the leading coefficient is –1.

MultipleChoice 5. The graph of a fourth degree polynomial function of the

form P(x) = ax 4 + bx 3 + cx 2+ dx + e is shown.

The values a and e must satisfy

A. a > 0, e < 0

B. a < 0, e > 0

C. a > 0, e > 0

D. a < 0, e < 0

Polynomial Functions and Equations Lesson #8: Graphing Polynomial Functions - Part Three 241

Answer Key 1 . - multiplicity refers to the number of times a zero repeats

- number of zeros refers to how many distinct zeros the function has

2 . a) i ) two zeros, the zero –1 has a multiplicity of 3 or 5 o7 ..., the zero 2 has a multiplicity of 1.i i ) two zeros, the zero 1 has a multiplicity of 2 or 4 or 6 ..., the zero 3 has a multiplicity of 1.i i i ) one zero, the zero 0 has a multiplicity of 3 or 5 o7 ... .i v ) two zeros, the zero –1 has a multiplicity of 2 or 4 or 6 ...,

the zero 2 has a multiplicity of 2 or 4 or 6 ...

b) ii), iii) c ) iii)

3 . a) 4 b) 4 c ) 5 4 . P(x) = –(x + 2) 4(x - 1) 5 . D


Polynomial Functions and Equations Lesson #9:Polynomial Functions with a

Leading Coefficient other than ±1

Warm-Up Review

In lesson 6 we introduced the Unique Factorization Theorem which states that every polynomial function of degree n ≥ 1 can be written as the product of a leading coefficient c, and n linear factors to get

P(x) = c(x - a1)(x - a2)(x - a3)...(x - an)

The above theorem implies the following two important points for polynomial functions of degree n ≥ 1:

Point #1Every polynomial function can be written as a product of its factors with a leading coefficient.

Point #2Every polynomial function has the same number of factors as its degree. The factors may be real or complex.

Note In lessons 6 - 8 we have considered only polynomial functions where the leading coefficient, c, was either 1 or –1.

In this lesson we will consider polynomial functions where the leading coefficient, c, can be any real number.

Class Ex. #1 The graph of a third degree polynomial function, P(x),

-2 -1 1 2 3 4

10

-10

-20

x

y

(2, –24)

is shown. The graph has integral x-intercepts and passes through the point (2, –24).

Find the equation of the polynomial function, P(x), writing the answer in factored form.

Class Ex. #2 The graph represents a polynomial function of

-5 5

20

x

P(x)

–8

–2 2

lowest possible degree.

The intercepts are shown. Determine the equation of the polynomial function.

Class Ex. #3 Find the equation of a fourth degree polynomial function which passes through (1, –12) and is tangent to the x-axis at (2, 0) and at (–3, 0).


244 Polynomial Functions and Equations Lesson #9: Leading Coefficient other than ±1

Class Ex. #4 A fourth degree polynomial, P(x), passes through the point (1, 2) and has

zeros –1, 0, 32 , and 2.

a) Determine the equation of P(x) in factored form.

b) Write the polynomial using only integral linear factors.

Class Ex. #5 A polynomial equation has the following three roots:

• –2 is a root with a multiplicity of 1

•13 is a root with a multiplicity of 2

• 1 is a root with a multiplicity of 3

The graph of the corresponding polynomial function has a y-intercept of 23 .

a) Determine the equation in factored form.

b) Write the equation using only integral linear factors.


Polynomial Functions and Equations Lesson #9: Leading Coefficient other than ±1 245

Assignment

-3 -2 -1 1 2 3

10

5

-5

x

P(x)1. The graph of the polynomial function shown

has integral intercepts.

Determine the equation of the function in factored form.

2. The graph passes through the point (1, –6) and has

-5 5

5

-5

x

P(x)integral x-intercepts. Determine the equation in factored form of the polynomial function, P(x), represented by the graph.

-5 -1x

y

(–6, 1)

3. The graph of a third degree polynomial function is shown.

The graph passes through the point (–6, 1).

If the polynomial function has zeros–5 and –1, determine the equation of the function in factored form.


4. The graph of a polynomial function of degree 4 has x-intercepts –2, –1, 0, and 1. If the graph passes through the point (–3, –48), determine the equation of the function in factored form.

5. Find the equation of the cubic function whose graph is tangent to the x-axis at the origin and passes through the points (2, 0) and (4, –16).

6. Find the equation of a quartic function whose graph has a point of inflection at the origin and passes through (4, 0) and (–1, 10).

7. Find the equation of a quartic function whose graph is tangent to the x-axis at both (2, 0) and (–3, 0) and passes through (1, 12).


-1 1 2 3

60

30

-30

-60

x

P(x)8. The graph shown has x-intercepts of –1, 1, 2.5, and 3 and a y-intercept of 60.

a) Determine the equation of the graph in factored form.

b) Write the equation using only integral linear factors.

9. The graph below has x-intercepts –32 , 0, and 2 and passes through the point (1, 3). Find the

equation of the graph using only integral linear factors.

-5 5

5

-5

x

P(x)

(1,3)


10. The design of a route for a cross country ski course was drawn on a Cartesian plane. The route is tangent to the x-axis at (1, 0) and (–3, 0). It crosses the x-axis at (–5, 0) and also passes through the point (–2, 9). Find the degree 5 function that will meet these conditions.

11. Find the equation of a degree 5 polynomial whose graph has a point of inflection at (3, 0),

is tangent to the x-axis at ËÊÁÁ

12 , 0

ˆ˜ , and passes through (2, 1). Give the answer using only

integral linear factors.

12. P(x) = –3x 3 + bx 2 + cx + d is an integral polynomial function with MultipleChoice zeros 2, –1, and 4. A sketch of y = P(x) is shown.

5x

P(x)At which of the following points does the graph of P(x) cross the y-axis?

A. (0, –8)

B. (0, –15)

C. (0, –16)

D. (0, –24)


13. If the zeros of a polynomial are –1, 12 and

23 , then the polynomial could be

A. 12x 3 - 2x 2 + 10x - 4B. 6x 3 + x 2 - 5x + 2C. 18x 3 + 3x 2 - 15x - 6D. 30x 3 - 5x 2 - 25x + 10

14. P(x) = ax 3 + bx 2 + cx + d, a > 0, is an integral polynomial functionNumerical Response with 2 and 5 as its zeros. The graph of y = P(x) is shown.

–5

–5

5

5x

P(x)

If the maximum y-intercept is at the point (0, –m), then m is _____ .(Record your answer in the numerical response box from left to right)

Answer Key 1 . P(x) = –2(x + 2)(x + 1)(x - 2) 2 . P(x) = x(x + 2)(x + 1)(x - 2)3 . P(x) = – 1

5 (x + 1)(x + 5)2 4 . P(x) = –2x(x + 2)(x + 1)(x - 1)

5 . P(x) = – 12 x2(x - 2) 6 . P(x) = 2x3(x - 4)

7 . P(x) = 34 (x - 2)2(x + 3)2

8 . a) P(x) = –8(x + 1)(x - 1) ËÊÁx - 5

2 (x - 3) b) P(x) = –4(x + 1)(x - 1)(2x - 5)(x - 3)

9 . P(x) = – 35 x(2x + 3)(x - 2) 1 0 . P(x) = 1

3 (x + 5)(x - 1)2(x + 3)2

1 1 . P(x) = – 19 (2x - 1)2(x - 3)3

1 2 . D 1 3 . D 1 4 . 2 0


Polynomial Functions and Equations Lesson #10:Polynomial Inequalities

Solving Polynomial Inequalities by Graphing

5

x

y

–1 2 3

Class Ex. #1 The graph of the polynomial function with equation y = x3 - 4x2 + x + 6 is shown.

Complete the following

a) The solution to the inequalityx3 - 4x2 + x + 6 ≥ 0 is _____________ or _____________ .

b) The solution to the inequalityx3 - 4x2 + x + 6 £ 0 is _____________ or _____________ .

Class Ex. #2 Solve the inequality x4 + 4x3 - 17x2 - 24x + 36 > 0.


Solving Polynomial Inequalities Algebraically

Recall the method used for solving quadratic inequalities on page #183.

The method for solving x2 + 2x - 8 > 0 is shown below.

Step 1: Factor the quadratic expression. (x + 4)(x - 2) > 0

Step 2: Determine the zeros of the quadratic expression. –4 and 2

Step 3: Use a chart which shows the sign of each factor (+, 0, –) to the left and right of each of the zeros.

x Æ –4 Æ 2 Æ

x + 4 – 0 + + +

x - 2 – – – 0 +

Product + 0 – 0 +

Step 4: Since the original inequality symbol is > 0, look for the solution under the + in the product row. The solution is x < –4 or x > 2.

Class Ex. #3 Complete the solution to solve the inequality –4x3 - 10x2 + 56x + 30 < 0.

In factored form, the inequality is –2(x + 5)(x - 3)(2x + 1) < 0.

x Æ –5 Æ –12 Æ 3 Æ

–2 – – – – – – –

x + 5 –

x - 3 –

2x + 1 –

Product +

The solution to –4x3 - 10x2 + 56x + 30 < 0 is _______________________________ .

252 Polynomial Functions and Equations Lesson #10: Polynomial Inequalities

Class Ex. #4 Consider the polynomial function P(x) = x4 - 8x3 + 15x2 + 8x - 16.

a) Show that 1 is a zero of P(x).

b) Algebraically factor P(x).

c) Solve the inequality P(x) ≥ 0.


Polynomial Functions and Equations Lesson #10: Polynomial Inequalities 253

Assignment

1. The graph of y = P(x) is shown with

-2 -1 1 2 3 4

10

-10

-20

x

yintegral x-intercepts.


a) P(x) = 0

b) P(x) £ 0

c) P(x) ≥ 0

-5 5

20

x

P(x)

–8

–2 2

40

y 2. The graph of y = x4 + 4x3 - 16x - 16 is shown.

The x-intercepts are integers.


a) x4 + 4x3 - 16x - 16 = 0

b) x4 + 4x3 - 16x - 16 < 0

c) x4 + 4x3 - 16x - 16 > 0


3. Use a graph to solve each of the following inequalities.

a) (x - 3)(5x + 2)(x + 9) < 0 b) x3 - 2x2 - 11x + 12 ≥ 0

c) 3x3 + 28x2 + 51x + 14 > 0 d) 2x4 + x3 - 38x2 - 79x - 30 £ 0

4. Solve each of the following inequalities algebraically.

a) x3 - 5x2 - x + 5 > 0

b) x3 + 11x2 + 23x - 35 < 0


c) 3x3 - 9x2 - 12x + 36 ≥ 0

5. Verify the solution to Class Ex. #2 algebraically.“Solve the inequality x4 + 4x3 - 17x2 - 24x + 36 > 0.”


6. The function f is defined by f(x) = x3 - 2x2 - 5x + 6. The function g is defined by x - 1.

a) Show that (f o g)(x) = x3 - 5x2 + 2x + 8.

b) Algebraically solve the inequality (f o g)(x) > 0.

7. Write a cubic polynomial inequality in factored form which has the solution 2 £ x £ 5 or x ≥ 8.

8. The solution to a cubic polynomial inequality is shown on the number line.MultipleChoice

52x

–3

The inequality which has the solution shown is

A. (x - 3)(x + 2)(x + 5) £ 0B. (x - 3)(x + 2)(x + 5) ≥ 0C. (x + 3)(x - 2)(x - 5) £ 0D. (x + 3)(x - 2)(x - 5) ≥ 0


9. The inequality 2x4 - 3x3 - 9x2 + 10x < 0 has a solution a < x < b or c < x < d. Numerical Response The value of a + b + c + d, to the nearest tenth, is _____ .


Answer Key

1 . a) –1, 0, 3 b) x £ –1 or 0 £ x £ 3 c ) –1 £ x £ 0 or x ≥ 3

2 . a) –2, 2 b) –2 < x < 2 c ) x < –2 or x > 2

3 . a) x < –9 or – 25 < x < 3 b) –3 £ x £ 1 or x ≥ 4

c ) –7 < x < –2 or x > – 13 d) –3 £ x £ –2 or – 1

2 £ x £ 5

4 . a) –1 < x < 1 or x > 5 b) x < –7 or –5 < x < 1c ) –2 £ x £ 2 or x ≥ 3

5 . x < –6 or –2 < x < 1 or x > 3

6 . b) –1 < x < 2 or x > 4

7 . (x - 2)(x - 5)(x - 8) ≥ 0

8 . C 9 . 1 . 5


Absolute Value, Radical, and Rational Equations Lesson #1:

Absolute Value Equations - Part One

Warm-Up #1 Piecewise Functions

Most of the functions you have met so far have been

x

y

5

5

–5

–5

defined by a single formula, eg. f(x) = 2x - 1.

Some functions, however, are defined by applying different formulas to different parts of the domain.

eg. f(x) = Ó

Ï

ÔÔÔÔÔÌ

1 if x < –1

x2 if –1 £ x £ 1x if x > 1

Sketch the graph of this function on the grid.

Functions like this, which are defined in pieces, are called piecewise functions.

Warm-Up #2

x

y

5

5

–5

–5

One of the simplest piecewise functions is the function defined by

f(x) = ÓÏÔÔÔÌ

x if x ≥ 0–x if x < 0

Sketch the graph of this function on the grid.

This piecewise function is called the absolute value function and is written as f(x) = ΩxΩ

Absolute Value Functions

The absolute value function f(x) = ΩxΩ is defined piecewise as

f(x) = ΩxΩ = ÓÏÔÔÔÌ

x if x ≥ 0–x if x < 0

Notice that Ω5Ω = 5 since 5 > 0, and, Ω–5Ω = –(–5) = 5 since –5 < 0

Note The absolute value of a number will never be negative.

Warm-Up #3

Every absolute value function can be defined in pieces.

For example, f(x) = Ωx - 3Ω can be written piecewise as

f(x) = ÓÏÔÔÔÌ

x - 3 if x - 3 ≥ 0–(x - 3) if x - 3 < 0

which simplifies to

f(x) = ÓÏÔÔÔÌ

x - 3 if x ≥ 3–x + 3 if x < 3

Note The absolute value of a quantity will always be the same quantity if the quantity is positive and the opposite quantity if the quantity is negative.

Writing an absolute value function in piecewise form is an integral part of determining the solution to absolute value equations or inequalities.

Class Ex. #1 State the value of the following.

a) Ω4Ω b) Ω–6Ω c) Ω3 - 5Ω d) –Ω7Ω - Ω–7Ω

Class Ex. #2 Define the following absolute value functions as piecewise functions.

a) f(x) = Ω3x + 2Ω b) g(x) = Ω4 - xΩ


260 Absolute Value, Radical, and Rational Equations Lesson #1: Absolute Value Equations - Part One

Solving Absolute Value Equations Algebraically

Ωx + 3Ω = 7 and Ω2x - 3Ω - Ωx + 4Ω = 8 are examples of absolute value equations

Note There are different ways in which to determine algebraically the solution to absolute value equations but the method below has the advantage that a virtually identical method can be used to determine the solution to absolute value inequalities in future math courses.

Step 1: Find the value(s) of the variable which will make the expression within the absolute value symbol(s) equal to zero.

Step 2: Divide the domain into smaller subdomains using the value(s) found in Step 1.

Step 3: Write the absolute value expressions in piecewise form using the piece that is appropriate for each sub domain.

Step 4: Solve the resulting equation in each subdomain.

Step 5: Check that the solution to each equation is in the subdomain and combine all valid solutions.

Class Ex. #3 Alan has started to solve the equation Ω2x + 3Ω = 8 using the steps above. Complete each solution.

Alan’s solution

Ω2x + 3Ω = 8

2x + 3 = 8

–32x < –

32

x ≥ –32

subdomain subdomain

Ω2x + 3Ω = 8Solve Solve

–(2x + 3) = 8

Is the solution in the subdomain? Is the solution in the subdomain?

Final solution: x = __________

number line

Absolute Value, Radical, and Rational Equations Lesson #1: Absolute Value Equations - Part One 261

Class Ex. #4 Jimmy has started to solve the equation Ω3 - xΩ = 2x + 1 using the steps above. Complete each solution.

Jimmy’s solution

solve Ω3 - xΩ = 2x + 1

subdomain subdomain

Is the solution in

the subdomain?

Is the solution in

the subdomain?

Final solution: x = __________

solve Ω3 - xΩ = 2x + 1

number line



Assignment

1. Which of the following statements are true and which are false?

a) Ω–7Ω = Ω7Ω b) Ω3 - 6Ω = –3 c) Ω2Ω - Ω4Ω = Ω–2Ω

d) ΩΩ5Ω - Ω–32ΩΩ = 27 e) ΩxΩ = –x, if x < 0 f) Ω–xΩ = –x, if x ≥ 0

g) Ω2x - 1Ω = 2x - 1, if x < 12 h) Ω3x + 4Ω = –3x - 4, if x < –

43

i ) Ω2 - 5xΩ = 2 - 5x, if x ≥ 25 j) Ωx - 7Ω = –x - 7, if x < 7

2. Write the following absolute value functions as piecewise functions

a) f(x) = Ω2x + 1Ω b) g(x) = Ω4x - 1Ω

c) f(x) = Ω2 - xΩ d) g(x) = Ω4 - 2xΩ

3. Solve each of the following equations algebraically.

a) Ωx + 5Ω = 10 b) Ω3x - 1Ω = 4


c) Ω2x + 1Ω = x d) Ω1 - 4xΩ = 6x

4. Algebraically, find the solution to the following equations.

a) Ω7x - 2Ω + 6 = 3x b) Ω4 - xΩ = –2x - 10

c) 3Ωx - 8Ω = 2x + 7 d) Ω2x - 8Ω - 2 = 4x


5. Solve for x in each of the following.

a) 5x - 2 = –Ωx - 6Ω b) Ω Ω3x - 2

4 = 1

6. The complete solution to ΩxΩ - 4 = 10 isMultipleChoice

A. 14B. 6 and 6C. 6 and 14D. –14 and 14


Answer Key 1 . a) T b) F c ) F d) T e ) T

f ) F g ) F h) T i ) F j ) F

2 . a) f(x) = Ó

Ï

ÔÔÔÔÌ

2x + 1 if x ≥ – 12

–2x - 1 if x < – 12

b) g(x) = Ó

Ï

ÔÔÔÔÌ

4x - 1 if x ≥ 14

–4x + 1 if x < 14

c ) f(x) = Ó

ÏÔÔÔÌ

2 - x if x £ 2

–2 + x if x > 2d) g(x) =

Ó

ÏÔÔÔÌ

4 - 2x if x £ 2

–4 + 2x if x > 2

3 . a) –15 or 5 b) –1 or 53 c ) no solution d) 1

10

4 . a) no solution b) –14 c ) 175 or 31 d) 1

5 . a) –1 b) – 23 or 2 6 . D



Absolute Value Equations - Part Two

Solving Absolute Value Equations Using a Graphing Calculator

Intersection Method

To solve the equation Ω2x + 5Ω = 3 by the

10

10

–10

–10x

yintersection method use the following procedure.

1. Graph Y1 = Ω2x + 5Ω.

2. Graph Y2 = 3.

3. Find the x-coordinate(s) of the point(s) of intersection using the intersect feature of the calculator.

x-intercept Method

To solve the equation Ω2x + 5Ω = 3 by the

10

10

–10

–10x

yzero method use the following procedure.

1. Rearrange the original equation with all terms on the left hand side and 0 on the right side to get Ω2x + 5Ω - 3 = 0.

2. Graph Y1 = Ω2x + 5Ω - 3.

3. Use the zero feature of the calculator to find the x-intercept(s).

Class Ex. #1 Verify the solutions to Class Ex. #3 and #4 from the previous lesson using a graphing calculator.

a) Ω2x + 3Ω = 8 b) Ω3 - xΩ = 2x + 1

10

10

–10

–10x

y

10

10

–10

–10x

y

Class Ex. #2 Solve the equation Ω2x - 3Ω - Ωx + 4Ω = 8

10

10

–10

–10x

yusing a graphing calculator.


Extension - Optional

The following class examples are beyond the scope of this course. We include the method here for students intending to take higher level math courses.

Class Ex. #3 Julie has started to solve the equation Ω2x - 3Ω - Ωx + 4Ω = 8 algebraically.Complete Julie’s solution below

2x - 3 = 0 fi x = 3 2

x + 4 = 0 fi x = –4

Ω2x - 3Ω = –2x + 3

Ωx + 4Ω = –x - 4

Solve Ω2x - 3Ω - Ωx + 4Ω = 8

–2x + 3 - (–x - 4) = 8

–2x + 3 + x + 4 = 8

–x + 7 = 8

–x = 8 - 7–x = 1

x = –1

Ω2x - 3Ω =

Ωx + 4Ω = Ωx + 4Ω =

Ω2x - 3Ω =

x > 3 2 –4 ≤ x ≤ 3 2 x < –4

3 2–4

subdomain subdomain

Is the solution in the subdomain? No

subdomain

Solve Ω2x - 3Ω - Ωx + 4Ω = 8

Final Solution: x = ____________

number line

268 Absolute Value, Radical, and Rational Equations Lesson #2: Absolute Value Equations - Part Two

Class Ex. #4 Complete the solution to the equation Ωx + 4Ω = Ωx - 2Ω

number line

subdomain subdomain

not true, no solution in this subdomain

subdomain

Final Solution x = _______

x < –4

–4 2

Ωx + 4Ω = x + 4

Ωx - 2Ω = x - 2

Ωx + 4Ω =

Ωx - 2Ω =

Ωx + 4Ω =

Ωx - 2Ω =

x + 4 = x - 2

solve Ωx + 4Ω = Ωx - 2Ω

4 = –2

\


Assignment

1. Describe clearly how to use the method of intersection to solve the equationΩx + 3Ω = 4. State the solution.

2. Describe clearly how to use the x-intercept method to solve the equation Ωx - 2Ω = x + 1. State the solution.

Absolute Value, Radical, and Rational Equations Lesson #2: Absolute Value Equations - Part Two 269

3. Solve each of the following equations graphically. Sketch and label each graph.

a) Ωx + 4Ω = Ωx - 2Ω b) Ω1 - 4xΩ = 6x

x

y

x

y

c) Ω3x + 1Ω = Ωx - 2Ω d) Ωx + 4Ω - Ω2xΩ = 0

x

y

x

y

4. Graphically, find the solution to the following equations.

a) ΩxΩ + Ω2 - xΩ = 3 b) Ω4 - xΩ = –2x - 10

x

y

x

y

c) 3Ωx - 8Ω = 2x + 7 d) Ωx2 - 4Ω = x2 - 4

x

y

x

y


5. The positive root, to the nearest hundredth, of the equation Numerical Response Ω

x2 + 3Ω - Ω2x - 1Ω = x2 is _____ .


Extension Question - Optional

6. Solve each of the following equations algebraically.

a) Ω4x - 1Ω = Ωx - 3Ω

Absolute Value, Radical, and Rational Equations Lesson #2: Absolute Value Equations - Part Two 271

b) Ω2x + 1Ω - Ωx - 2Ω = 2

Answer Key

1 . • Graph Y1 = Ωx + 3Ω • Graph Y2 = 4• Find the x-coordinate(s) of the point(s) of intersection using the intersect feature of the calculator.• Solution is x = –7 or 1.

2 . • Graph Y1= Ωx - 2Ω - x - 1 • Use the zero feature of the calculator to find the x-intercept(s).• Solution is x = 1

2 .

3 . a) x = –1 b) x = 110 c ) x = – 3

2 or 14 d) x = – 4

3 or 4

4 . a) x = – 12 or 5

2 b) x = –14 c ) x = 175 or 31 d) x £ –2 or x ≥ 2

5 . 1 . 3 9

6 . a) x = – 23 or 4

5 b) x = –5 or 1



Radical Equations

Radical Equation

A radical equation is an equation which contains a radical.

In this lesson we will solve radical equations graphically and algebraically.

Warm-Up #1 Graphing Radical Functions

Sketch the following radical functions using a graphing calculator. Answer the questions which follow.

x

y

5

5

–5

–5

a) f(x) = x

i ) What is the domain of f?

i i ) What is the range of f?

b) f(x) = – x

x

y

5

5

–5

–5



c) f(x) = x + 3

x

y

5

5

–5

–5



Class Ex. #1 Consider the radical equation x + 1 = 4.

5 10 15

5

-5

20

a) Describe how to use a graphing calculator to find the solution to the equation by using the x-intercept method .

b) Use the method in a) to sketch the radical equation x + 1 = 4 with a window x:[ –3, 20, 1] y:[–5, 5, 1]. Label the displayed graph on the grid.

c) State the solution to the equation.

d) Verify the solution.

Class Ex. #2 Consider the equation x + x - 3 = 5.

5 10

5a) Describe how to use a graphing calculator to find

the solution to the equation by finding a point of intersection.

b) Complete the following statement.“The grid provided shows the window x:[ , , ] y:[ , , ].”

c) Use the method in a) and the window in b) to solve the equation giving the root(s) to the nearest tenth. Sketch and label the displayed graph on the grid.


274 Absolute Value, Radical, and Rational Equations Lesson #3: Radical Equations

Solving Radical Equations Algebraically

In this section we will learn how to solve radical equations like:

• 3x + 7 = 7

• 3 + x - 1 = x

• x + x - 3 = 5

An equation like x - 4 - x + 1 = –1 is beyond the curriculum, but is covered in the extension section of this lesson.

Use the following method to solve radical equations algebraically.

Step 1: Isolate the radical term. If there are two radical terms, isolate the more complex term.

Step 2: Square both sides of the equation.

Step 3: If the resulting equation contains a radical term, repeat steps 1 and 2. Solve the resulting equation.

Step 4: Verify all answers because the squaring in step 2 may result inextraneaous roots.

Class Ex. #3 Solve the following radical equation and verify the solution.

3x + 7 = 7

Note Radical equations must be verified so that we do not allow extraneous roots in the solution.

Absolute Value, Radical, and Rational Equations Lesson #3: Radical Equations 275

Class Ex. #4 Solve the following radical equation.

3 + x - 1 = x


x + x - 3 = 5



Extension - Optional

The following class examples are beyond the scope of this course. We include the method here for students intending to take higher level math courses.

Class Ex. #6 Billy was given the radical equation 3a + 4 - a + 1 = 3 to solve. Billy solved the equation incorrectly. His work is shown below.

3a + 4 - a + 1 = 3

3a + 4 = 3 + a + 1

Ë 3a + 4 2

= Ë3 + a + 1 ¯2

3a + 4 = 9 + a + 1

2a = 6

a = 3

a) Explain where Billy made his error.

b) Show the correct work.



3 + x - 2 = 2x + 5



Assignment

1. Solve the following radical equations graphically. Answer to the nearest hundredth. Sketch and label the displayed graph on the grid.

a) 3x - 7 = x - 5 b) x + 5 - 2 x = 2

c) 2(5x - 1) + 3 = 0 d) 4p + 5 = 2 + 2p - 1

2. Describe how to solve the following radical equations by the method indicated. State the solution of the equation to the nearest tenth.

a) 6x + 4 = 3x - 1, b) 7x - 1 = x + 4 , by the x-intercept method by the intersection method.


3. Solve the following radical equations algebraically.

a) x - 7 = 8 b) 2y + 3 = 4

c) 3x - 25 = 6 d) 4 + x - 2 = x

e) 19a + 6 - 2a = 3 f) x = 2 2x - 4


4. Solve the following radical equations algebraically.

a) x + 5 = 2x + 1 b) x + x - 4 = 4

c) 2a + 1 - 5 = – a d) 2x = 5x + 9 - 3


5. Consider the two rectangles shown.

x

2x

3x

2x + 1

A B

CD

GH

FE

a) Determine the exact length of diagonal BD.

b) Determine the exact length of diagonal FH.

c) If BD is 1 unit longer than FH, determine the length and width of each rectangle.


6. When solving the equation x - 3 = x - 1 , the extraneous root is MultipleChoice

A. –2B. 2C. –5D. 5

7. The solution to the equation 2 x - x + 4 = 3, to the nearest tenth, is _____ . Numerical Response


Extension Question - Optional

8. Solve the following radical equations.

a) x + 11 - x - 9 = 2 b) x + 3 + 2 = x + 11


c) 4p + 5 = 2 + 2p - 1 d) 3 - a - 3 = – 2a + 3

Answer Key 1 . a) 9.70 b) 0.01 c ) no solution d) 1.00 or 5.00

2 . a) • Graph Y1 = 6x + 4 - 3x + 1 b) • Graph Y1 = 7x - 1 • Use the zero feature of the calculator • Graph Y2 = x + 4

to find the x-intercept(s) • Find the x-coordinate(s) of the point(s) of intersection• Solution is x = 1.5 using the intersect feature of the calculator

• Solution is x = 0.83 . a) 71 b) 13

2 c ) 1823 d) 6 e ) 3

4 or 1 f ) 4

4 . a) 144 b) 254 c ) 4 d) 0 or 8

5 . a) 5x + 1 b) 3x c ) rectangle ABCD has dimensions 7 by 3rectangle EFGH has dimensions 6 by 3

6 . B 7 . 1 2 . 4

8 . a) 25 b) –2 c ) 1 or 5 d) –1 or 3



Rational Functions

Rational Functions

A rational function is a function which takes the form f(x) = n(x)d(x) where n(x) and d(x) are

polynomial functions and d(x) π 0. The degree of d(x) needs to be greater than zero or the function f(x) is simply a polynomial function.

Examples of rational functions are f(x) = x + 2x - 1

, g(x) = 1

x2- 9, h(x) =

x2 - 4x - 2

etc.

Since rational functions are expressed as fractions, the denominator cannot equal zero. The domain of a rational function cannot include the zeros of d(x) and so the graph of a rational function will not be a continuous curve. There must be some kind of discontinuity in the graph. In this lesson we will learn about two types of discontinuity - asymptotes and point discontinuity.

Graphing Rational Functions

Warm-Up #1 Graphing the function f(x) ==== 1

x ---- 1

Consider the rational function f(x) = 1

x - 1 .

a) Since division by zero is not defined the domain of the function is __________________.

b) Complete the table of values then plot and join the points on the grid. x –5 –2 –1 0 0.5 0.8 0.9 0.99 1 1.01 1.1 1.2 1.5 2 5

f(x)

c) Describe what happens to the value of the

x

y

5

5

–5

–5

function as ΩxΩ gets larger and larger.

d) If x > 1, describe what happens to the value of the function as x gets closer and closer to 1.

e) If x < 1, describe what happens to the value of the function as x gets closer and closer to 1.

f) Verify the graph using a graphing calculator.

From the warm-up we notice the following:

• As ΩxΩ increases in value, the graph of f(x) gets closer and closer to the x-axis (the line with equation y = 0) but will never reach the x-axis.

• As x gets closer and closer to 1, the graph of f(x) gets closer and closer to the line x = 1 but will never reach the line x = 1.

Asymptotes

A line that a curve approaches more and more closely is called an asymptote.

With reference to Warm-Up #1:

• The line y = 0 is called a horizontal asymptote of the graph of f(x).

• The line x = 1 is called a vertical asymptote of the graph of f(x). This occurs because 1 is a zero of the denominator of the rational function.

Finding Equations of Asymptotes of Rational Functions Graphically

Class Ex. #1 Consider the function f(x) =

3x - 2 .

5-5

5

-5

x

y

a) State the domain of the function.

b) Use a graphing calculator to graph the function, but do not sketch on the grid.

c) Write the equation of the vertical asymptote using the information from a) and b).

d) Show the vertical asymptote on the grid with a dashed line.

e) Complete the graph on the grid.

f) State the equation of the horizontal asymptote.

g) State the range of the function.

286 Absolute Value, Radical, and Rational Equations Lesson #4: Rational Functions


2xx + 1 .

5-5

5

-5

x

y

a) State the domain of the function.

b) Use a graphing calculator to graph the function, but do not sketch on the grid.

c) Write the equation of the vertical asymptote using the information from a) and b).

d) Show the vertical asymptote on the grid with a dashed line.

e) Estimate the equation of the horizontal asymptote. Verify the equation by using the table feature of the calculator with very large values of x, eg. 100, 1000, 10000, etc.

f) Complete the graph on the grid using a dashed line for the horizontal asymptote.

g) State the range of the function.


x2

x - 2 .

5-5

5

-5

x

y

a) Draw the vertical asymptote on the grid and state its equation.

b) Sketch the graph of the function on the grid.

c) Does this function have a horizontal asymptote?


Absolute Value, Radical, and Rational Equations Lesson #4: Rational Functions 287

Note The graphing calculator has limitations when graphing some rational functions. The suggestions below may help to give a clearer representation of the graph of the function.

• If the calculator is in connected mode a line will connect two parts of the graph. This line is an approximation of the vertical asymptote. The horizontal asymptote line does not appear.

• If the calculator is set in dot mode neither the vertical asymptote nor the horizontal asymptote appears.

• Using zoom decimal (ZDecimal), or a multiple of the zoom decimal window, may produce a clearer representation of the graph.

Note The asymptotes are not part of the graph of the function, and are shown with dashed lines in

order to give greater understanding to the behaviour of the function.


Finding Equations of Asymptotes of Rational Functions Algebraically

Class Ex. #1 - #3 show examples of the following rules which can be used to algebraically determine the equations of vertical and horizontal asymptotes.

These rules apply for rational functions of the form f(x) = n(x)d(x) provided that n(x) and d(x)

have no factors in common. Situations where the numerator and denominator have a factor in common will be dealt with under point discontinuity.

Vertical Asymptotes

Algebraically we can find the equations of vertical asymptotes of rational functions by finding the zeros of the denominator because the graph is undefined at those value(s). The equation(s) will be x = the zero value(s).

Horizontal Asymptotes

The graph of f(x) has a horizontal asymptote under the following conditions:

• If the degree of n(x), is less than the degree of d(x), then the line y = 0 is a horizontal

asymptote. See Class Ex. #1 where f(x) = 3

x - 2 .

• If the degree of n(x), is equal to the degree of d(x), then the line y = ab is a horizontal

asymptote, where a is the leading coefficient of n(x) and b is the leading coefficient of

d(x). See Class Ex. #2 where f(x) = 2x

x + 1 .

• If the degree of n(x), is greater than the degree of d(x), then the graph has no horizontal

asymptote. See Class Ex. #3 where f(x) = x2

x - 2 .


Class Ex. #4 Algebraically determine the equations of the asymptotes of the graph of the function

f(x) = x2 + x - 62x2 - x - 3

. Verify using a graphing calculator.

Warm-Up #2 Point Discontinuity

Consider the functions f(x) = x + 2 and g(x) = x2 + 3x + 2

x + 1 .

a) State the domain of each function.

b) Use a graphing calculator to graph each function using zoom decimal. Show each graph on a separate grid.

f(x) = x + 2 g(x) = x2 + 3x + 2

x + 1

x

y

5

5

–5

–5x

y

5

5

–5

–5

c) Explain the difference between the two graphs.

d) Factor the numerator of g(x) to show that g(x) is identical to f(x) except for a domain restriction.


Point of Discontinuity

If the numerator and denominator of a rational function have a factor in common, then the graph of the rational function has a “hole” in it.

The point where the break in the graph occurs is called a point of discontinuity and the function is said to have point discontinuity This is illustrated in Warm-Up #2.

The point of discontinuity is represented on a graph by an open circle.

The coordinates of the point of discontinuity can be determined by factoring the rational expression and substituting the value of x for which the function is undefined into the factored form.

Note When graphing a function with a point of discontinuity, the “hole” in the graph will usually NOT be seen unless the calculator is set to zoom decimal (ZDecimal), or a multiple of the zoom decimal window.


2x2 + 7x + 3x + 3 .

a) Sketch the graph of the function f(x) and determine the coordinates of the point of discontinuity.

x

y

b) State the domain and range of f.



Assignment

1. In each of the following examples use a graphing calculator to determine;i ) the equation of the vertical asymptote

i i ) the equation of the horizontal asymptote i i i) the domain and range of the function

iv) the x and y-intercepts of the graph of the function.

a) f(x) = 1

x - 4 b) f(x) = 4x + 5x + 2

x

y

x

y

c) f(x) = x2 + 92x - 3

d) f(x) = x2 + 4x + 4

x2 + 3x - 10

x

y

x

y


2. Algebraically determine the equations of the asymptotes of the graph of each of the following functions.

a) f(x) = 2

x + 3 b) f(x) = x2 + 5x + 6

x + 6

c) f(x) = 4x

1 - 4x d) f(x) = x

x2 - 4

3. Consider the function f(x) = x2 + 2x - 8

x - 2 .

a) Sketch the graph of the function f(x) and determine the point of discontinuity.

x

y



4. Consider the function f(x) = 3x2 - 10x + 3

3x - 1 .

a) Sketch the graph of the function f(x) and determine the point of discontinuity.

x

y


5. For each of the graphs of the following rational functions, algebraically determine the equation of any asymptotes or the coordinates of any points of discontinuity.

a) f(x) = x

(x - 1)(2x - 7) b) f(x) = 2x - 15 - 3x

c) f(x) = (3x - 1)(x + 4)3x2 + 4x + 1

d) f(x) = (3x - 1)(x + 4)3x2 + 10x - 8


6. The horizontal asymptote of the function g(x) = –7x + 23x + 2 isMultiple

ChoiceA. y = 0

B. y = –7

3

C. y = –3

7

D. x = –2

3

7. The function f(x) = x2 - 4x - c

x + 2 has a point of discontinuity. The value of c is _____ .Numerical Response


Answer Key 1 . a) b) c ) d)i ) x = 4 x = –2 x = 3

2 x = –5, 2i i ) y = 0 y = 4 none y = 1

i i i ) xΩx π 4, x Œ ¬ xΩx π –2, x Œ ¬ ÓÏÔÔÌxΩx π 3

2 , x Œ ¬ ÔÔ xΩx π –5, 2, x Œ ¬

yΩ y π 0, y Œ ¬ yΩ y π 4, y Œ ¬ yΩy £–1.85 or y ≥ 4.85, y Œ ¬ yΩy π 1, y Œ ¬ i v ) x-intercept = none x-intercept = – 5

4 x-intercept = none x-intercept = –2

y-intercept = –14 y-intercept = 5

2 y-intercept = –3 y-intercept = – 25

2 .

a) b) c ) d)

Vertical Asymptote x = –3 x = –6 x = 14 x = ±2

Horizontal Asymptote y = 0 none y = –1 y = 0

3 . a) (2, 6) b) Domain: xΩx π 2, x Œ ¬ Range: yΩ y π 6, y Œ ¬

4 . a) ËÊÁ13 , – 8

3 b) Domain: ÓÏÔÔÌxΩx π 1

3 , x Œ ¬ ÔÔ Range: ÓÏÔÔÌyΩ y π – 8

3 , y Œ ¬ ÔÔ5 .

a) b) c ) d)

Vertical Asymptote x = 1, x = 72 x = 5

3 x = –1, x = – 13 x = 2

3

Horizontal Asymptote y = 0 y = – 23 y = 1 y = 1

Point of Discontinuity - - - ËÊÁ–4, 13

14

6 . B 7 . 1 2



Rational Equations

Rational Equation

A rational equation is an equation in which at least one of the terms is a rational expression with a variable in the denominator.

Solving Rational Equations Using A Graphing Calculator

Recall the two methods (the intersection method and the x-intercept, method from Lesson #2 in “Absolute Value Equations” page 267) to use a graphing calculator to solve equations.

Class Ex. #1 Solve the following rational equations graphically.

a) 3x + 1 +

1x - 1 = 2 b) 2x2 + x - 15

x + 3 = 15

x

y

x

y

Solving Simple Rational Equations Algebraically

The following strategies should be considered when solving rational equations algebraically:

1. If the equation consists of a single rational expression on each side use cross-multiplication to simplify the equation.

2. If the rational equation has more than one term, on either side, consider multiplying each term in the equation by the lowest common multiple of the denominators.

3. Use previous skills for solving linear or quadratic equations.

4. Always be aware there are domain restrictions when dealing with rational functions and check your solutions accordingly.

Class Ex. #2 Solve the following rational equations algebraically.

a) x2 - 5x - 6x + 1 = 2 b) x +

2x = 3 c) x

x2 - 4 =

2x + 2

Class Ex. #3 A student hiking in the mountains covered 10 km of open terrain at a certain average speed and covered 2 km of rugged terrain at an average speed 3 km/h slower. The total time taken was 3 hours.

a) If we let x km/h be the average speed over open terrain, write expressions for the time taken for each part of the hike.

b) Form an equation in x and solve it to find the student’s average speed for each part of the hike.


296 Absolute Value, Radical, and Rational Equations Lesson #5: Rational Equations

Assignment

1. Solve the following rational equations using a graphing calculator. Answer to the nearest hundredth where necessary.

a) x + 3x + 1 =

x + 75x + 1 b) 8

x - 5 = x2

x

y

x

y

c) xx + 2

+ x

x - 2 =

16x2 - 16

d) 2x - 4 +

6x + 2 =

12

x

y

x

y

2. Solve the following rational equations algebraically.

a) 4x + 2 = 3 b) 3

2x - 1 = 4

x + 7

Absolute Value, Radical, and Rational Equations Lesson #5: Rational Equations 297

3. Solve the following rational equations algebraically.

a) x - 1x + 1 =

2x15 b) 4x

3x + 4 - 10

x + 6 = 0

c) x + 32x + 1 =

x + 75x + 1 d) x + 3

x2 + 4x + 3 = 1

e) 2x +

16 - x = 1 f) 8

x - 5 = x2


4. A student taking part in a cross-country endurance race was required to cycle for 70 km then run for 7 km. Her average cycling speed was five times as fast as her average running speed. She completed the course in 5 hours 15 minutes. Find the student’s average cycling speed.

5. Ann and Mike had to attend a conference at the West Edmonton Mall. Ann took a charter bus and rode 400 km to Edmonton. Mike drove his own car and travelled 368 km to Edmonton. On his journey he was delayed by road construction which resulted in his average speed for the journey being 8 km per hour slower than the average speed of the bus. If they both left home at the same time, and arrived in Edmonton at 10:30 a.m., at what time did Mike leave home?

6. Two numbers differ by three and their quotient is 34 . Form a rational equation in a single

variable and solve it to determine the numbers.

Absolute Value, Radical, and Rational Equations Lesson #5: Rational Equations 299

7. Amy, Becky, and Christine are playing a math game involving positive numbers.When Amy chooses a number, Becky has to choose the number one more than Amy’s number, and Christine has to choose the number one more than Becky’s number. The numbers are such that the reciprocal of the smallest number is equal to the sum of the reciprocals of the other two numbers. Determine Amy’s number.

8. The complete solution to the equation 5 = x2 + x - 6

x + 3 isMultipleChoice

A. x = –3 onlyB. x = 7 onlyC. x = –3 or 7D. x = 3 or –7

9. The positive root of the equation x - 2x + 4 =

x - 32 , to the nearest tenth, is _____ .Numerical

Response(Record your answer in the numerical response box from left to right)

Answer Key 1 . a) x = –2.41 or 0.41 b) x = –11.4 or 1.4 2 . a) x = – 2

3 b) x = 5c ) x = –4.75, –1.19, 1.19, or 4.75 d) x = 2 or 16

3 . a) x = 32 , 5 b) x = – 5

2 , 4 c ) x = – 43 , 1

d) x = 0 e ) x = 3, 4 f ) –5 ± 41

4 . 20 km/hr 5 . 6:30 a.m. 6 . 9, 12

7 . 2 8 . B 9 . 3 . 4



Inverse of a Rational Function

Warm-Up Review

Recall the following steps to find the inverse of a function. Step 1 Step 2 Step 3 Step 4

Replace f(x) by y

Interchange x and y to obtain the inverse.

Solve for yReplace y by f –1(x)

(see below)

• If the inverse of f(x) is a function, then the inverse function is denoted by f –1(x).• If the inverse of f(x) is not a function, then the notation f –1(x) should not be used.


2x - 4 .

a) State the equations of the asymptotes of the graph of f(x).

b) State the domain and range of f(x).

c) Use the procedure above to determine the equation of the inverse function.

d) State the equations of the asymptotes of the graph of f –1(x).

e) State the domain and range of the inverse function

f) Comment on your observations from the results of a), b), d), and e).

g) Graph y = f(x) and y = f –1(x) on

5-5

5

-5

x

ythe same grid.

Class Ex. #2 Find the inverse of the function f where f(x) =

x - 2x + 3 stating any domain restrictions on the

inverse function.


302 Absolute Value, Radical, and Rational Equations Lesson #6: Inverse of a Rational Function

Assignment

1. Consider the function f(x) = 2x

x + 1 .

a) State the equations of the asymptotes of the graph of f(x).

b) State the domain and range of f(x).

c) Determine the equation of the inverse function.

d) State the equations of the asymptotes of the graph of f –1(x).

e) State the domain and range of the inverse function

f) Graph y = f(x) and y = f –1(x) on

5-5

5

-5

x

ythe same grid.

Absolute Value, Radical, and Rational Equations Lesson #6: Inverse of a Rational Function 303

2. In each of the following functions whose equation is given:

• find the equation of the inverse function• sketch a graph of the original function and the inverse.

• determine the domain, range and equations of any asymptotes of the inverse function.

a) y = 1

x - 3

x

y

b) y = 2x

x - 3

x

y

c) y = 2x

3 - 4x

x

y


3. Find the inverse of the following functions, stating any domain restrictions for the inverse function.

a) f(x) = 1

x - 7 b) f(x) = x

x + 3

c) f(x) = x - 2

x d) f(x) = 5x - 34x + 1

4. If f –1(x) is the inverse of the function f(x) = 1

x + 2 , x π –2, then f –1(x) equalsMultipleChoice

A. 1 - 2x

x , x π 0

B. 1

x + 2 , x π - 2

C. 1

x - 2 , x π - 2

D. x + 2

5. If f(x) = x

x - 2 , then the domain and the range of the inverse function are respectivelyA. x π –2; y π –1B. x π 2; y π 1 C. x π 1; y π 2D. x π –1; y π –2

Absolute Value, Radical, and Rational Equations Lesson #6: Inverse of a Rational Function 305

Answer Key 1 . a) vertical asymptote: x = –1, horizontal asymptote: y = 2

b) domain: xΩx π –1, x Œ ¬ range:yΩy π 2, y Œ ¬

c ) f –1(x) = x

2 - xd) vertical asymptote: x = 2, horizontal asymptote: y = –1e ) domain: xΩx π 2, x Œ ¬ range:yΩy π –1, y Œ ¬

2 . a) y = 1 + 3x

x, domain: xΩx π 0, x Œ ¬ vertical asymptote: x = 0

range: yΩy π 3, y Œ ¬ horizontal asymptote: y = 3

b) y = 3x

x - 2, domain: xΩx π 2, x Œ ¬ vertical asymptote: x = 2

range: yΩy π 3, y Œ ¬ horizontal asymptote: y = 3

c ) y = 3x

4x + 2, domain: Ó

ÏÔÔÌxΩx π – 1

2 , x Œ ¬ ÔÔ vertical asymptote: x = – 12

range: ÓÏÔÔÌyΩy π 3

4 , y Œ ¬ ÔÔ horizontal asymptote: y = 34

3 . a) f –1(x) = 1 + 7x

x, x π 0 b) f –1(x) =

3x1 - x

, x π 1

c ) f –1(x) = 2

1 - x, x π 1 d) f –1(x) =

–3 - x4x - 5

, x π 54

4 . A 5 . C


Mathematical Reasoning Lesson #1:Inductive Reasoning, Conjectures, and Counterexamples

Warm-Up Pascal’s Triangle

The triangular array of numbers shown is known as Pascal’s Triangle (named after Blaise Pascal who developed the triangle and its applications in the 17th century. The Chinese developed such an array seven centuries earlier in the 10th century).

• Complete the next two rows of Pascal’s Triangle.

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 1

1 6 15 20 15 6 15

Inductive Reasoning

Inductive reasoning is a type of reasoning in which we arrive at a conclusion, generalization, or educated guess based on experience, observations, or patterns. In the Warm-Up, the next two rows of Pascal’s Triangle were completed by your observing the patterns from the previous rows.

Conjecture

The conclusion, generalization, or educated guess which is arrived at by inductive reasoning is called a conjecture.

Note Conjectures may or may not be true.

Class Ex. #1 Consider the following22 + 1 = 5 (prime number)42 + 1 = 17 (prime number)62 + 1 = 37 (prime number)102 + 1 = 101 (prime number)202 + 1 = 401 (prime number)362 + 1 = 1297 (prime number)

a) Write a conclusion based on the information.

b) The answer to a) is a _____________ based on _______________ ______________.

Counterexample

A counterexample is an example which shows that a conjecture is not true (false).

Class Ex. #2 Provide a counterexample to show that the conclusion in Class Ex. #1 is false.

Class Ex. #3 A student constructed the following three triangles and measured the interior angles.

37°

53°

110°

30°

40°

45°

65°

70°i) ii) iii)

What conjecture can be made from this information?

Note • A theorem is a statement which can be proved using logical or deductive reasoning as we shall see later in Lesson 2 of this unit. A theorem cannot be proved using inductive reasoning because we can never be certain that the conclusion is always true.

• The conjecture in Class Ex. #3 is a theorem called the Angle Sum of a Triangle Theorem.

• Notice we have not proved this theorem.

Class Ex. #4 In the first diagram –ACD is an exterior angle and –CAB and –CBA are the interior opposite angles to –ACD.

A

C DBI

KL

J

EF

G

H

35° 120°85°

127°

38°

53°

89°

115°52°

63°

65°60°

Write a conjecture about the relationship between the exterior angle of a triangle and the sum of the two interior opposite angles.

Note The conjecture in this example is a theorem called the Exterior Angle Theorem. We have not proved this theorem because we have used inductive reasoning.

308 Mathematical Reasoning Lesson #1: Inductive Reasoning, Conjectures, & Counterexamples

Class Ex. #5 Dawn graphed the equation y = x x using her graphing calculator and the window x: [–5, 5, 1] y:[–4, 4,1]. After observing the screen, she made the conjecture y = x x has the domain D = xΩx > 0, x Œ ¬. Is her conjecture true? If not, give a counterexample.


Assignment

1. Consider the following pattern of triangles.

Figure 1 Figure 2 Figure 3

a) Determine the number of small triangles in each figure and use this information to make a conjecture.

b) Draw two more examples that demonstrate your conjecture.

2. Consider the sequence of multiplications shown.

3 ¥ 9 = 2733 ¥ 9 = 297

333 ¥ 9 = 2997

3333 ¥ 9 = 29997

a) Use this information to make a conjecture.

b) Show two more examples that demonstrate your conjecture.

Mathematical Reasoning Lesson #1: Inductive Reasoning, Conjectures, & Counterexamples 309

3. Consider a pattern of multiplications.

1 ¥ 1 = 1 11 ¥ 11 = 121 111 ¥ 111 = 12321

1111 ¥ 1111 = 1234321

a) Predict the answers to 11111 ¥ 11111 = ________________________

1111111 ¥ 1111111 = ____________________________

b) Use this information above to make a conjecture.

4. Consider a pattern of multiplications.

11 ¥ 11 = 12111 ¥ 11 ¥ 11 = 1 331

11 ¥ 11 ¥ 11 ¥ 11 = 14 641


b) Provide a counterexample to prove your conjecture is false.

5. Consider the following computations for subtracting a number from the reverse of the number. In each case the larger number is subtracted from the smaller number.

41 - 14 = 27 Æ 2 + 7 = 9

312 - 213 = 99 Æ 9 + 9 = 18 Æ 1 + 8 = 9

9451 - 1549 = 7902 Æ 7 + 9 + 0 + 2 = 18 Æ 1 + 8 = 9

98652 - 25689 = 72963 Æ 7 + 2 + 9 + 6 + 3 = 27 Æ 2 + 7 = 9




6. Consider the pattern of multiplications.

1 2 3 4 5 6 7 9 ¥ 9 ¥ 1 = 1 1 1 1 1 1 1 1 1

1 2 3 4 5 6 7 9 ¥ 9 ¥ 2 = 2 2 2 2 2 2 2 2 2

1 2 3 4 5 6 7 9 ¥ 9 ¥ 3 = 3 3 3 3 3 3 3 3 3



c) Provide a counterexample to show your conjecture is false.

7. In each case a conjecture is given. Provide one example which supports each conjecture and one counterexample which shows the conjecture is false.

a) The sum of two consecutive prime numbers is even.

b) If a quadrilateral has four equal sides, it has four equal angles.

c) x2 = x.

d) A point lies in the second quadrant if its x coordinate is negative.

e) The square of a number is greater than or equal to the number.


8. The diagrams below show the number of distinct regions which can be formed when points on a circle are joined.

a) Complete only the first three entries in the table below.

Number of points on the circle 2 3 4 n

Number of regions

b) Complete the following conjecture and complete the fourth entry in the table.

“If there are n points on a circle, the number of regions equals _________ .”

c) Show that n = 5 supports the conjecture by completing the diagram.

d) Complete the diagram for n = 6 and use the result to test the conjecture.

e) Complete the following statement.

“The value of n = 6 is a _____________________ to the conjecture in b)


Questions #9 - #11 refer to the following A Fibonacci sequence of numbers is a sequence in which the sum of two consecutive termsgives the next term, eg. starting with the sequence 1, 1 . . ., the third term is 1 + 1 = 2, etc.Consider the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, . . .

9. a) Write the first 15 terms of the sequence.

1, 1, 2, 3, 5, 8, 13, 21, ____, ____, ____, ____, ____, ____, 610

b) Heidi picks four consecutive Fibonacci numbers. She multiplied the outside two and the inside two and subtracted the answers. She tried this procedure with the following sets of four numbers. Complete her work.2, 3, 5, 8 Æ 2 ¥ 8 =16, 3 ¥ 5 = 15 Æ 16 - 15 = 1

8, 13, 21, 34

____, ____, ____, 610

c) Use this information to make a conjecture.

d) Show one more example that demonstrates your conjecture.

10. a) Write any ten consecutive Fibonacci numbers.

b) Find the sum of the ten numbers.

c) Multiply the fourth last number by 11. What do you notice?

d) Repeat parts a) - c) by selecting two further sets of ten consecutive Fibonacci numbers.

e) Make a conjecture based on your observations.

11. Viktor made a conjecture that for any three consecutive Fibonacci numbers, the absolute difference of the product of the first and last numbers and the square of the middle number is 1. Provide three examples which support this conjecture.


12. Stacey makes the following conjecture for all real numbers a, b, and c.MultipleChoice If a < b then ca < cb.

Which of the following values of c provides a counter example to Stacey’s conjecture?

A. 2B. 0.5C. –2D. p

Answer Key Figure 4 Figure 5

42 = 16 52 = 25

1 . a) 1, 4, 9. b) Answers may vary

The number of triangles is the square of the figure number.

2 . a) Multiplying a numeral with n 3’s by 9 will result in a numeral with first digit 2 followed by n - 1 9’s, followed by 7

b) Answers may vary 33333 ¥ 9 = 299997, 333333 ¥ 9 = 2999997

3 . a) 123454321, 1234567654321b) Multiplying a numeral with n 1’s by itself results in the numeral 123 ... (n - 1) n (n - 1) ... 321.

4 . a) Multipling 11 ¥ 11 ¥ 11 ¥ ... to n factors results in a numeral which is the (n + 1)th row of Pascal’s Triangle

b) 11 ¥ 11 ¥ 11 ¥ 11 ¥ 11 ¥ 11 = 116 = 1771561 which is not the 7th row of Pascal’s Triangle.

5 . a) When a number is subtracted from the reverse of the number, and the digits in the answer are repeatedly added until a single digit is produced, the answer is 9

b) Answers may vary 634 - 436 = 198 1 + 9 + 8 = 18 1 + 8 = 98437 - 7248 = 1089 1 + 0 + 8 + 9 = 18 1 + 8 = 9

6 . a) When the number 12345679 is multiplied by 9 and then by the whole number n, the result is the number nnnnnnnnn.

b) Answers may vary 12345679 ¥ 9 ¥ 5 = 555555555, 12345679 ¥ 9 ¥ 8 = 888888888c ) Answers may vary 12345679 ¥ 9 ¥ 10 = 1111111110

7 . Answers may vary a) 3 + 5 = 8, 2 + 3 = 5 b) Square, rhombus c ) 52 = 5, (–6)2 π –6

d) (–3, 4), (–3, –4) e ) 62 ≥ 6, (0.1)2 ≥ 0.1

8 . a) 2, 4, 8 b) 2n - 1 c ) There are 16 regions, so n = 5 supports the conjecture. d) There are 31 regions, so n = 6 does not support the conjecture. e ) counterexample

9 . a) 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610 b) 8, 13, 21, 34 Æ 8 ¥ 34 = 272, 13 ¥ 21 = 273 Æ 273 - 272 = 1

144, 233, 377, 610 Æ 144 ¥ 610 = 87840, 233 ¥ 377 = 87841 Æ 87841 - 87840 = 1c ) When four consecutive Fibonacci numbers are chosen, the absolute difference of the product of the

outside two and the product of the inside two is 1 d) Answers may vary

1 0 . a), b), c), d), answers may varye ) the sum of 10 consecutive Fibonacci numbers equals the fourth last number multiplied by 11

1 1 . Answers may vary 2, 3, 5 Æ 2 ¥ 5 = 10, 32 = 9 Æ 10 - 9 = 15, 8, 13 Æ 5 ¥ 13 = 65, 82 = 64 Æ 65 - 64 = 1

1 2 . C 55, 89, 144 Æ 55 ¥ 144 = 7920, 892 = 7921 Æ 7921 - 7920 = 1


Mathematical Reasoning Lesson #2:Deductive Reasoning

Warm-Up #1 Review of Inductive Reasoning

Recall that inductive reasoning is a type of reasoning in which we arrive at a conjecture based on experience, observations, or patterns.

Inductive reasoning, however, does not guarantee that the conjecture is true in all cases, no matter how many examples we have to support a conjecture. Just because we cannot find a counterexample does not mean that one does not exist.

Inductive reasoning can never be used to prove a conjecture.

Inductive reasoning can play a part in a discovery of mathematical truths, but some other form of reasoning is required to make the proof.

Warm-Up #2

Suzy dropped off a film at “Quick One Hour Photo”. Quick One Hour Photo advertises that your film will be developed within one hour or there is no charge.

Suzy dropped off a film at 10:00 am and it wasn’t developed until 11:15 am

What can Suzy deduce from this information?

Deductive Reasoning

Deductive reasoning (also called logical reasoning) is the logical process of using true statements to arrive at a conclusion. In Warm-Up #2 Suzy used deductive reasoning to conclude her film would be developed free of charge.

Class Ex. #1 Write a conclusion which can be deduced from each pair of statements.

a) Every whole number is an integer. Six is a whole number.

b) Water freezes below 0°C. The temperature is –15°C.

Theorem

A theorem is a statement which can be proved using logical or deductive reasoning.

Class Ex. #2 In Lesson 1, Class Ex. 3, we used inductive reasoning to write a conjecture about the relationship between the exterior angle of the triangle and the sum of the two interior opposite angles. This conjecture is a theorem called the Exterior Angle Theorem. Use deductive reasoning to prove the Exterior Angle Theorem.

a

b c x

Class Ex. #3 a) Complete the chart using inductive reasoning to make a conjecture.

Instruction

Test Case 1

Test Case 2

Test Case 3

choose a number less than 10

Add 7

Multiply by 2

Subtract the original number

Subtract 2


b) Prove your conjecture using deductive reasoning. Let x represent the original number chosen.

Instruction General Case

choose a number less than 10

Add 7

Multiply by 2


Subtract 2


c) We have proved that whatever original number is chosen the final answer is _____ .

d) Does the original number have to be less than 10?

316 Mathematical Reasoning Lesson #2: Deductive Reasoning

Note In this lesson we will be considering statements involving odd numbers or even numbers.

Note that:

• Every even number is of the form 2n, where n Œ N.

• Every odd number is of the form 2n ---- 1, where n Œ N.

Class Ex. #4 Consider the following statement:

“When two odd numbers are added, their sums are always even”

a) Use inductive reasoning ( three cases) to suggest the statement is true.

b) Use deductive reasoning to complete the proof of the statement above.

Let the numbers be 2n - 1 and 2m - 1.


Mathematical Reasoning Lesson #2: Deductive Reasoning 317

Assignment

1. Write a conclusion which can be deduced from each pair of statements.

a) Leona lives in 100 Mile House. 100 Mile House is in British Columbia.

b) Joan is taller than Stefan. Stefan is taller than Patrick.

c) The sides of a rhombus are equal. PQRS is a rhombus.

d) Prime numbers have two factors. 13 is a prime number.

2. a) Complete columns “Choice 1”, “Choice 2”, and “Choice 3”, only.

Instruction Choice 1 Choice 2 Choice 3 General CaseChoose a number less than 10 nAdd on 4Double itAdd on 7Add on the original numberDivide by 3Subtract the original numberAdd on the number of the month you were born in.Add on 4Multiply by 100Add on the number of the day you were bornMultiply by 100Add the last two digits of the year you were bornSubtract 90 000

b) Make a conjecture based on your answers in columns “Choice 1”, “Choice 2”, and “Choice 3”.

c) Use deductive reasoning to complete column “General Case” and show that no matter which number you choose to start with, the conjecture is true.


3. Use deductive reasoning to prove the following statements.

a) The sum of any three consecutive even numbers is divisible by six.

b) The sum of any three odd numbers is an odd number.

c) A product of any two odd numbers is an odd number.

d) The difference of the squares of two consecutive even numbers is divisible by four.

Mathematical Reasoning Lesson #2: Deductive Reasoning 319

4. In the last lesson, we made a conjecture about Fibonacci numbers, which stated that the sum of any 10 consecutive Fibonacci numbers was equal to 11 times the fourth last number.

Use deductive reasoning with the Fibonacci sequence starting x, y, x + y, x + 2y, . . to prove the conjecture.

Answer Key 1 . a) Leona lives in British Columbia. b) Joan is taller than Patrick.

c ) The sides of PQRS are equal. d) 13 has two factors.

2 . a) Answers may vary, Choice 1 was for birth date May 12, 1986

Instruction Choice 1 General CaseChoose a number less than 10 4 nAdd on 4 8 n + 4Double it 16 2n + 8Add on 7 23 2n + 15Add on the original number 27 3n + 15Divide by 3 9 n + 5Subtract the original number 5 5Add on the number of the month you were born in. 10 5 + MAdd on 4 14 9+MMultiply by 100 1400 900 + 100MAdd on the number of the day you were born 1412 900 + 100M + DMultiply by 100 141200 90000 + 10000M + 100DAdd the last two digits of the year you were born 141286 90000 + 10000M + 100D + YSubtract 90 000 51286 10000M + 100D + Y

b) The answer is the birth date MM/DD/YY.


Mathematical Reasoning Lesson #3: Connecting Words - ”And”, “Or”, & “Not”

Statement

In mathematics we deal with statements. A statement is a sentence that is either true or false. A sentence which may be judged true by one person and false by another is not considered a statement - it is an opinion.

Consider the following:

• Wayne Gretzky played for the New York Rangers during the 1998-99 hockey season. (This statement is true - it is a fact)

• Wayne Gretzky played for the Edmonton Oilers during the 1996-97 hockey season.(This statement is false)

• Wayne Gretzky will always be known as Canada’s greatest hockey player. (This is not a statement - it is an opinion).

• Was Wayne Gretzky born in Canada? (This is not a statement - it is a question.).

Compound Statement

A compound statement is a statement formed by combining two or more statements using the words and, or, or not.

Class Ex. #1 Consider the following statements:

i ) Jan Arden is a Canadian singer.i i ) Jan Arden has a restaurant in Calgary.

Combine the above statements to form a compound statement.

The Negation of a Statement

The negation of a statement is the exact opposite of the statement. Often the word not is used to form the negation of a statement.

Class Ex. #2 A student made the following false statement: “All triangles are isosceles.”

Which of the statements below is the negation of the above statement?

A. All triangles are not isosceles.B. Not all triangles are isosceles.

C. All triangles are scalene or equalateral.

“Or” as Inclusive in Mathematics

When “or” is used in everyday language, it can be inclusive or exclusive.

Note However in mathematics, “or” is always inclusive.

The use of “or” in ordinary English

Consider the following examples for the use of “or” in ordinary English usage:

i ) “In order for Drew to have the necessary prerequisites to go to college next term, he needs a pass in Grade 12 Physics or a pass in Grade 12 Chemistry this semester”. This statement means that Drew will have the necessary prerequisites if;

• he passes Grade 12 Physics alone, or

• he passes Grade 12 Chemistry alone, or • he passes both Grade 12 Physics and Grade 12 Chemistry.

This is an example of the inclusive use of “or” in everyday language.

i i ) John asked Helen how she was getting to school tomorrow. She replied “I will take the bus or drive myself”.

This means that Helen either takes the bus or she drives, but not both.

This is an example of the exclusive use of “or” in everyday language.

The use of “or” in Mathematics

Consider the following example for the use of “or” in Mathematics:

i ) Ivan rolled a pair of dice and the sum of the two numbers was an odd number or a prime number.

This means that the sum was • an odd number, or

• a prime number, or • both an odd number and a prime number.

This is an example of the inclusive use of “or” in mathematics.

Class Ex. #3 a) List the whole numbers less than 20 that are divisible by 3.b) List the whole numbers less than 20 that are divisible by 5.

c) List the whole numbers less than 20 that are divisible by 3 and 5.

d) List the whole numbers less than 20 that are divisible by 3 or 5.


322 Mathematical Reasoning Lesson #3: Connecting Words - “And”, “Or”, & “Not”

Graphing Single Variable Inequalities on a Number Line

Recall the following for graphing single variable inequalities on a number line where the domain is the set of real numbers:

• an open circle on the number line means that the solution to the inequality does not include that particular number.

• a solid circle on the number line means that the solution to the inequality does include that particular number.

Class Ex. #4 Write a compound statement using inequalities and the word and or or to describe each solution set. In the cases where the word and is used, write the compound statement as a single statement.

-4 -3 -2 1 0 1 2 3 4 5 6 7

a)

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5

b)

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5

c)

-3 -2 -1 0 1 2 3 4 5 6 7 8

d)

Class Ex. #5 Graph the following inequalities on a number line, where x Œ ¬.

a) x > 0 and x < 2 b) x £ –3 or x > 5

c) x > 2 and x > 4 d) x > 2 or x > 4

e) x > 2 and x < –6 f) x £ 5 or x ≥ 0


Mathematical Reasoning Lesson #3: Connecting Words - “And”, “Or”, & “Not” 323

Assignment

1. Which of the following sentences are statements?

a) Vancouver is the capital of British Columbia.

b) Edmonton is a better place to live then Calgary.

c) Mars has two moons.

d) Is Grade 11 math the best subject in high school?

e) Po is one of the Teletubbies.

f) The New York Yankees are the best baseball team.

g) A parallelogram has four equal sides.

h) How many equations are required to form a system of equations?

i ) Not all rectangles are squares.

j) Pat is a girl.

2. For each of the statements in question #1, write the negation of the statement.

3. In each of the following cases, form a compound statement from the statements given.

a) • 7 is a prime number • 7 is an odd number.

b) • Hydrogen is a gas • Helium is a gas.

c) • Juno is a German Shepherd • Juno is not a cat.


4. Consider the following compound statements. In each case state all the values of the variable which make the compound statement true. The variables are defined on the set of natural numbers.

a) a is less than10, and a is a prime number less than 25.

b) b is less than10, or b is a prime number less than 25.

c) c is an odd number less than 10, or c is a factor of 10.

d) d is an odd number less than 10, and d is a factor of 10.

e) e is a factor of nine, or e is an even prime number.

f) f is a factor of nine, and f is an even prime number.

g) g is an even number less than 40, and g is divisible by 7.

h) h is an odd number, and h2 is less than 30.

i ) i is a multiple of 5, and i is a factor of 30, and i is an even number.

5. a) Give an example of the inclusive use of “or” in everyday language.

b) Give an example of the exclusive use of “or” in everyday language.

c) Give an example of the inclusive use of “or” in mathematics.

d) Give an example of the inclusive use of “and” in mathematics.


6. Write a compound statement using inequalities and the word and or or to describe each solution set. In the cases where the word and is used, write the compound statement as a single statement.

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4

a)

1 2 3 4 5 6 7 8 9 10 11 12

b)

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5

d)

-3 -2 -1 0 1 2 3 4 5 6 7 8

c)

-3 -2 -1 0 1 2 3 4 5 6 7 8

f)

-9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2

e)

7. Graph the following inequalities on a number line, where x Œ ¬.

a) x £ –2 or x > 6 b) x > –2 and x < 3

c) x > 4 or x < –2 d) x £ –1 and x > –3

e) x < 5 and x £ 4 f) x > 1 or x > 3

g) x < 5 or x > 7 h) x < 5 and x > 7

i) x and ≥ 1x < –4 j) x £ 4 or x ≥ –2


8. Graph the solution for the following compound inequalities.

a) 2x + 4 £ 8 and –10x ≥ 50 b) –3x + 7 < –x - 3 or 2x < 6

9. The graph shown represents the solution to which of the following compound inequalities?MultipleChoice

-7 -1

A. x > –7 or x < –1B. x < –7 or x > –1C. x < –7 and x > –1D. x > –7 and x < –1

10. Which of the following compound inequalities can be written as 2 £ x £ 6?

A. x ≥ 2 or x £ 6B. x £ 2 or x ≥ 6C. x ≥ 2 and x £ 6D. x £ 2 and x ≥ 6

11. The solution to the compound inequality x - a ≥ 2.2 or x + 2 ≥ 8 is x ≥ 5. Numerical Response The value of a, to the nearest tenth, is _____ .



Answer Key

1 . a), c), e), g), i), j)

2 . a) Vancouver is not the capital of British Columbia.c ) Mars does not have two moons.e ) Po is not one of the Teletubbies.g ) A parallelogram does not have four equal sides.i ) All rectangles are squares.j ) Pat is not a girl.

3 . a) 7 is a prime number and an odd number. b) Hydrogen and helium are gases.c ) Juno is a German Shepherd and is not a cat.

4 . a) 2, 3, 5, 7 b) 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 17, 19, 23 c ) 1, 2, 3, 5, 7, 9, 10 d) 1, 5e ) 1, 2, 3, 9 f ) no solution g ) 14, 28 h) 1, 3, 5 i ) 10, 30

5 . Answers may vary a) For a class party you can bring cakes or juice.b) Tommorrow I will wear black shoes or brown shoesc ) Depending on the context of the question, the solution to x2 = 9 is 3 or –3.d) A square has four equal angles and four equal sides

6 . a) x > –5 and x < 1 Æ –5 < x < 1 b) x £ 4 or x > 10 c ) x < 3 or x ≥ 6d) x > –3 and x £ 3 Æ –3 < x £ 3 e ) x ≥ –6 and x £ –2 Æ –6 £ x £ –2 f ) x < –1 or x = 3

7 .

2 6

a)

–2 3

b)

–2 4

c)

–3 –1

d)

4

e)

1

f)

5 7

g)

no solution

–4 1

i)

–2 4

j)

h)

8 .

–5

a) b)

3 5

9 . D 1 0 . C 1 1 . 2 . 8


Mathematical Reasoning Lesson #4:Venn Diagrams - Part One

In the last lesson we defined the use of “and”, “or”, and “not” in mathematics and applied them to number lines and everyday usage. In this lesson we will apply the use of “and”, “or”, and “not” to sets and Venn Diagrams.

SetA set is a well defined collection of objects which can be defined by listing its members or by using set notation.

eg. P = the set of even numbers less than 10 = 2, 4, 6, 8

Warm-Up Review

Recall and complete the following class example from the last lesson.

a) List the set of whole numbers less than 20 that are divisible by 3.

b) List the set of whole numbers less than 20 that are divisible by 5.

c) List the set of whole numbers less than 20 that are divisible by 3 and 5.

d) List the set of whole numbers less than 20 that are divisible by 3 or 5.

Class Ex. #1 Consider the following two sets.

A = whole numbers less than 20 that are divisible by 3B = whole numbers less than 20 that are divisible by 5

The intersection of the two sets is the set A and B (sometimes written A « B). It is the set of elements which are members of both sets A and B.

a) List the intersection of the sets. A and B =

The union of two sets is the set A or B (sometimes written A » B). It is the set of elements which are members of set A or set B, or both.

b) List the union of the sets. A or B =

Venn Diagrams

3. 6.9.

12. 18.

15.5.

1.

10.

4.

2.

7.

11.

16.13.

14.

17.

19.

A B

8.

Relationships between sets can be shown in a Venn diagram. The Venn diagram to the right represents the information from Class Ex. #1.

The elements of each set are marked with a dot to distinguish them from the number of elements in each set (see Class Ex. #5).

Class Ex. #2 In each of the following Venn diagrams, shade the region representing the given set.a) The intersection of the two sets , i.e. the set A and B.

A B

b) The union of the two sets , i.e. the set A or B.

A B

c) The complement of the set A , i.e. the set not A.

A B

Class Ex. #3 Use the Venn diagram to list the members of the following sets.

2.9.8.

10.7.

29.

11.

16.

P Q

a) P =

b) Q =

c) P and Q =

d) P or Q =

e) not P =

f) not Q =

330 Mathematical Reasoning Lesson #4: Venn Diagrams - Part One

Class Ex. #4 Use the information in Class Ex. #3 to list the following sets

a) not P and not Q =

b) not P or not Q =

c) not (P or Q) =

d) not (P and Q) =

e) Which two pairs of sets are identical?

Note Venn diagrams can also be used to represent the number of members in each set and not the individual elements as the next class example illustrates.

Class Ex. #5 The diagram displays the number of students who are

712

S Y

3

members of Students’ Council (S) and the number of students who are on the Yearbook Committee (Y)

How many students are:

a) on Students’ Council?

b) on both Students’ Council and Yearbook Committee?

c) on the Yearbook Committee but not on Students’ Council?

d) on Students’ Council or Yearbook Committee?

Class Ex. #6 In a homeroom of 20 students, 15 take Math, 12 take Social, and 10 take Math and Social. Show this information in a Venn Diagram. How many students take neither Math nor Social?


Mathematical Reasoning Lesson #4: Venn Diagrams - Part One 331

Assignment

1. Consider the Venn diagram shownM N

p.

q.t.

s.

r.v.

u.x.

w

List the elements of the following sets:

a) M =

b) M and N

c) M or N

d) not M

e) not N

f) not (M and N)

2. Using the Venn diagram in question #1, which sets are represented by:

a) r, s, t, v b) p, q c) u, w, x

3. Consider the set of prime numbers less than 20. A BLet A = 3, 5, 7, 11, 19 and B = 2, 3, 7, 13.

a) Complete the Venn diagram to illustrate this information.

b) List the members of the following sets:

i ) A and B i i ) A or B

i i i) not A iv) not(A or B)

4. The diagram displays the number of girls who are

915

S V

2

members of the school soccer team (S) and the school volleyball team (V).

How many girls are:

a) on both teams?

b) on the soccer team and not on the volleyball team?

c) on only one team?

d) on the soccer team or on the volleyball team?


5. All the students in a class of 35 take Physics or Chemistry or both. 29 take Chemistry, and 15 take Physics. How many take both?

6. Of the students in grade 11 at a certain high school, 76 are enrolled in physical education, 24 are enrolled in music, and 10 are enrolled in both physical education and music. If there are 15 students in grade 11 who are not enrolled in physical education or music, how many students are in grade 11?

7. In a school survey it was found that 140 students had a cell phone or a personal compact disc player. If 86 students had a cell phone and 70 students had a personal compact disc player, how many students had both?

8. The Venn diagram shows the number of students who did English homework, Math homework, or Social homework on the weekend. None of the students did homework for any other subject. How many students did:

English Math

Social

11 7 21

14

a) English homework

b) only English homework

c) English homework and Social homework

d) English homework or Math homework

e) Math homework and not Social homework

f) Homework for only one subject.

Mathematical Reasoning Lesson #4: Venn Diagrams - Part One 333

9. If P = quadrilaterals which have 4 equal sides and Q = quadrilaterals which have 4 MultipleChoice equal angles, then a trapezoid is a member of which of the following sets?

A. P and QB. P and not QC. not P and QD. not P and not Q

10. In a survey of 400 households, 285 had 2 televisions, and 320 had a microwave oven. Numerical Response If 63 households did not have 2 televisions or a microwave oven, the percentage of

households in the survey which had both 2 televisions and a microwave oven, to the nearest tenth, is _____ .


Answer Key 1 . a) p, q, t b) t c ) p, q, r, s, t, v

d) r, s, u, v, w, x e ) p, q, u, w, x f ) p, q, r, s,u, v, w, x

2 . a) N b) M and not N c ) not M and not N or not (M or N)

3 . a)A B

5.11.

19.

3.7.

2.13.

17.

b) i ) 3 and 7 i i ) 2, 3, 5, 7, 11, 13, 19 i i i ) 2, 13, 17 i v ) 17

4 . a) 2 b) 15 c ) 24 d) 26

5 . 9 6 . 105 7 . 16

8 . a) 18 b) 11 c ) 0 d) 39 e ) 28 f ) 46

9 . D 1 0 . 6 7 . 0


Mathematical Reasoning Lesson #5:Venn Diagrams - Part Two

Class Ex. #1 The Venn diagram displays the results of a survey of 100

6

P

C D

3

11

68 2 17

2

students in a high school regarding technology in their personal lives.

P represents the number of students with a cellular phone C represents the number of students with a computer

D represents the number of students with a CD player

a) How many students own only a cellular phone?

b) How many students own a cellular phone?

c) How many students own a computer and a personal CD player?

d) How many students own a computer or a personal CD player?

e) How many students own all three items?

f) What does the number 2 outside the three circles represent?

g) What does the number 2 inside the circles represent?

Class Ex. #2 In each of the following Venn diagrams, shade the region indicated.

a) A and B b) A or BA

B

C

A

B

C

c) A and not B d) A and not B and not C

A

B

C

A

B

C

Class Ex. #3 In Big Hill High School, 185 grade eleven students were surveyed to determine which soft drinks they liked to drink. 115 drank coke, 92 drank root beer, 100 drank orange, 43 drank coke and orange, 52 drank root beer and coke, 57 drank root beer and orange, and 25 drank all three. Show this information in a Venn Diagram and answer the following questions.

Venn Diagram

How many students:

a) drank only coke? b) drank coke or root beer?

c) did not like to drink any of the three drinks?

Class Ex. #4 150 grade 11 students were asked which of the following 3 television programs they watch regularly - “Friends”, “Survivor”, and “Crocodile Hunter”. 102 students watched “Friends”, 70 watched “Survivor” and 40 watched “Crocodile Hunter”. 25 students watched both “Friends” and “Survivor”, 27 watched “Friends” and “Crocodile Hunter”, and 30 watched “Survivor” and “Crocodile Hunter”.

Let x = the number of students who watch all three programs.

Complete each section of the Venn diagram in

x

Friends

Survivor Crocodile Hunter

terms of x starting from the inside out and hence form an equation in x which can be solved to give the number of students who watched all three programs.


336 Mathematical Reasoning Lesson #5: Venn Diagrams - Part Two

Assignment

1. Use the Venn diagram to list the elements of the following sets:

8.

A

B C

5.

9.2.

1. 7. 6.

3.

10.

4.

a) A

b) A and B

c) A and B and C

d) B or C

e) A or B or C

f) not C

g) A and not B

h) C and not A and not B

2. In each of the following Venn diagrams, shade the region indicated.

a) R or T b) R and not S

R S

T

R S

T

c) R and S and not T d) not R and not S and not T

R S

T

R S

T

Mathematical Reasoning Lesson #5: Venn Diagrams - Part Two 337

3. The partially completed Venn diagram displays the

5

P

B W810

35

results of a 145 teenagers regarding their favorite fast food.

• P = 75 represents the number of teenagers who liked pizza.

• B = 60 represents the number of teenagers who liked burgers.

• W = 68 represents the number of teenagers who liked wraps.

a) Complete the Venn Diagram.

b) How many teenagers liked:

i ) pizza and burgers and wraps? i i ) burgers and not wraps?

i i i)burgers and pizza? iv) only burgers and pizza?

v) burgers or pizza? vi) none of the three types of fast food?

4. The students from Mr. Hennesey’s grade 11 class were surveyed. 19 students take Math, 10 students take Math and Physics, 14 students take only Chemistry, 12 students take Chemistry and Math, and 3 students take all three subjects. 2 students do not take any of these subjects and all Physics students take Math.

a) Show this information in a Venn diagram.

b) How many students:

i) are there in Mr. Hennesey’s homeroom? i i ) take only Math?

i i i)take Chemistry and not Math? iv) take Chemistry or Physics?


5. All the students in grade 11 were surveyed about their participation in the following extra-curricular activities - sports (S), music (M), or drama (D).

170 students participated in sports of whom 139 participated only in sports. 52 students participated in drama, of whom 18 participated in only drama. 48 students participated in music, of whom 12 participated only in music. 5 students participated in all three activities and 20 students did not participate in any of the three activities.

a) Draw a Venn diagram and illustrate as much of the above information as you can.

b) In order to complete the Venn diagram, Kalon represented the number of students who participated in only sports and drama by x, the number of students who participated in only sports and music by y, and the number of students who participated in only drama and music by z.

Form three equations in the variables and solve the system to determine the values of x, y, and z.

c) How many students are in grade 11?


6. To cater for a school party, all of the 115 students involved brought at least one of the following items: sandwiches (S), chips (C), or lemonade (L). 54 brought sandwiches, 70 brought lemonade, 19 brought chips and lemonade only, 22 brought sandwiches and lemonade only, 24 brought lemonade only, and 15 brought sandwiches only. How many students brought only chips?


The Venn diagram illustrates the number of girls who played on various school sports teams.

Volleyball Soccer

Basketball

Field Hockey

10 111

22 3

2

3

15

7. The number of girls who played soccer and volleyball was MultipleChoice

A. 1B. 3C. 8D. 29

8. The number of girls who played on exactly two teams was

A. 6B. 8C. 9D. 11


Use the following information to answer questions #9 and #10 Of the 21 teachers in a high school who teach Biology, Chemistry, or Physics, or some combination of these, no one teaches both Biology and Physics. 8 teach Biology, of whom 5 do not also teach Chemistry. 7 teach Physics.

9. The number of teachers who teach Chemistry or Physics is _____ .Numerical Response (Record your answer in the numerical response box from left to right)

10. If the number of teachers who teach either Biology or Chemistry is the same as the number of teachers who teach either Chemistry or Physics, then the number of teachers who teach Chemistry is _____ .(Record your answer in the numerical response box from left to right)

Answer Key 1 . a) 2, 3, 5, 8, 9 b) 2, 8 c ) 8 d) 1, 2, 3, 6, 7, 8, 9, 10

e ) 1, 2, 3, 5, 6, 7, 8, 9, 10 f ) 1, 2, 4, 5 g ) 3, 5, 9 h) 6, 10

2 .

a) b) d)c)

3 . a)

5

P

B W810

35

52

10

5

20

b) i ) 5 i i ) 20 i i i ) 15i v ) 10 v ) 120 vi) 5


4 . a)

M P

C

0 7

30

0

14

9

2

b) i ) 35 i i ) 0 i i i ) 14 i v ) 33

5 . a)

S M

D

139 y

5z

12

18

x

20

b) x = 12, y = 14, z = 17

c ) 237

6 . 18 7 . B 8 . C

9 . 1 6 1 0 . 1 1


Mathematical Reasoning Lesson #6:If/Then Statements

Note To be eligible to vote in Canada, a person must meet the following criteria:• a Canadain citizen • at least 18 years of age • name must be on list of electors

Assume that for this lesson when the phrase “Canadian citizen 18 years of age or older” is used that the person’s name is on the list of electors.

Conditional Statement

A conditional statement is a statement which is written using “if” and “then”. A conditional statement has two parts to it:

• a hypothesis (the part following “if”), and, • a conclusion (the part following “then”).

A conditional statement, sometimes called an “if - then” proposition, may be true or false.

eg. If a Canadian citizen is 18 years of age or older, then the person is able to vote.eg. If a person lives in Calgary, then the person must have been born in Alberta.

Converse

A conditional statement has a converse which may or may not be true. This occurs when the hypothesis and the conclusion are interchanged.

eg. If a Canadian citizen is able to vote, then the person is 18 years of age or older.eg. If a person was born in Alberta, then the person must live in Calgary.

Class Ex. #1 Write the following statement as a conditional statement and identify the hypothesis and the conclusion.

“An acute angled triangle has three angles which are each less than 90°.”

Class Ex. #2 Write the converse of the conditional statement in Class Ex #1. Is the converse true?

Class Ex. #3 Write down a true conditional statement for which the converse is not true.

Biconditional Statement

A biconditional statement is a statement in which the conditional statement and its converse are true.

When this occurs the conditional statement and its converse can be combined in an “if and only if” statement.

eg. A Canadian citizen is allowed to vote if and only if the person is 18 years of age or older.

Class Ex. #4 A conditional statement and its converse are given.

“If a triangle is obtuse angled, then the triangle has one angle between 90° and 180°.”

“If a triangle has one angle between 90° and 180°, then the triangle is obtuse angled.”

If both statements are true, write a biconditional statement.

Contrapositive Statement

The contrapositive of a conditional statement is formed by taking the converse of the conditional statement and negating both the hypothesis and the conclusion of the converse.

eg.

Conditional StatementIf a Canadian citizen is 18 years of age or older, then the person is able to vote.

ConverseIf a Canadian citizen is able to vote, then the person is 18 years of age or older.

ContrapositiveIf a Canadian citizen is not able to vote, then the person is not 18 years of age or older.

Class Ex. #5 Write the contrapositive of the following conditional statement.

“If a triangle is obtuse angled, then the triangle has one angle between 90° and 180°.”

Converse

Contrapositive

344 Mathematical Reasoning Lesson #6: If / Then Statements

Class Ex. #6 Consider the following “if - then” proposition:

“If z > 2, then z2 > 4.”

a) Is the original proposition true? If not, give a counterexample.

b) Write the converse. Is it true? If not, give a counterexample.

c) Write the contrapositive. Is it true? If not, give a counterexample.

Class Ex. #7 Consider the following conditional statement:

“If Peter lives in Alberta, then Peter lives in Red Deer.”

a) Is the conditional statement true?

b) Write the converse. Is it true?

c) Write the contrapositive. Is it true?

Note Class Ex. #6 and #7 are examples of the following general rules.

A conditional statement and its contrapositive are equivalent statements. This means that:

• If the conditional statement is true, then so is the contrapositive. • If the conditional statement is false, then so is the contrapositive.


Mathematical Reasoning Lesson #6: If / Then Statements 345

Assignment

1. Write each statement as a conditional statement.

a) A quadrilateral which is a square has diagonals which bisect each other.

b) Students who attend high school in Kelowna attend high school in British Columbia.

c) A polygon with 6 sides is a hexagon.

d) A composite number has only one factor.

e) Triangles with three equal angles are congruent to each other.

f) A quadratic function has two distinct zeros.

g) A cubic function is a polynomial function of degree 3.

h) The highest mountain in the world is in Nepal.

2. Which of the conditional statements in question #1 are true and which are false?

3. Write the converse of each of the conditional statements in question #1.


4. Which of the converse statements in question #3 are true and which are false?

5. Write the contrapositive of each of the conditional statements in question #1.

6. Which of the contrapositive statements in question #5 are true and which are false?

7. Comment on any relationships you observe between the answers to questions #2, #4, and #6.

8. State the conditions for a statement to be biconditional. Where appropriate, write the statements in question #1 as biconditional statements.


9. Consider the following conditional statement:

“If x > 3, then x2 > 3 ”

a) Is the conditional statement true? If not, give a counterexample.



10. Consider the following “if - then” proposition:

“If –3x < –6, then x < 2 ”

a) Is the proposition true? If not, give a counterexample.



11. Give an example of the following:

a) an “if - then” statement where the converse is true.

b) an “if - then” statement where the converse is false.

c) an “if - then” statement where the contrapositive is true.

d) an “if - then” statement where the contrapositive is false.


12. Consider the following conditional statement:MultipleChoice

“If the cube root of a number is negative, then the number is negative.”

The statement which is true is

A. only the conditional statementB. only the contrapositive of the statementC. only the converse of the statementD. all of the above

13. Consider the following conditional statement:

“If a number is a multiple of 3, then it is divisible by 6.”

The statement which is true is

A. the conditional statementB. the contrapositive of the statementC. the converse of the statementD. none of the above


Answer Key 1 . a) If a quadrilateral is a square, then the diagonals of the quadrilateral bisect each other.

b) If students attend high school in Kelowna, then they attend high school in British Columbia.c ) If a polygon has six sides, then it is a hexagon.d) If a number is composite, then it has only one factor.e ) If triangles have three equal angles, then they are congruent to each other.f ) If a function is a quadratic function, then it has two distinct zeros.g ) If a function is a cubic function, then it is a polynomial function of degree three.h) If a mountain is the highest in the world, then it is in Nepal.

2 . True - a), b), c), g), h) False - d). e), f)

3 . a) If a quadrilateral has diagonals which bisect each other, then it is a square.b) If students attend high school in British Columbia, then they attend high school in Kelowna.c ) If a polygon is a hexagon, then it has six sides.d) If a number has only one factor, then the number is composite.e ) If triangles are congruent to each other, then they have three equal angles.f ) If a function has two distinct zeros, then the function is a quadratic function.g ) If a polynomial function is degree three, then it is a cubic function.h) If a mountain is Nepal, then it is the highest mountain in the world.

4 . True - c), g) False - a). b), d), e), f), h)

5 . a) If a quadrilateral has diagonals which do not bisect each other, then it is not a square.b) If students do not attend high school in British Columbia, then they do not attend

high school in Kelowna.c ) If a polygon is not a hexagon, then it does not have six sides.d) If a number does not have only one factor, then the number is not composite.e ) If triangles are not congruent to each other, then they do not have three equal angles.f ) If a function does not have two distinct zeros, then the function is not a quadratic function.g ) If a polynomial function does not have degree three, then it is not a cubic function.h) If a mountain is not in Nepal, then it is not the highest mountain in the world.

6 . True - a), b), c), g), h) False - d). e), f)

7 . The answers to question #2 and question #6 are identical, i.e., if a conditional statement is true, then so is the contrapositive. If a conditional statement is false, then so is the contrapositive.

8 . A statement is biconditional if both the statement and the converse is true.c) A polygon is a hexagon if and only if it has six sides.g)A polynomial function is a cubic function if and only if it has degree three.

9 . a) True b) If x2 > 3, then x > 3 - False, x = 2 is a counter example.c ) If x2 > 3, then x > 3 - True

1 0 . a) False - x = 3 is a counterexampleb) If x < 2, then –3x < –6, False - x = 1 is a counterexample.c ) If x < 2, then –3x < –6, False - x = 3 is a counterexample.

1 1 . a) If a number is divisible by 3, then it is divisible by 6. Answers may vary.b) If a number is divisible by 6, then it is divisible by 3. Answers may vary.c ) If a number is divisible by 6, then it is divisible by 3. Answers may vary.d) If a number is divisible by 3, then it is divisible by 6. Answers may vary.

1 2 . D 1 3 . C


Mathematical Reasoning Lesson #7:Direct Proof

Direct Proof

Direct proof, or direct reasoning, begins with given information and uses deductive reasoning to reach a conclusion.

Sometimes we start a direct proof by using a known theorem as in the next example.

Class Ex. #1 In the diagram, –PQR = 48° and –PRS = 114°. S

R

Q P48°

114°

Use direct reasoning to prove that DPQR is an isosceles triangle. The method using a two column approach has been started below.

Statement Reason

angle sum of a triangle is 180°

–PRQ = _____°

Class Ex. #2 Gail was asked to prove the following P

Q R

“In DPQR prove that –P + –Q + –R = 180°.”

She has started the proof using a two column approach. Complete her proof.

Gail’s SolutionP

Q R

S T

–1

–2 –3

–4 –5Draw a triangle PQR and a line segment ST through P parallel to QR. Give each angle a number as indicated in the diagram.

To prove: –1 + –2 + –3 = 180°

Statement Reason

–2 = –4

–3 = –5

straight line

Class Ex. #3 Complete the two column proof to show that the E F

H G

D

diagonals of a parallelogram bisect one another.

Statement Reason

EH = FG

–HED = –FGD

DHED is congruentto DFGD

Corresponding sidesof congruent triangles are equal

i)

ii)

iv)

iii)

v)

vi)


Assignment

1. Complete the proof below to show that when two lines

p

qr s

intersect the opposite angles are equal, i.e. in the diagram prove that –r = –s.

Statement Reason

–p + –s = 180°

352 Mathematical Reasoning Lesson #7: Direct Proof

2. Complete the proof to show the following:

x

y z

t“ Given that the diagonals of a square intersect at right angles, prove that one diagonal bisects the other.

By using the information in the diagram, prove that t = z.

Statement Reason

each side = x units

Pythagorean Theorem

3. Triangle ABC is an isoceles triangle with AC =BC.

A

B

C

D

EIf –CAE = –CBD prove that –EAB = –ABD

Mathematical Reasoning Lesson #7: Direct Proof 353

4. In the diagram, DOPS and DOQR are both isosceles triangles.

P

O

Q

R

SProve that –POQ = –SOR.

354 Mathematical Reasoning Lesson #7: Direct Proof

Circle Geometry Lesson #1:Circles and Chords

Warm-Up #1 Basic Terminology Review

Equidistant - an equal distance from.

Perpendicular Lines - intersect at right angles.

Bisect - divide into two equal parts.

Perpendicular Bisector - a line which bisects another line at right angles.

Warm-Up #2 Circle Terminology

Circle - a circle is defined as the set of all points equidistant from a fixed point.

Chord - a line segment which joins two points on the circumference of a circle.

Segment - a chord which divides a circle into two segments.

Diameter - a chord which passes through the centre of a circle.

Radius - a line segment which joins the centre of a circle to any point on the circumference of the circle (the radius is half the length of the diameter).

Arc - part of the circumference of a circle.

Sector - a region of a circle bounded by two radii and an arc.

Class Ex. #1 In the diagram C is the centre of each circle and different parts of the circle are labelled.

Figure 1 Figure 2 Figure 3

CY

Z

C

S

R

C

B

A

D

E

a) List the names of:i ) a diameter i i ) four radii iii) two chords

iv) a major arc v) a minor arc

b) Shade the following regions with the given pattern:i ) a major segment i i ) a minor segment iii) a major sector iv) a minor sector

Properties of Chords on a Circle

The perpendicular bisector of a chord passes through the centre of a circle.

Extensions to the above property:

• If a line is drawn from the centre of a circle and is perpendicular to a chord, then the line bisects the chord.

• The line drawn from the centre of a circle to the midpoint of a chord is perpendicular to the chord.

Note The following Warm-Up can be done by:• using a software program such as Geometer’s Sketchpad,

or • using the diagrams below with a compass set.

Confirming Properties of Chords on a CircleWarm-Up #3

Confirm the following properties of chords on a circle:a) The perpendicular bisector of a chord passes through the centre of a circle.

C C

b) If a line is drawn from the centre of a circle and is perpendicular to a chord, then the line bisects the chord.

C C

c) The line drawn from the centre of a circle to the midpoint of a chord is perpendicular to the chord.

C C

356 Circle Geometry Lesson #1: Circles and Chords

The Perpendicular Bisector Theorem

Any point on a perpendicular bisector of a line segment is equidistant from the endpoints of the segment.

Class Ex. #2 Use the method below to prove the Perpendicular Bisector Theorem.

DB

E

A

Solution:Given line segment BD with midpoint A. Let E be any point on the perpendicular bisector of chord BD.Prove DEAB and DEAD are congruent and show that EB = ED.

The converse of the Perpendicular Bisector Theorem (which is also true) states the following:Any point equidistant from the endpoints of a line segment lies on the perpendicular bisector of the segment.

Class Ex. #3 Use the converse of the Perpendicular Bisector Theorem to prove that

C

DB A

the perpendicular bisector of chord BD passes through the centre C of the circle.

Solution:

Circle Chord Properties

The following circle chord properties can be used in problem solving.

• The perpendicular bisector of a chord passes through the centre of a circle.• If a line is drawn from the centre of a circle and is perpendicular to a chord, then the line

bisects the chord.• The line drawn from the centre of a circle to the midpoint of a chord is perpendicular to

the chord.• Two chords are equidistant from the centre of a circle if and only if the chords are of

equal length.

Circle Geometry Lesson #1: Circles and Chords 357

Class Ex. #4 Find the value of x in each of the following diagrams.

C

8

10

a)

x

C

b)

Cx

8 30

c)

x°

Class Ex. #5 Use congruent triangles and the properties of chords on a circle to

C

x 8

10 determine the value of x to one decimal place. C is the centre of the circle.

Class Ex. #6 The diagram show the cross section of a horizontal cylindrical pipe. Water is lying in the pipe and is 20 cm deep at the middle of the section. If the width of the water surface is 80 cm, calculate the diameter of the pipe.



Assignment

1. State the circle chord property which can be established from each diagram. O is the centre of the circle in each case.

O

B

Aa)

O

B

Ab)

O

B

A

D

Cc)

2. Explain whether each statement is true or false. O is the centre.

O

B

A

C

D

O

xy

a) b)

AD = DB x = y


3. In the following diagrams, O is the centre of the circle. Find the values of x and y rounding to the nearest tenth where necessary.

O y

x

12

13a) b)

O

9

x12

Ox 3

7

10

c)

O

x

15

d)

O

64

e)

xO15

17

f) x

y

g)

O O6

x

x

14

8

h)


4. In order to hang a circular mirror with a radius of 8 3 cm, a triangular bracket is glued to the back of it.

Calculate the exact value of the area of the triangular bracket if each edge of the triangular bracket is 24 cm in length.

5. A circular water pipe has a diameter of 50 cm. If the width of the water surface in the pipe is 36 cm, find the maximum depth of the water to the nearest tenth of a cm. Explain why there are two possible answers.

6. PQ and RS are two parallel horizontal chords 6 cm and 8 cm long in the upper half of a MultipleChoice circle. If the radius of the circle is 5 cm, the distance in cm between the chords is

A. 1B. 2C. 3D. 4


7. In a circle chord AB is twice the length of chord CD. The distance of AB from the centre of the circle

A. is twice the distance of CD from the centreB. is half the distance of CD from the centreC. is the same as the distance of CD from the centreD. cannot be found from the given information.

8. A chord 7.6 cm long lies in a circle with radius 4.8 cm. The distance, to the nearest tenth of Numerical Response a cm, between the chord and the centre of the circle, is _____ .


9. A circular poker table is supported by a wooden square bracing inscribed in the circle. If the diameter of the table is 2.4 m, then the length of each side of the square bracing, to the nearest tenth of a metre, is _____ .


10. The circular base, of a large hemispherical dome is shown.

60 m

16 m

The diameter is 60 m and the chords shown are equal in length and form the support beams of the dome.

To the nearest tenth of a metre, the two support beams are _____ metres apart.



11. A square is inscribed in a circle of diameter 30 cm with centre C.

C

Four lines are drawn between the midpoints of the adjacent sides of the square which form a smaller square. The perimeter of the smaller square, to the nearest tenth of a cm, is _____ .


Answer Key 1 . a) A line drawn from the centre of the circle, perpendicular to a chord, bisects the chord.

b) The line drawn from the centre of a circle to the midpoint of a chord is perpendicular to the chord.c ) Two chords equidistant from the centre of a circle are of equal length.

2 . a) False - because the perpendicular bisector of the chord AB should pass through the centre of the circleb) True - because chords equidistant from the centre of the circle are of equal length and the line drawn

from the centre of the circle, perpendicular to a chord, bisects the chord.

3 . a) x = 12, y = 5 b) x = 6.7 c ) x = 5.7 d) x = 7.5 e ) x = 5 f ) x = 8, y = 33.0 g ) x = 3.9 h) x = 2.5

4 . 144 3 cm2.

5 . 7.7 cm or 42.3 cm.There are two answers because the water level could be above or below the middle of the pipe

6 . A 7 . D

8 . 2 . 9 9 . 1 . 7

1 0 . 5 7 . 8 1 1 . 6 0 . 0


Circle Geometry Lesson #2:Circles and Angles

Warm-Up #1 Basic Terminology Review

Supplementary Angles - angles which add to 180°

Obtuse Angle - an angle between 90° and 180°

Reflex Angle - an angle between 180° and 360°


Central Angle - An angle with its vertex at the centre of a circle and with two radii forming the arms.

Inscribed Angle - An angle with its vertex on the circumference of a circle and with two chords forming the arms.

Subtended - enclosed by or surrounded by.

Note Recall that a chord divides a circle into two arcs. The smaller arc is a minor arc and the larger arc is a major arc. Unless otherwise stated, any arc referred to will be a minor arc.

Class Ex. #1 In the following diagrams C is the centre of the circle. Fill in the blanks.

C

BA

D

a) –ADB is an _______________ angle _______________ by

the _______________ AB.

C

BA

b) –ACB is a _______________ angle _______________ by

the _______________ AB.

C

BA

c) Reflex –ACB is a _______________ angle _______________ by

the _______________ AB.

Class Ex. #2 a) Name two angles on the circumference subtended by:

O

A

B C

D

E

i ) arc BC i i ) arc DE iii) arc AED

b) Name a chord which subtends the following angles:

i ) –CEA i i ) –BED

c) Which arc subtends the following inscribed angles?

i ) –CED i i ) –ACE iii) –AEC iv) –AED

Class Ex. #3 O is the centre of the circle.

OA

B

C

EDa) Name the central angle subtended by:

i ) arc BC i i ) arc EB iii) arc EC

iv) arc CD v) major arc CD

b) Which arc subtends:

i ) –EOC ? i i ) –CBA ? iii) –BDE ?


366 Circle Geometry Lesson #2: Circles and Angles

Note The following Explorations can be done by:• using a software program such as Geometer’s Sketchpad,

or • by using the diagrams below with a protractor.

Properties of Angles in a Circle

Exploration 1

a) Use a protractor to confirm the following property of angles in a circle. O is the centre of each circle:

The measure of the central angle is equal to twice the measure of the inscribed angle subtended by the same arc.

i) ii) iii)

O

AC

D

O

A

C

D

O

A

C

D

b) Verify the above property by calculating the measure of –AOB and comparing it with the measure of –ACB.

O

BA

C

D

30° 25°

c) Prove the above property in the general case below.

O

BA

C

D

x° y°

Circle Geometry Lesson #2: Circles and Angles 367

Exploration 2

a) In the diagrams below, where O is the centre, name three angles subtended by the arc, or chord CA.

O

A

C

DR

S

i)

O

A

C

DR

S

ii)

b) Use a protractor to confirm the following property of angles in a circle.

Inscribed angles subtended by the same arc (or chord) are congruent.

c) Use the property in exploration 1 to prove the above property for the general case below where –COA = x°.

O

AC

RS

y°x° z°


Exploration 3

a) Use a protractor to confirm the following property of angles in a circle. O is the centre of each circle:

The angle inscribed in a semi circle is a right angle.

O

A

B C

D

E

O

A

B

C

Di) ii)

b) Use the property in exploration 1 to prove an angle subtended by the diameter on a circle is a right angle.

O

A

B C

Properties of Angles in a Circle

• The measure of the central angle is equal to twice the measure of the inscribed angle subtended by the same arc.

• Inscribed angles subtended by the same arc (or chord) are congruent.

• The angle inscribed in a semi circle is a right angle.


Class Ex. #4 Find the missing values in each. O is the centre.

S

O

E

F

G

130°

f°

g°

a)

S

O

E

FG

120°

x° + 60°

x°

b)

OR

T U

c)

150°45°t°

r°

r°

Class Ex. #5 In the diagram O is the centre of the circle and angle PSQ = 48°.

OR

P

S

QDetermine the measure of angle QOR.


Assignment

Unless otherwise stated, O is the centre of the circles for all the questions in this assignment.

1. Consider the following diagram.

OB

C

D

E

Aa) Name two angles at the circumference subtended by:

i ) arc ED i i ) arc CB iii) arc ABC

b) Name a chord which subtends the following angles.

i ) –DBA i i ) –EBC

c) Which arc subtends the following inscribed angles?

i ) –DBC i i ) –ADB iii) –ABD iv) –ABC


2. Consider the following diagram.

O

A

E

D

B

C

a) Name the central angle subtended by:

i ) arc ED i i ) arc EB iii) arc BD

iv) arc CD v) major arc CD vi) major arc BD

b) Which arc subtends:

i ) –BOD ? i i ) –BEA ? iii) –BDE ?

3. For each of the following diagrams• determine which of the statements are true,• describe the property of inscribed angles for those statements which are true, and, • for those statements which are false, rewrite the given statement so that it is true.

O

AC

D

O

ACD

a) b) c)

O

AC

D

B

–COA = 2–CDA –COA = –CDA –CBA = 12 –CDA

OA

CD

E

Bd)

O

A

C DB

E

e)

–ABD = –ACD = –AED –ABD = –ACD = –AED


4. Find the value of the missing variables.

O

AC

D

O

AC

D

B

O AC

D

a) b) c)

100°

x°

40°

x°

83° y°

x°

5. Find the value of the variables. The centre of each circle is marked with a solid point.

38°

x° 32°

x°

y°

z°

a) c)b)

275°x°

68°x°

y°z°

u°v°

w°

f)

x° y°z°

d)75°

310°x°

y°

e)

x°

y°

z°

40°

x° + 20°

y°z°

110°

g) h)

y°z°

v° + 10°

w°

74°

i)


6. O is the centre of the circle shown.MultipleChoice If the angles GOI and GHI differ by 32°, then the size of GHI is

O

I

G

H

A. 16°B. 32°

C. 64°

D. 67°

7. The solid point is the centre of the circle shown. Of the following relationships, the one which is false is

y° z°

v° w°

x°

A. v = w

B. v = 1

2 x

C. v + w = x

D. y = z = 2v

8. In the diagram O is the centre of the circle.

A

B

C

D

O

112°

If –AOC is 112°, then –ABC is

A. 90°B. 124°

C. 137°

D. 248°

9. The sketch shows a tunnel with a steel frame

A

B C

D62°

bolted underneath it to give it more support.A, B, C, and D represent the points at which the frame is bolted. The arch of the tunnel of the bridge is a semi-circle, with AD as the diameter.The size of –ACB is

A. 28°B. 31°

C. 34°

D. 42°

10. In the diagram PR = QR. If –QOS = x°, then –PRQ is equal to

T

O

SQ

P R

A. 2x°

B. 180° - 12 x°

C. 180° - x°D. 180° - 2x°


Use the following information to answer numerical response questions #11 - #15.

O

DC

B

A

E

For the diagram:• –BAD = 65°

• OB intersects the midpoint of chord AC

• AD and BE are diameters.

11. The measure of –ADB, to the nearest degree, is _____ .Numerical Response (Record your answer in the numerical response box from left to right)

12. The measure of –BOC, to the nearest degree, is _____ .(Record your answer in the numerical response box from left to right)

13. The measure of –BOD, to the nearest degree, is _____ .(Record your answer in the numerical response box from left to right)

14. The measure of –CED, to the nearest degree, is _____ .(Record your answer in the numerical response box from left to right)

15. The measure of –ACE, to the nearest degree, is _____ .(Record your answer in the numerical response box from left to right)


16. The points L, M, and N are on the circumference of the circle with centre C. The measure of –LMN = 55° and –MCN = 86°. The measure of –MCL, to the nearest degree, is _____ .(Record your answer in the numerical response box from left to right)

17. Isosceles triangle ABC, with AB = AC, is inscribed in a circle centre O. If angle OBC = 25°, then the measure of angle ACO, to the nearest tenth of a degree, is _____ .



Answer Key 1 . a) i ) –EAD, –EBD, –ECD i i ) –CDB, –CEB i i i ) –AEC, –ADC

b) i ) DA i i ) ECc ) i ) DC i i ) AB i i i ) AD i v ) major arc AC

2 . a) i ) –EOD i i ) –EOB i i i ) –BOD i v ) –COD v ) reflex –COD v i ) reflex –BOD b) i ) BD i i ) BA i i i ) BE

3 . a) True - The measure of the central angle is equal to twice the measure of the inscribed angle subtended by the same arc.

b) False - –COA = 2–CDA c ) False - –CBA = –CDAd) True - The angle inscribed in the semi circle is a right angle.e ) False - –ABD = –ACD π –AED

4 . a) x = 50° b) x = 57°, y = 40° c ) x = 90°

5 . a) x = 76° b) x = 42.5° c ) x = 32°, y = 64°, z = 32° d) x = 90°, y = 90°, z = 210°e ) x = y = 25° f ) u = 34°, v = 56°, w = 34°, x = 34°, y = 56°, z = 56°g ) x = 40°, y = 80°, z = 90° h) x = 35°, y = 70°, z = 55° i ) v = 27°, w = 37°, y = 106°, z = 53°

6 . B 7 . D 8 . B 9 . A 1 0 . C

1 1 . 2 5 1 2 . 5 0 1 3 . 1 3 0

1 4 . 4 0 1 5 . 6 5 1 6 . 1 6 4

1 7 . 3 2 . 5


Circle Geometry Lesson #3:Cyclic Quadrilaterals


Concyclic - Points which lie on the circumference of the same circle.

Cyclic Quadrilateral - A quadrilateral whose vertices are concyclic.

Note Recall that:

• a straight angle has a measure of 180°.

• the sum of the angles in a quadrilateral is 360°.

• two angles whose sum is 180° are called supplementary angles.

Class Ex. #1 Draw a diagram in which four points are concyclic.

Class Ex. #2 The following question was asked on a circle geometry unit test

B

O

A

C

D“ Is the four sided figure shown a cyclic quadrilateral? Explain.”

Kevin answered that it was a cyclic quadrilateral because the diagram shows a four sided figure contained within a circle.

Explain why the teacher marked Kevin’s answer wrong.

Class Ex. #3 O is the centre of the circle and BD is the diameter.

O

A

B

C

ED

a) Name three cyclic quadrilaterals.

b) Name two quadrilaterals which are NOT cyclic quadrilaterals.


Note The following Warm-Up can be done:

• by using a software program such as Geometer’s Sketchpad, or

• by using the diagrams below with a compass set.

Exploration 1 Exploring the Sum of the Opposite Angles of a Cyclic Quadrilateral

a) Use a protractor to measure the opposite angles of the following cyclic quadrilaterals.

C

OD

BA

C

O

D

B

A

i)

C

ODB

Aii) iii)


The sum of the opposite angles of a cyclic quadrilateral is _____ .

c) In order to prove the result in the general case, complete the

S

O

R

Q

P

x°

y°

following for the given diagram.

i ) Express in terms of x and y.

–QOS = _____ reflex –QOS = _____

i i ) The sum of –QOS and reflex –QOS in degrees is _____ .

i i i) The sum of –QOS and reflex –QOS in terms of x and y is __________ .

iv) The sum of x and y is _____ .

d) Write the converse of the statement in b).

e) Given that both the statement in b) and the converse are true, write a biconditional statement.

378 Circle Geometry Lesson #3: Cyclic Quadrilaterals

Exploration 2 Comparing the Exterior Angle to the Interior Opposite Angle

a) Use a protractor to measure the given exterior angle and the interior opposite angle in each of the following cyclic quadrilaterals.

C

OD

BA

E C

O

D

BA

K

C

O

D

B

A

E

i) ii) iii)

b) What do you notice when you compare the exterior angle and the interior opposite angle in a cyclic quadrilateral?

c) Prove the above result in the general case shown in the diagram.

x° y°

z°

Properties of Cyclic Quadrilaterals

• The opposite angles of a cyclic quadrilateral are supplementary.

• The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Circle Geometry Lesson #3: Cyclic Quadrilaterals 379

Class Ex. #4 Find the missing values in each. O is the centre.

O

D

G

F

E

HO

D

GF

E

a) b) c)

140°35°

x°y°

z° x°

y° 95°

105°

z°

O

D

GF

E

x°

y°

z°

110°45°

Class Ex. #5 In the diagram O is the centre of the circle and

O

R

TU

VW

–UVW = 82° and –TOV = 144°.

Name the angles of the cyclic quadrilateral and state the measure of the angles.


Assignment

In this assignment O is the centre of the circle unless otherwise stated.

1. a) Name three cyclic quadrilaterals.

O

DE

C

B

A F

b) Name two quadrilaterals which are NOT cyclic quadrilaterals.


2. In the following diagrams determine the values of a — j.

O

a) b) c)

78°

63°a°b°

c° 110°d°

e°

f°

230°

100°

g°

h°i°

j°

3. On each of the diagrams below, write the sizes of as many angles as possible.

D

E

F

G

C

O

D

B

C

60°

110°

A 114°

a) b)

57°

T

O

P

QS

R

U

c) d)

LH

I

K

J

40°

80°


4. In the diagram, PQR and STV are double chords. P

Q R

ST

V

Angle SPQ = 70° and angle PST = 84°Determine the size of the remaining angles in the diagram.

5. In the diagram, ABCD is a parallelogram. Prove that DABE is isosceles by showing that two of its angles are equal.

A B

E

CD

6. On the diagram there are twelve acute or obtuse angles whose measures are not given.Determine the measure of each of these angles.

KL

M

N

P30°62°50°

Q


7. The diagram shows a triangle with its three altitudes intersecting at G.

a) Mark the six right angles in the diagram.

A

B

C

D

E

F

G

b) Explain why AEGF is a cyclic quadrilateral.Find two others like AEGF.

8. If –DOF is 150° and –ODE is 60°, MultipleChoice then –DEF and–EFO, respectively, are

O

D

F

E

150°

60°

A. 30° and 120°B. 120° and 30°C. 105° and 45°D. 150° and 60°

9. In the diagram P, Q, R, and S are concyclic points.

P

Q

R

S

a°

2a°

a equals

A. 30B. 60C. 90D. 120

10. ABCD is a cyclic quadrilateral. BA and CD produced meet at K. Angle BCD = 80° and angle ADK = 70°. The size of angle CKB is

A. 10°B. 30°C. 140°D. unable to be determined

from the given information


11. ABCD is a cyclic quadrilateral. –A = (3x + 5)°, –B = (2x + 5)°, and –C = (x - 1)°.Numerical Response The measure of –D, to the nearest degree, is _____ .


Answer Key 1 . a) Any three of AEDB, AEDC, ECBA, EBCD b) EDCO, EDCF

2 . a) –a = 117°, –b = 102°, –c = 63 b) –d = 70°, –e = 55°, –f = 55°c ) –g = 25°, –h = 65°, –i = 40°, –j = 10°

3 .

D

E

F

G

C

O

D

B

C

60°

110°

A 114°

57°

T

O

P

QS

R

U

a)

c)

b)

d)

LH

I

K

J

40°

80°

110°

70°

120° 57°

246°

123°

90°

57° 57°

123° 123°

60°

120°

80°

100°

60°

PQ R

ST

V

70° 96° 84° 110°

110° 84°

70° 96°

KL

M

N

P30°62°50°

42° 20°

50°

42° 88°

68° 110°

110°

70° 70°

20° 68°

4. 6.

7 . b) Opposite angles are supplementary. CEGD, BDGF 8 . C 9 . B 1 0 . A 1 1 . 8 7


Circle Geometry Lesson #4:Circles and Tangents

Warm-Up #1 Intersection of a Line and a Circle

The following cases are illustrated in the diagrams. l

C• There is no point of intersection

C

Al • There is one point of intersection. In this case the line l is tangent to the circle. Point A is a point of tangency.

C

l • There are two points of intersection. In this case the line l is a secant.

Note The following Warm-Ups can be done by:

• using a software program such as Geometer’s Sketchpad, or

• by using the diagrams below with a compass set.

Angle between Tangent and RadiusWarm-Up #2

a) Use a protractor to measure

O

H

I

N

P

Q

A

O

B

C

M

a) b)the angles between the tangent and the radius at the points of tangency.

b) What do you notice about the angle between the tangent and the radius at the point of tangency?

Comparing Lengths of Tangents from an External PointWarm-Up #3

a) Use a ruler to

O

H

I

N

P

Z O

H

I

Z

a) b)measure each of the length of the tangent segments to the circle from the external point.

b) What do you notice about the length of the tangent segments to a circle from an external point?

Warm-Up #4 Angle between Tangent and Chord

A. • –a is an angle between the tangent and a chord.

• Name the inscribed angle on the opposite side of the chord.

O

–a–b–c

–d

–e

• Measure both of these angles.

• What do you notice?

B. • Name another angle between the tangent and a chord.

• Name the inscribed angle on the opposite side of the chord.

• Measure both of these angles.

• What do you notice?

Properties of Tangents to a Circle

• A tangent to a circle is perpendicular to the radius at the point of tangency (as confirmed in Warm-Up #2).

• The tangent segments to a circle from any external point are congruent (as confirmed in Warm-Up #3)

Extension to the above property - the two tangents from the external point together with the two radii form a

kite shape, consisting of two congruent triangles.

• the angle between a tangent and a chord is equal to the inscribed angle on the opposite side of the chord (as confirmed in Warm-Up #4)

Extension to the above property - the angle between a tangent and a chord is one half of the central angle

subtended by the same chord at the centre of the circle.

386 Circle Geometry Lesson #4: Circles and Tangents

Class Ex. #1 Prove the tangent segments to a circle from any external point are congruent.

Class Ex. #2 Find the values of a - g using the properties of tangents to a circle. O is the centre of each circle. Answer to one decimal place where necessary.

O

65°a°

b° O

c8

7O

A

B

C

D

63°

15°

a) b) c)

d°

e° f ° 48°

g°

Class Ex. #3 Find the measures of the indicated angles. O is the centre of each circle.

O

A

B

O w°x°

y°z°

70°

a) b)

48°

x° y°

Circle Geometry Lesson #4: Circles and Tangents 387

Class Ex. #4 In the diagram, PR is tangent to the circle at Q.

OR

PQ

SIf the radius of the circle is 10 cm, and the length of RQ is 18 cm, determine the length of RS to the nearest tenth of a cm.


Assignment In this assignment O is the centre of the circle unless otherwise stated.

1. In the diagram, OACB is a tangent-kite. Name:

C

O

B

A

a) one pair of congruent triangles.

b) two pairs of equal lines

c) three pairs of equal angles

d) a pair of right angles

2. Using the diagram in question #1, if OA = 3 cm and OC = 8 cm, calculate the lengths of the tangents AC and BC to one decimal place.


3. MN is tangent to the circle.

ON

M

MN = 48 cm and ON = 50 cm.

Find the length of the radius of the circle.

4. In the diagram, OABC is a tangent-kite and

O

A

B

Cangle ABC = 48°.

Name two isosceles triangles and calculate the measure of each angle in the diagram.

5. In the diagram KL and ML are

L

O

K

M

N

tangents to the circle.

a) If angle NKL = 70°, calculate the measure of each angle in the diagram.

b) If OL = 5.4 cm and OK = 1.8 cm calculate the length of the tangents to the nearest tenth of a cm.


6. In each diagram tangents to the circle are shown. Find the values of a - i using the properties of tangents to a circle.

Oa°

b°c°

50°

30°

O

P

T

Q

A

B

C110° d°e°

f °

a) b)

O129°

g°

h°i°

c)

7. Tangents AB and AC are drawn from an external point A to a circle. Prove that the points O, B, A, and C are concyclic points.

8. A spherical scoop of ice cream is placed in a cone. The scoop touches the cone 15 cm from the apex of the cone. If the centre of the scoop is 15.4 cm from the apex of the cone, calculate the radius, to the nearest tenth of a cm, of the scoop of ice cream.


9. In the diagram, C and D are respectively the

C D

BE

centres of circles of radius 4 cm and 6 cm.

If CD = 15 cm, determine the length of the common tangent BE to the nearest tenth of a cm

P T

O

Q

10. In the diagram, PT is a tangent to the circle at T. MultipleChoice If –PTQ = 2 –OQT, the size of –OQT is

A. 20°B. 30°C. 45°D. 60°

Numerical Response 11. In the diagram, tangents AB, AC, and BC, touch the A

Y

Z

BCX

circle at Z, Y, and X respectively.

BZ = 10 cm, CX = 8 cm and AY = 6 cm.

The perimeter of triangle ABC, to the nearest tenth of a cm, is _____ .



12. In the diagram, AB is a tangent to the circle.

OA

B

C

If AB = 9 cm and AC = 4 cm, the length of OB, to the nearest hundredth of a cm, is _____ .


Answer Key 1 . a) DOAC @ DOBC b) OA = OB, AC = BC

c ) –OAC = –OBC, –AOC = –BOC, –ACO = –BCO d) –OAC and –OBC

2 . AC = BC = 7.4 cm 3 . 14 cm

4 . DAOC and DABC 5 . a) b) KL = ML = 5.1 cm

O

A

B

C

48°

L

K

M

20°

20°

20°

20°

70°

70°

70°

70°

66° 24°

24°

132° 66°

6 . a) –a = 130° , –b = 65°, –c = 65° b) –d = 30°, –e = 40°, –f = 70° c ) –g = 39°, –h = 78°, –i = 51°

8 . 3.5 cm 9 . 14.9 cm

1 0 . B 1 1 . 4 8 . 0 1 2 . 8 . 1 3


Circle Geometry Lesson #5:Polygons

Regular and Irregular Polygons

A closed plane figure with 3 or more sides is called a polygon. When all its sides are equal and all its angles are equal, it is called a regular polygon. A polygon which is not regular is called an irregular polygon.

Polygons have special names according to the number of sides, eg, pentagon (5-sides), hexagon (6-sides), etc.

Regular polygons can be constructed inside a circle as shown in Warm-Up #1.

Warm-Up #1 Calculating the Measure of the Interior Angle and the Sum of the Measures of the Interior Angles of a Regular Polygon

• The diagrams above show some regular polygons inscribed in a circle. Joining each vertex to the centre results in a number of isosceles triangles.

• Complete the table below to determine the measure of the interior angle of each polygon and the sum of the measures of the interior angles of each polygon.

Name of

Regular

Polygon

Number

of sides

Measure

of the

Central

Angle

Measure of

the Base Angle

of the

Isosceles

Tr iang le

Measure

of the

Interior

Angle of

the

Polygon

Sum of

the

Measures

of the

Interior

Angles in

Degrees

Sum of the

Measures

of the

Interior

Angles in

terms of

Right

Angles

Tr iang le

3360 ÷ 3

= 120° (180 -120) ÷ 2

= 30° 2 x 30°

= 60° 3 x 60°

= 180° 180 ÷ 90

= 2Square

4360 ÷ 4

= 90° (180 - 90) ÷ 2

= 45° 2 x 45°

= 90° 4 x 90°

= 360° 360 ÷ 90

= 4Pentagon

Nonogon

9Dodecagon

1 2n-sided

polygon

Note • Warm-Up #1 shows that the sum of the measures of the interior angles of an n-sided regular polygon is (2n - 4) right angles or 180(n - 2)°

Calculating the Sum of the Measures of the Interior Angles of an Irregular PolygonWarm-Up #2

The diagrams below show some irregular polygons. Choosing a vertex and joining to other vertices we can split the area of the polygon into non-overlapping triangles.

a) Choose a vertex on each of the irregular polygons below. Then join that vertex to other vertices to construct non overlapping triangles.

b) Complete the table below to determine the sum of the measures of the interior angles of each polygon.

Name of Polygon Number

of sides

Number of

Non-

Overlapping

Tr iangles

Sum of the

Measures of the

Interior Angles

in Degrees

Sum of the

Measures of the

Interior Angles in

terms of Right

Angles

Tr iang le

3 1 180° x 1 = 180° 180 ÷ 90 = 2Quadr i la tera l

4 2 180° x 2 = 360° 360 ÷ 90 = 4Pentagon

Octagon

Decagon

100 sided polygon

n-sided polygon

394 Circle Geometry Lesson #5: Polygons

Note • Warm-Up #2 shows that the sum of the measures of the interior angles of an n-sided irregular polygon is (2n - 4) right angles or 180(n - 2)°.

• Warm-Ups #1 and #2 show that the sum of the interior angles of an n-sided polygon does not depend on whether the polygon is regular or irregular.

We have the following property

The sum of the measures of the interior angles of an n-sided polygon is

(2n ---- 4) right angles or 180(n ---- 2)°

Extension to the above property

- Since there are n equal interior angles in a regular polygon, the measure of each angle is180(n - 2)°

n = ËÊÁÁ180 -

360n

ˆ˜

°

interior angle = ËÊÁÁ180 -

360n

ˆ˜

°

The interior angle of an n-sided regular polygon

measures ËÊÁÁ180 ---- 360

n ˆ˜

°

Extension to the above property

- Since the exterior angle is supplementary to the interior angle, the measure

of the exterior angle = (180 - interior angle),

i.e. exterior angle = 180° - ËÊÁÁ180 -

360n

ˆ˜

° = Ë

ÊÁÁ

360n

ˆ˜

°

exterior angle =ËÊÁÁ

360n

ˆ˜

°

The exterior angle of an n-sided regular polygon

measures ËÊÁÁ

360n

ˆ˜

°

It follows from the above property that

The sum of the measures of the exterior angles of an n-sided regular polygon is 360°

Circle Geometry Lesson #5: Polygons 395

Class Ex. #1 Calculate the sum of the measures of the interior angles of a polygon with the given number of sides. Answer in terms of right angles and degrees.

a) 9 b) 4x

Class Ex. #2 Determine the number of sides of a polygon whose interior angle sum equals 4140°

Class Ex. #3 Determine the number of sides of a regular polygon whose interior angle measures;

a) 170° b) k°

Class Ex. #4 Determine the number of sides of a regular polygon whose exterior angle measures

a) 60° b) k°

Class Ex. #5 Determine the values of x and a in the diagrams below.

115°

130°

x°110°

120°

125°a) 2a°

(3a - 40)°

(a + 12)°

3a°

b)(a + 20)°



Assignment

1. Complete the table below to determine the measure of the interior angle of each polygon and the sum of the measures of the interior angles of each polygon.

Name of

Regular

Polygon

Number

of sides

Measure

of the

Central

Angle

Measure of

the Base Angle

of the

Isosceles

Tr iang le

Measure

of the

Interior

Angle of

the

Polygon

Sum of

the

Measures

of the

Interior

Angles in

Degrees

Sum of the

Measures

of the

Interior

Angles in

terms of

Right

Angles

Hexagon

Octagon

Decagon

20 sided

polygon

100 sided

polygon

p-sided

polygon

2. Complete the table below to determine the sum of the measures of the interior angles of each polygon.


of sides

Number of

Non-

Overlapping

Tr iangles

Sum of the

Measures of the

Interior Angles

in Degrees

Sum of the

Measures of the

Interior Angles in

terms of Right

Angles

Hexagon

Heptagon

Nonagon

Dodecagon

20 sided polygon

p-sided polygon


3. Calculate the sum of the measures of the interior angles of a polygon with the given number of sides. Answer in terms of right angles and degrees.

a) 12 b) 15 c) 7y

4. Determine the number of sides of a polygon whose interior angle sum equals:

a) 720° b) 1260° c) 2880° d) 26 right angles

5. Determine the number of sides of a regular polygon whose interior angle measures;

a) 156° b) 175° c) 171° d) 2x°

6. Determine the number of sides of a regular polygon whose exterior angle measures

a) 15° b) 40° c) 1° d) 5k°

7. Determine the values of t and y in the diagrams below.

y°

y°

y° 135°

42°33°b)

94°

t°

127°

83°132°57°

114°

a)


8. Regular polygon PQRST is inscribed in a circle. Determine the measure of P

Q

RS

T

a) –TSP

b) –TSQ

c) –SPR

9. Determine the measures of the angles

C

D

E F

G

H

7x°(4x + 10)°

18x°(3x – 5)°

(7x + 5)°

(8x + 5)°

at C, D, E, F, G, and H.

10. The interior angle of a regular polygon is 5 times as large as the exterior angle. The Numerical Response number of sides in the regular polygon is _____ .



Answer Key 1 . 2 .


of sides

Number of

Non-

Overlapping

Tr iangles

Sum of the

Measures of the

Interior Angles

in Degrees

Sum of the

Measures of the

Interior Angles in

terms of Right

Angles

Hexagon6 4 720° 8

Heptagon7 5 900° 10

Nonagon9 7 1260° 14

Dodecagon

12 10 1800° 2020 sided polygon

20 18 3240° 36p-sided polygon

p p - 2 [180(p - 2)]° 2p - 4

Name of

Regular

Polygon

Number

of sides

Measure

of the

Central

Angle

Measure of

the Base Angle

of the

Isosceles

Tr iang le

Measure

of the

Interior

Angle of

the

Polygon

Sum of

the

Measures

of the

Interior

Angles in

Degrees

Sum of the

Measures

of the

Interior

Angles in

terms of

Right

Angles

Hexagon

6 60° 60° 120° 720° 8Octagon

8 45° 67.5° 135° 1080° 12Decagon

10 36° 72° 144° 1440° 1620 sided

polygon 20 18° 81° 162° 3240° 36100 sided

polygon 100 3.6° 88.2° 176.4° 17 640° 196p-sided

polygonp [180(p - 2)]° 2p - 4 Ë

ÊÁÁ360p

ˆ˜ ° Ë

ÊÁÁ180(p - 2)

p ˆ˜° Ë

ÊÁÁ90(p - 2)

p ˆ˜ °

3 . a) 20 right angles, 1800° b) 26 right angles, 2340° c ) (14y - 4) right angles, (1260y - 360)°

4 . a) 6 b) 9 c ) 18 d) 15

5 . a) 15 b) 72 c ) 40 d)180

90 - x

6 . a) 24 b) 9 c ) 360 d)72k

7 . a) 121° b) 110° 8 . a) 36° b) 72° c ) 36°

9 . –C = 105°, –D = 70°, –E = 270°, –F = 40°, –G = 110°, –H = 125°

1 0 . 1 2


Circle Geometry Lesson #6:Relations Between Arcs, Sectors, and Angles

Note The work in this lesson is an extension to the curriculum, but should be a benefit to students when they study arc length and radian measure in grade 12.

Warm-Up

a) How does the measure of –AOB compare to the measure

O

A

B

of one complete rotation?

b) How does the length of arc AB compare to the circumference of the circle?

c) How does the area of sector AOB compare to the area of the circle?

In general we have the following equal ratios:

–AOB360° =

arc ABcircumference =

area of sector AOBarea of circle

Relations Between Arcs, Sectors, and Angles

O

A

C

B

D

In the diagram, there are the following equal ratios.

–AOB–COD =

arc ABarc CD =

area of sector AOBarea of sector COD

Class Ex. #1 AB is the diameter of the circle. Find the numerical values of the following ratios:

O

C

B

D

A30°

a) arc ADarc DC

b) area of sector AODarea of sector COB

c) arc DBarc AC

Class Ex. #2 The tip of a minute hand of a clock is 8 cm from the centre.

How far, to the nearest tenth of a cm, does the tip of the minute hand move in five minutes?

Complete Assignment Quesitons #1 - #8

Assignment

1. In each of the diagrams below, calculate the indicated measure to one decimal place where necessary.

a) arc AB b) the circumference c) angle EOFof the circle

O

A

B

O

D

C

OF

E

135°55°

Area = 248cm2 4 cm

16 cm

15.2 cm

402 Circle Geometry Lesson #6: Relations Between Arcs, Sectors, and Angles

2. In the diagram, –AOB = 40° and –COD = 30°.

O

A

C

B

D

Arc AB = 8.5 cm and the area of sector COD = 38.8 cm2.

Calculate, to one decimal place:

a) the area of sector AOB

b) the length of arc CD.

3. In the diagram, arc PQ = 4.8 metres and arc RS = 7.8 metres. If –POQ = 32°, determine:

O

P

R

Q

S

a) the measure of – ROS to the nearest degree

b) the diameter of the circle to the nearest metre

c) the area of sector POQ to the nearest square metre.

Circle Geometry Lesson #6: Relations Between Arcs, Sectors, and Angles 403

4. Ron buys a ten inch pizza (a circular pizza of diameter ten inches) and Hermione buys a twelve inch pizza. Ron’s pizza is cut into six equal parts, and he eats four of them. Hermione’s pizza is cut into eight equal parts and she eats three of them.

a) What surface area of pizza, to the nearest square inch, did each eat?

b) Harry ate the rest of the pizzas. What surface area, to the nearest square inch, did he eat?

5. In the diagram, O is the centre of the circle with radius 5 cm.

O

C

A B

Angle AOB = 108° and angle ABC = 80°.

a) Determine the measure of: i ) –ACB i i ) –OAC

b) If the area of DOAB is 11.8 cm 2, show that the area of the shaded segment is approximately equal to the area of triangle OAB.


Questions #6 - #8 refer to the following information.

A

O

C

B

AB and BC are tangents at B and C to the circle centre O. Angle BOC = 150° and OB = 6 cm.

6. The measure of –BAC is MultipleChoice

A. 15°B. 30°C. 75°D. 210°

7. The area, in cm2, of minor sector BOC is

A. 5pB. 15pC. 21pD. 30p

8. The length, to the nearest tenth of a cm, of minor arc BC _____ .Numerical Response


Circle Geometry Lesson #6: Relations Between Arcs, Sectors, and Angles 405

Answer Key 1 . a) arc AB = 9.4 cm b) circumference = 104.7 cm c ) –EOF = 123.0°

2 . a) 51.7 cm2 b) 6.4 cm

3 . a) 52° b) 17 m c ) 143 m2

4 . a) RonÆ 52 in2 HermioneÆ 42 in2 b) 97 in2

5 . a) i ) 54° i i ) 44°

6 . B 7 . B 8 . 1 5 . 7


Coordinate Geometry and Trigonometry Lesson #1:Coordinate Geometry Review

Warm-Up #1 Review

- - - - - - - - - - - - - -A(x1, y1)

B(x2, y2)

y y2 1-

x x2 1-

MConsider two points, A(x1, y1) and B(x2, y2). Let M be the midpoint of AB. Recall the following formulas:

Slope Formula mAB = y2 - y1

x2 - x1

Distance Formula dAB = (x2 - x1)2 + (y2 - y1)2

Midpoint Formula The coordinates of M are ËÊÁÁ

x1 + x2

2 , y1 + y2

2 ˆ˜

Warm-Up #2 Equations of Lines

Slope y-intercept Form y = mx + b Point-Slope Form y - y1 = m(x - x1)

Warm-Up #3 Lines Associated with Triangles

Median - a line drawn from a vertex to the midpoint of the opposite side.

Altitude - a line drawn from a vertex perpendicular tothe opposite side.

Class Ex. #1 Complete the following statements:

a) Parallel lines have the ____________ slope.

b) ____________ lines have slopes which are negative reciprocals of one another.

c) Line one has slope m1 and line two has slope m2. If the lines are perpendicular then m1•m2 = ____________.

d) A ____________ line has a slope of zero.

e) A ____________ line has a slope that is undefined.

g) Lines rising from left to right have a ____________ slope.

h) Lines falling from left to right have a ____________ slope.

Class Ex. #2 A(–3, –1), B(5, 9), and C(6, 0) are vertices of a triangle.

x

yD is the midpoint of AB.

a) Determine algebraically if DC is perpendicular to AB.

b) What is the relationship between the length of the line DC and the length of the line AB?

c) Find the equation of the line parallel to AC passing through B. Express your answer in the general form Ax + By + C = 0.

d) i ) State the equation of the horizontal line through A.

ii) State the equation of the vertical line through A.


408 Coordinate Geometry and Trigonometry Lesson #1:Coordinate Geometry Review

Assignment 1. Write the following equations in general form Ax + By + C = 0, where A, B, C Œ I.

a) y = 4x - 7 b) y = 13 x + 2

2. Determine the slope of a line that is parallel to the given line.a) y = –2x + 5 b) 4x - 3y + 7 = 0 c) x = 5

3. Determine the slope of a line that is perpendicular to the given line.

a) y = 34 x + 8 b) 2x + 5y - 4 = 0 c) x = 5

4. A and B are the points (4, –2) and (–2, 6) respectively. Find the equation of the perpendicular bisector of AB.

5. Triangle PQR has vertices P(–3, 2), Q(0, 6) and R(–7, 5). Show that DPQR is isosceles, but not equilateral.

Coordinate Geometry and Trigonometry Lesson #1: Coordinate Geometry Review 409

6. Show that the triangle with vertices A(2, 6), B(–1, 4), and C(4, 3) is right-angled at A:

a) by using slopes b) by using the distance formula and the Pythagorean Theorem

7. The point A has coordinates (1, 6) and the line BC has A

B

C

x

yequation 2x - 5y - 1 = 0.

a) Find the equation of the line through A perpendicular to BC.

b) Find the coordinates of the point where the line in a) meets BC.


8. The diagram shows a right-angled triangle BAC with

B(-4, 0) A(6, 0)

C(6, 8)

M

x

ymedian AM.

Show that the length of the hypotenuse BC is twice the length of the median AM.

9. A rectangle EFGH has coordinates E(3, –4), F(3, 0) and G(7, 0).

a) Find the coordinate of H.

b) State the equation of: i ) EF i i ) EH

c) Show that the diagonals are equal in length.

d) S is the midpoint of EF and T is the midpoint of FG. Show that ST is parallel to EG.


10. Triangle ABC has coordinates A(1, 0), B(–4, 3), and C(0, –1).

x

y

–2–4 2

–2

2

a) Make a rough sketch of the triangle on the grid and draw median AD and altitude CE.

b) Find the equation of the median AD.

c) Find the equation of the altitude CE.

d) AD and CE intersect at F. Find the coordinates of F.

11. Consider the isosceles triangle shown. Show that the

B(–2x, 0)x

y

C(2x, 0)

A(0, 2y)medians CL and BM are equal in length.


12. In the diagram, the line PQ has equationMultipleChoice

–2x

y

3

P

QA. 2y = 3x

B. 3y = 2x + 3

C. 2y = 3x + 3

D. 2y = 3x + 6

13. Consider the following equations representing straight lines.

(1) y = 34 x (2) y =

43 x (3) 3y + 4x = 5 (4) 3y - 4x = 4

Which of the above lines are parallel to the line with equation 3y = 4x - 1?

A. Only (1) and (3)

B. Only (2) and (4)

C. Only (1), (2), and (4)

D. Only (4)

14. A(2, 4), B(–1, 0), C(6, 1) are the vertices of a triangle. Which of the following best describes the triangle?

A. Equlateral

B. Isosceles

C. Scalene

D. Right-angled

15. The line with equation y = mx + b has an x-intercept of 6 and a y-intercept of 3. Numerical Response The value of m + b, to the nearest tenth, is _____ .



Answer Key

1 . a) 4x - y - 7 = 0 b) x - 3y + 6 = 0 2 . a) –2 b)43

c ) undefined

3 . a) –43

b)52

c ) 0 4 . 3x - 4y + 5 = 0 5 . PQ = PR = 5, QR = 5 2

6 . a) mAB = 23

, mAC = –32

b) AB = AC = 13 , BC = 26

7 . a) 5x + 2y - 17 = 0 b) (3, 1) 8 . BC = 2 41 , AM = 41

9 . a) (7, –4) b) i ) x = 3 i i ) y = –4 c ) 4 2 d) slopes equal 1

1 0 . b) x + 3y - 1 = 0 c ) 5x - 3y - 3 = 0 d) ËÊÁÁ

23

, 19

ˆ˜

1 1 . Length = 9x2 + y 2 1 2 . D 1 3 . B 1 4 . B

1 5 . 2 . 5


Coordinate Geometry and Trigonometry Lesson #2:Distances Between Points and Lines

Warm-Up Horizontal Distance, Vertical Distance, and Shortest Distance Between a Point and a Line

• Draw the line l1 which represents the vertical distance between

A

B

C

the point C and the line AB.

• Draw the line l2 which represents the horizontal distance between the point C and the line AB.

• Draw the line l3 which represents the shortest distance between the point C and the line AB.

Finding the Horizontal & Vertical Distances Between a Point and a Line

Class Ex. #1 Find the horizontal distance and the vertical distance

(7, 4)

y = 2x + 5

5

5 10

10

15

20

x

y

A

between the point A(7, 4) and the line with equation y = 2x + 5.

Class Ex. #2 Find the horizontal distance and the vertical distance between the point P(–2, –3) and the line with equation 3x + 2y - 12 = 0.

Finding the Shortest Distance Between a Point and a Line

The following procedure can be used to calculate the shortest distance between a point and a line.

1. Sketch a diagram showing the point, the original line, and the shortest distance.

2. Find the slope of the original line and the slope of a perpendicular line.

3. Find the equation of the line perpendicular to the original line passing through the given point.

4. Find the point, P, of intersection of the original line and the perpendicular line.

5. Calculate the distance between P and the original point.

Note The shortest distance between a point and a line can also be found using trigonometry as we shall see in Lesson 4.

Class Ex. #3 Find the shortest distance between the point A(7, 4) and

(7, 4)

y = 2x + 5

5

5 10

10

15

20

x

y

A

the line with equation y = 2x + 5. Give the answer as an exact value.

416 Coordinate Geometry and Trigonometry Lesson #2: Distances Between Points and Lines

Class Ex. #4 Find the shortest distance, to the nearest hundredth, between the point (–2, –3) and the line with equation 3x + 2y - 12 = 0.


Coordinate Geometry and Trigonometry Lesson #2: Distances Between Points and Lines 417

Assignment

1. Find the horizontal distance and the vertical distance from the given point to the given line.

a) (2, 0), y = x + 3, b) (4, 3), y = 2x - 1

c) (0, 0), x + y = 9 d) (2, –1), 4x - y + 1 = 0


2. Find the shortest distance from the given point to the given line. Round the answers to the nearest tenth, if necessary.

a) (2, 0), y = x + 3 b) (4, 3), y = 2x - 1

c) (0, 0), x + y = 9 d) (2, –1), 4x - y + 1 = 0

Coordinate Geometry and Trigonometry Lesson #2: Distances Between Points and Lines 419

3. DPQR has vertices P(3, 0), Q(1, 6), R(–4, 1).

a) Determine the equation of PQ.

b) Find the length of the altitude from R to PQ. Express the answer as a radical in simplest form.

c) Calculate the area of the triangle.

Answer Key

1 . a) dh = 5 units, d v = 5 units b) dh = 2 units, d v = 4 unitsc ) dh = 9 units, d v = 9 units d) dh = 2.5 units, d v = 10 units

2 . a) 3.5 units b) 1.8 units c ) 6.4 units d) 2.4 units

3 . a) y = –3x + 9 b) 2 10 c ) 20 sq. units


Coordinate Geometry and Trigonometry Lesson #3:Distances Between Parallel Lines

Warm-Up Horizontal Distance, Vertical Distance, and Shortest DistanceBetween Parallel Lines

A

B

C

D• Draw the line l1 which represents the vertical distance between the parallel lines.

• Draw the line l2 which represents the horizontal distance between the parallel lines.

• Draw the line l3 which represents the shortest distance between the parallel lines.

Finding the Horizontal and Vertical Distance Between Parallel Lines

Class Ex. #1 Find the horizontal distance and the vertical distance between the parallel lines with equations y = 3x + 4 and y = 3x - 5.

Finding the Shortest Distance Between Parallel Lines

Class Ex. #2 Find the shortest distance, to the nearest hundredth, between the parallel lines with equations y = 3x + 4 and y = 3x - 5.

Class Ex. #3 From Class Ex. #2 write down in your own words the general procedure needed to determine the shortest distance between two parallel lines.


Assignment

1. Find the horizontal distance and the vertical distance between the parallel lines with equations y = 2x + 4 and y = 2x - 3.

2. Find the horizontal distance and the vertical distance between the parallel lines with equations y = –x + 4 and y = –x + 10.

422 Coordinate Geometry and Trigonometry Lesson #3: Distances Between Parallel Lines

3. Find the horizontal distance and the vertical distance between the parallel lines with equations 4x - y - 8 = 0 and 4x - y - 12 = 0.

4. Find the shortest distance between the parallel lines with equations y = 2x + 4 and y = 2x - 3. Round the answer to the nearest tenth.

5. Find the shortest distance between the parallel lines with equations y = –x + 4 and y = –x + 10. Express the answer as a radical in simplest form.

Coordinate Geometry and Trigonometry Lesson #3: Distances Between Parallel Lines 423

6. The horizontal distance between the parallel lines with MultipleChoice equations y = 3x + 2 and y = 3x - 7 is

A. 53

B. 3

C. 5

D. 9

7. The shortest distance, to the nearest hundredth, between the parallel lines with Numerical Response equations 4x - y - 8 = 0 and 4x - y - 12 = 0 is _____ .


Answer Key

1 . dh = 3.5, d v = 7 units 2 . dh = 6, d v = 6 units 3 . dh = 1, d v = 4 units

4 . 3.1 units 5 . 3 2 units 6 . B 7 . 0 . 9 7

424 Coordinate Geometry and Trigonometry Lesson #3: Distances Between Parallel Lines

Coordinate Geometry and Trigonometry Lesson #4:Coordinate Geometry and the Circle

In this lesson we will use coordinate geometry to solve circle problems.

Class Ex. #1 P and Q are the points (10, 6) and (4, –2) respectively.

a) Determine the coordinates of the centre and the radius of the circle whose diameter is PQ.

b) Verify that A(7, 7) and B ËÊÁÁ

115 ,

35

ˆ˜ are the ends of a chord of this circle

that is parallel to PQ.

c) Describe how to find the distance between the chord AB and the diameter PQ.

Class Ex. #2 The line with equation 4x + 3y + 7 = 0 is tangent to a circle with centre C(3, 2) at the point P.

a) Find the equation of the radius CP.

b) Find the coordinates of P.

c) Determine the length of the diameter of the circle.

426 Coordinate Geometry and Trigonometry Lesson #4: Coordinate Geometry and the Circle

Class Ex. #3 a) Find the point(s) of intersection of the line y = –2x + 10 and the circle x2 + y2 = 20.

b) Explain why the line is a tangent to the circle.


Assignment

1. P, Q and R are the points (–3, 5), (2, –5) and (4, 1) respectively.

a) Find the equation of the line PQ.

b) Find the coordinates of S, the point of intersection of PQ and the y-axis.

c) Prove that RS is perpendicular to PQ.

d) Find the coordinates of the centre of the circle which passes through Q, R and S.

Coordinate Geometry and Trigonometry Lesson #4: Coordinate Geometry and the Circle 427

2. A line segment has endpoints A(3, –4) and B(4, 3).

a) By replacing the coordinates into the equation, show that AB is a chord of the circle with equation x2 + y2 = 25.

b) Determine the equation of the line that passes through the midpoint of chord AB and the centre of the circle (0, 0).

c) Show that the line in b) is perpendicular to the chord AB.

3. P and Q are the points (2, 2) and (4, 8) respectively.

a) Find the equation of the perpendicular bisector of PQ.


b) C, a point in the first quadrant equidistant from both axes, is the centre of a circle passing through P and Q.

i ) Use the equation in a) to find the coordinates of C.

i i ) Find the radius of the circle to the nearest tenth.

4. a) Find the point(s) of intersection of the line y = x - 4 and the circle x2 + y2 = 16.

b) Is the line y = x - 4 a tangent to the circle? Explain.

5. Show that the line y = x + 2 is a tangent to the circle x2 + y2 = 2 and find the point of contact.

Coordinate Geometry and Trigonometry Lesson #4: Coordinate Geometry and the Circle 429

6. The line y + 2 = 0 is a tangent to the circle with equation x2 + y2 + 8x - 2y + k = 0. MultipleChoice The value of k is

A. 8

B. 11

C. 16

D. 59

7. The line y = 2x + 8 intersects the circle x2 + y2 + 4x + 2y - 20 = 0 at the points (a, b) Numerical Response and (c, d), where a > c. The value of a + b - c - d is _____ .


Answer Key

1 . a) 2x + y + 1 = 0 b) S(0, –1) c ) mRS = 12

, mPQ = –2 d) (3, –2)

2 . b) x + 7y = 0 c ) mline = –17

, mAB = 7

3 . a) x + 3y - 18 = 0 b) i ) C ËÊÁÁ

92

, 92

ˆ˜ i i ) 3.5 units

4 . a) (0, –4), (4, 0) b) No, because there are two intersection points

5 . (–1, 1) 6 . A 7 . 1 2


Coordinate Geometry and Trigonometry Lesson #5:Equation of a Circle

The equation of the circle, as derived from the distance formula, is a useful tool to solve problems related to this topic.

Equation of a Circle with Centre at the Origin

Let P(x, y) be any point on the circumference of a

x

y

O

P(x, y)circle with centre at the origin and radius r.

Complete the following using the distance formula to determine the equation of the circle.

a) d = (x2 - x1)2 + (y2 - y1)2

r =

b) The equation of a circle, centre at the origin, with radius r is _____________________

Equation of a Circle with Centre Not at the Origin

Let P(x, y) be any point on the circumference of the

x

y

C(h, k)

P(x, y)

O

circle with centre C(h, k) and radius r.

Complete the following using the distance formula to determine the equation of the circle.

a) d = (x2 - x1)2 + (y2 - y1)2

b) The equation of a circle, centre C(h, k) and radius r is _____________________

Class Ex. #1 Find the equation of the circle with centre C given the following information.a) C(0, 0), radius 5 b) C(0, 0), diameter 12

c) C(–2, 4), radius 7 d) C(0, –3), diameter 9

Class Ex. #2 Find the coordinates of the centre and the radius of each of the following circles.

a) (x - 7)2 + y2 = 36 b) (x - 2)2 + (y + 12)2 = 20

Class Ex. #3 Find the equation of the circle with centre C(–2, 5) and passes through the point P(1, 9).


Assignment

1. Find the equation of the circle with centre C given the following information.

a) C(0, 0), radius 9 b) C(0, 0), diameter 14

c) C(–1, 5), radius 3 3 d) C(-2, 0), diameter 12

e) C(–e, f), radius g f) C ËÊÁÁ

14 , –

13

ˆ˜ , radius 1

2. Find the coordinates of the centre and the radius of each of the following circles.

a) x2 + y2 = 49 b) (x + 3)2 + (y - 3)2 = 16

c) (x - 5)2 + y2 = 81 d) (x + 0.5)2 + (y - 0.9)2 = 0.04

e) (x - b)2 + (y + c)2 = d 2 f) 3x2 + 3y2 = 48

432 Coordinate Geometry and Trigonometry Lesson #5: Equation of a Circle

3. Find the equation of the circle with centre C(–8, 0) and passes through P(–2, 8).

4. Find the equation of the circle with centre C(0, 0) and has y intercept of –4.

5. Find the equation of the circle with centre C(–3, 5) and has x intercept of –2.

6. A circle has equation (x - 1)2 + (y - 2)2 = 8.

a) Determine the coordinates of the centre, C, and the radius of the circle.

b) The tangent to the circle at the point P(3, 4) on its circumference is drawn.i ) Find the equation of the tangent at P.

i i ) Verify that the point Q(7, 0) lies on the tangent.

i i i) Find the equation of the circle which passes through the points C, P, and Q.

Coordinate Geometry and Trigonometry Lesson #5: Equation of a Circle 433

7. The equation of a tangent to the circle x2 + y2 = 13 at the point (–2, 3) is MultipleChoice

A. y - 3 = 23 (x + 2)

B. y + 3 = 2

3 (x - 2)

C. y - 3 = –3

2 (x + 2)

D. y + 3 = –3

2 (x - 2)

8. The quadrant(s) in which a point on the circumference of the circle (x - 3)2 + (y + 4)2 = 10 can lie is/are

A. the fourth only B. the first and fourth onlyC. the third and fourth onlyD. some other combination of quadrants

9. The distance, to the nearest tenth, from the point (5, –3) to the centre of the circle with Numerical Response equation (x + 4)2 + (y + 7)2 = 25 is _____ .


Answer Key

1 . a) x2 + y 2 = 81 b) x2 + y 2 = 49 c ) (x + 1)2 + (y - 5)2 = 27

d) (x + 2)2 + y 2 = 116

e ) (x + e)2 + (y - f)2 = g2 f ) ËÊÁÁx -

14

ˆ˜2

+ ËÊÁÁy +

13

ˆ˜2

= 1

2 . a) C(0, 0), r = 7 b) C(–3, 3), r = 4 c ) C(5, 0), r = 9d) C(–0.5, 0.9), r = 0.2 e ) C(b, –c), r = d f ) C(0, 0), r = 4

3 . (x + 8)2 + y 2 = 100 4 . x2 + y 2 = 16 5 . (x + 3)2 + (y - 5)2 = 26

6 . a) C(1, 2) radius = 2 2 b) i ) y = –x + 7 i i i ) (x - 4) 2 + (y - 1) 2 = 10

7 . A 8 . C 9 . 9 . 8

434 Coordinate Geometry and Trigonometry Lesson #5: Equation of a Circle

Coordinate Geometry and Trigonometry Lesson #6:Trigonometry Review

Trigonometry

Trigonometry means “three angle measurement”. It is a branch of mathematics which deals with the measurement of angles and sides of triangles.

Primary Trigonometric Ratios

Complete the following

side opposite to q

side adjacent to q

q

hypotenusesine fi sin q =

cosine fi cos q =

tangent fi tan q =

These ratios are called the Primary Trigonometric Ratios and can be remembered by the acronym SOHCAHTOA.

Class Ex. #1 Use the diagram to determine 29

34 x°

a) sin x°

b) x°

Class Ex. #2 In each diagram find the length of the indicated side to the nearest tenth.

a) b)x

34 50° 37°83

m

Class Ex. #3 In the diagram, ABCD represents a horizontal rectangular lawn. AB = 12m, BC = 8m and AE is a vertical pole of length 2m.

E

M

A B

CD

At a certain time of day, the tip of the shadow cast by the pole on the lawn is exactly at M, the midpoint of DC.

Calculate;

a) the length of AM

b) the measure of –AME to the nearest degree.

In Lesson 2 we found the shortest distance between a point and a line using algebraic methods. In this example we solve the same problem using trigonometry.

Class Ex. #4 Complete the following procedure to find the

P(7, 4)

y = 2x + 5

5

5 10

10

15

20

x

y

B

C

A

shortest distance (to the nearest tenth) between the point (7, 4) and the line with equation y = 2x + 5.

a) The diagram shows the horizontal distance (PA) the vertical distance (PB) and the shortest distance (PC) from the point to the line.

Determine the lengths of PA and PB.

436 Coordinate Geometry and Trigonometry Lesson #6: Trigonometry Review

b) Let –BAP = q °. Use DABP to determine the value of q to the nearest tenth of a degree.

c) Use this value of q in DACP to determine the length of PC to the nearest tenth.

d) Compare the answer with Class Ex. #3 in Lesson 2.


Assignment

1. Use the diagram to determine

x°15

27a) cos x°

b) x°

2. Use the diagram to determine

15 x°9

a) tan x°

b) x°

3. In each diagram, calculate the value of x to two decimal places.

9.2

61°

x

8

x°

5.4

a) b)

Coordinate Geometry and Trigonometry Lesson #6: Trigonometry Review 437

4. Use trigonometry to find the shortest distance from the given point to the given line. Round your answers to the nearest tenth, if necessary.

a) (2, 0), y = x + 3,

b) (4, 3), y = 2x - 1


5. AD is an altitude of right angled triangle BAC with dimensions as shown. MultipleChoice The length of AD in cm is

30°

1 cm

2 cm

A

B CD

A. 33

B. 2 3

3

C. 3

2

D. 1

2

Questions #6 and #7 refer to the following information

PQ

R

A B

C D

A right pyramid on a square base of side 12 cm is 10 cm high.

6. The angle, to the nearest degree, between a sloping face and the base, is _____ .Numerical Response


7. The angle, to the nearest degree, between a sloping edge and the base, is _____ .


Coordinate Geometry and Trigonometry Lesson #6: Trigonometry Review 439

Answer Key

1 . a)59

b) 56° 2 . a)43

b) 53°

3 . a) 16.60 b) 42.45° 4 . a) 3.5 b) 1.8

5 . C 6 . 5 9 7 . 5 0


Coordinate Geometry and Trigonometry Lesson #7:Law of Sines and Law of Cosines

Warm-Up Reviewa b

c AB

C

For any triangle DABC, we have the following laws:

Law of Cosines

a2 = b2 + c2 - 2bccos A

Law of Sines

asin A =

bsin B =

csin C

sin Aa =

sin Bb =

sin Cc

or

cos A = b2 + c2 - a2

2bc

Note To solve for an angle or a side in a triangle using trigonometry, three pieces of information are required, one of which must be the length of a side.

Use the following guide to determine which law to use in any given situation:

• In a right-angled triangle use SOHCAHTOA

• Use Law of Cosines if you are given either; i) all three sides, orii) two sides and the contained angle.

• In all other cases use the Law of Sines.

Class Ex. #1 Mr. Post’s two metre high fence has almost been blown

fencerope

25°

down by the wind. As a temporary measure, he wants to tie a rope from the top of the fence to a peg one metre from the base of the fence. The fence has moved so that it is leaning 25° to the vertical as shown. Determine, to the nearest tenth of a metre, the minimum length of rope required if he allows 50 cm for knots.

Class Ex. #2 PQ is a chord of a circle with centre C and radius 8 cm.

C

PQ

R

Angle CPQ is 35°. Chord PQ is extended to the point R such that CR = 12 cm.

a) Find –CRP to the nearest tenth.

b) Find the length of PR.

c) Find the length of the chord PQ.


Assignment

1. The first hole at a golf course is 210 metres long from the tree to the hole in a direct line. Andrew Duffer hit his first shot at an angle of 15° off the direct line to the hole. His second shot landed in the hole.

15°

105°

hole

flag

tree

If the angle between his first shot and his second shot was 105°, how long was his second shot?

442 Coordinate Geometry and Trigonometry Lesson #7: Law of Sines and Law of Cosines

2. In the diagram, chord PQ = 13.6 cm, and radius CP = 10.5 cm. PQ

R

C

Determine the measure of angle PRQ, to the nearest degree.

3. In the diagram, O is the centre of a circle of radius 3.25 cm. QR = 2.5 cm and RS = 4.7 cm. Calculate, to the nearest whole number:

P

Q

S

R

O

a) the length of PQ

b) the measure of –QPR

c) the measure of –SQR.

Coordinate Geometry and Trigonometry Lesson #7: Law of Sines and Law of Cosines 443

4. Violet and Thomas have formed their own student painting 6.5 m

8.0 m

4.3 m

A B

Ccompany. They have been given a contract to paint a small triangular portion of the sides of 100 houses. The area to be painted is shown on the diagram. The paint they are going to use covers 10m2 per litre. Violet had found a formula for the area of a triangle in one of her teacher’s old math books. Thomas doubted the formula worked.

a) Follow the instructions below to find the area of the triangle

Thomas Violet

i ) Find base angle CAB i ) Find base angle CAB to two decimal places. to two decimal places.

i i ) Determine the vertical height i i ) Use the formula A = 12 bc sin A to find

to two decimal places. the area of the triangle to one decimal place.

i i i)Use the formula A = 12 bh to find

the area of the triangle to one decimal place.

b) Determine the amount of paint required.


5. In the diagram, the chords AB and CD intersect at E.

A

B

C

D

E

EB = 12 cm, EA = 16 cm, ED = 8 cm, and –BED = 120°.

a) Use the law of cosines to calculate the length of BD to two decimal places.

b) Determine the measure of –ACD to the nearest degree.

6. An oil company drilling off shore has pipelines from platform North

A

B

D 125°

50°

Alpha and platform Beta to the same shore station Delta. Platform Alpha is 180 km on a bearing of 50° from Delta and platform Beta is 250 km on a bearing of 125° from Delta. Calculate the distance between platform Alpha and platform Beta to the nearest km.

Coordinate Geometry and Trigonometry Lesson #7: Law of Sines and Law of Cosines 445

Numerical Response 7. Circle researcher Tory Tate has discovered a bizarre

Tory

circle in a farmer’s field. The circle is tangent to the road that runs along side of it. While standing outside of the circle, Tory uses his surveying equipment and calculates the distance from himself to the centre of the circle to be 18 m. The distance from Tory to the point of tangency between the circle and the road is two thirds that length. If the angle between these two lines is 55°, the radius of the circle, to the nearest tenth of a metre, is _____.


8. At 5 p.m the distance between the tip of the minute hand on a clock and the tip of the hour hand is 17.4 cm. If the minute hand is 10 cm long, the length of the hour hand, to the nearest tenth of a centimetre, is _____. (Record your answer in the numerical response box from left to right)

Answer Key1 . 56.3 m 2 . 40° 3 . a) 6 cm b) 23° c ) 47°

4. a) Thomas i) 32.45° ii) 3.49 m iii) 14.0 m2 Violet i) 32.45° ii) 14.0 m2 b) 140 L

5 . a) 17.44 cm b) 23° 6 . 268 km

7 . 1 4 . 8 8 . 8 . 0


Personal Finance Lesson #1:Income

Gross and Net Income

Many people in Canada are employed on a full time or part-time basis for which they receive an income.

Gross income is the amount of money earned through employment before deductions.

Net income (or “take-home” pay) is the amount of money after deductions.

Types of Income

Types of Income

Salary Hourly Pay GratuitiesCommission

SalaryA person who receives an annual salary is usually paid on a monthly basis and will not receive extra income for working extra hours, eg. teacher.

Hourly PayA person who is paid by the hour usually works for a fixed number of hours per week and will receive overtime (usually at a greater rate per hour) for any extra hours worked, eg. cashier

CommissionA person who is paid according to how much product they sell. A person can be paid in straight commission, graduated commission, or by salary and commission, and may have to meet a predetermined sales quota before any commission is paid, eg. salesperson. A variation of commission is piecework, where a person is paid according to the number of items they make.

GratuitiesA person receives “tips” or gratuities for the work they do. This may be in addition to an hourly rate of pay, eg. waitress.

Note Some people will earn income through a combination of the above types.

Class Ex. #1 Verne is an elementary teacher and has an annual salary of $52 320. What is his gross monthly income?

Class Ex. #2 Mr. Alpine is an electronics salesperson who is paid a graduated commission. He is paid 5% of the first $4000 of sales, 6% of the next $6000 and 10% on any further sales. Last month his total sales amounted to $13 520. Calculate his gross income for the month.

Class Ex. #3 Betty Lou works full time at the local convenience Monday 9:00 a.m. - 5:30 p.m.Tuesday 9:00 a.m. - 5:30 p.m.Wednesday 6:00 p.m. - 10:00 p.m.Thursday 9:30 a.m. - 6:30 p.m.Friday 9:00 a.m. - 5:30 p.m.Saturday 9:00 a.m. - 5:30 p.m.Sunday o f f

store. She is paid $12.00 per/hour for a 36 hour work week. Overtime is paid at a rate of time and a half.Last week she worked the hours shown in the table. On a full work day, she receives a one hour unpaid lunch break. Calculate her gross income for the week.

Class Ex. #4 Ms. Baitress works once a week as a server at Sloppy Burgers restaurant. Her income consists of an hourly salary ($6.00/hr) plus tips. She noticed that her tips usually average to about 15% of the total food revenue of her assigned wing of the restaurant. If she works from 7:00 p.m. to 11:00 p.m. on a night when her wing of the restaurant has earned $2500 in food revenue, calculate her gross income for the night.


Assignment

1. Kelsey is a research engineer. She has an annual salary of $65 300. Determine her monthly gross income.

2. Elizabeth works at the dry cleaners. She is paid $1.25 for every shirt she dry cleans, $1.75 for every pair of slacks, and $2.25 for every suit. Last week she laundered 45 shirts, 22 pair of slacks and 12 suits. Calculate Elizabeth’s gross income for the week.

448 Personal Finance Lesson #1: Income

3. Dominic works as a headwaiter for a downtown restaurant. Last week he worked five 8 hour shifts at $15.75 per hour. His gratuities for those shifts were as follows:

$18.25, $15.50, $21.50, $37.50, and $44.25.

Calculate Dominic’s weekly gross income.

4. Determine the gross weekly income for the hours worked at the given rate.Overtime is double-time for more than 36 hours worked.

a) 40 h at $8.75 b) 52.5h at $10.75

5. Determine the gross weekly income for each employee.

a) Tara-Lee works as a sales clerk for a major clothing chain. She earns $7.75 per hour. Last week she worked for 40 hours.

b) When Madison works overtime at the ice cream parlour, she receives time-and-a-half pay for overtime. This week she worked a regular shift of 25 hours and 11 hours of overtime. Her hourly rate is $5.50 .

c) Alyssa works as a waitress which pays her $6.35/h for each shift. This week she worked four 7-hour shifts and her tips averaged $11 a shift.

d) Zachary works in a bicycle shop, where he receives $350 a week plus 2.5% commission on bicycles sold. Last week his bicycle sales totalled $2500.

e) Abigail sells snowboards, which pays $7.50/h for a 30-hour workweek, plus 3% commission on sales, which during the snow season averages $4500 per week.

Personal Finance Lesson #1: Income 449

6. Joshua as a Senior Fund Manager is paid a graduated commission. He is paid 3.5% of the first $3000 of sales, 5% of the next $5000 and 12% on any further sales. Last month his total sales amounted to $25 500. Calculate his gross income for the month.

7. Mr. Martinez works as an Assistant Manager. He is

Monday 07:00 to 18:30

Tuesday 08:00 to 14:30

Wednesday 15:00 to 21:00

Thursday 09:30 to 19:00

Friday 10:00 to 18:00

Saturday 08:00 to 18:00

Sunday Day Off

paid $13.00 per hour for 40 hour workweek. Overtime is paid time-and-a-half. This week he will work the hours shown. Calculate his gross income for the week.

8. Mrs. Rodriguez has been offered a job as a technical sales representative with Cyber MultipleChoice Software where four different salary options were presented. The payment option which

would generate the highest gross income is

A. A yearly salary of $65 800B. A weekly wage, payable for 52 weeks, of $25 per hour for a forty hour work week

with overtime, averaged at eight hours per week, calculated at time and a half.C. A monthly income of $5 750.

D. $25 per hour for a forty-eight hour week plus 2% commission on sales expected to average around $8 200 per week.

9. Mrs. Parker works as a Chief Administrative Officer with a gross annual income Numerical Response of $47 500. She works 46 weeks in a year. If she works an average of 40 hours per week,

her hourly rate of pay in dollars per hour, to the nearest dollar, is _____ .


Answer Key

1 . $5441.67 2 . $121.75 3 . $767 4. a) $385 b) $741.75 5. a) $310 b) $228.25 c) $221.80 d) $412.50 e) $360 6 . $2455 7 . $744.25 8 . D 9 . 2 6

450 Personal Finance Lesson #1: Income

Personal Finance Lesson #2:Payroll Deductions Part One - CPP, EI, etc.

Net Income

Job postings in the newspaper or on the internet usually list the income an employee will earn. This income is the gross income and is not the income the prospective employee will receive on their paycheck. The amount the employee actually receives (or the “take-home” pay) is the net income - the income left after deductions are taken off the gross income.

Payroll Deductions

There are many different types of payroll deductions which vary from year to year. The flowchart illustrates some of them.

Some Types of Payroll Deductions

GovernmentDeductions

Union, or,Professional Dues RPP, or,

RRSP

Dental PlansHealth Plans

Life Insurance Premiums

Government Deductions

• Canada Pension Plan (CPP)

The Canada Pension Plan is designed to provide an employee with pension income when he/she retires. The government collects money from both employee and employer at the same rate and puts it towards the employee’s future pension. In the year 2000 the government collects CPP contributions from an employee at the rate of 3.9% of gross annual income between $3500.00 and $37 600.00. The government does not collect CPP from anyone who has a gross income of $3500 or less.

• Employment Insurance (E.I. Premiums) Employment Insurance is an insurance plan which provides the employee with some income during periods of unemployment. The government collects EI Premiums from both the employee and the employer (the employer contributes premiums at 1.4 times the amount charged to the employee). In the year 2000 the government collects EI premiums from the employee at the rate of 2.4% of the employee’s gross yearly income up to $39 000.

• Income Tax Income tax is a tax on the taxable income of an employee and is collected by the federal government to run the country. The employee pays income tax at both federal and provincial levels. We will discuss this topic in the next lesson.

Other Types of Payroll Deductions

• Union or Professional Dues • Child Care Expenses

• RPP (Registered Pension Plan) • RRSP (Registered Retirement Savings Plan) • Dental Plans, Health Plans, Life Insurance Premiums, etc.

Class Ex. #1 Calculate the annual CPP contribution for the following employees:a) Sheila who has an annual salary of $29 500

b) A student who works 15 weeks part time for $175 per week

c) A civil servant who earns $4760 per month.

d) A service manager who earns $37 600 per year.

Note The maximum CPP contribution for the year 2000 is ________

Class Ex. #2 Calculate the annual EI Premium for the following employees:

a) Ranjit who earns $35 236 per year.

b) A technician with a gross monthly salary of $4025

Note The maximum EI Premiums for the year 2000 is ________

452 Personal Finance Lesson #2: Payroll Deductions Part 1 - CPP, EI, etc.

Tax-Exempt Deductions

Tax-Exempt Deductions are deductions which are subtracted from a person’s gross income to determine the person’s taxable income.

The following are tax-exempt deductions:

• Union or Professional Dues• RPP• RRSP• Child Care Expenses

Taxable Income

Taxable Income is income on which income tax is calculated.

Taxable Income ==== gross income ---- tax-exempt deductions.

Class Ex. #3 Helen is a chemical engineer who works for a research chemical company. She has a gross monthly income of $5750. She pays union dues of $65 per month, contributes $500 per month to a Registered Retirement Savings Plan, has child care expenses of $200 per month, and her dental health care plan is $30 per month. Calculate her annual taxable income.

Personal Finance Lesson #2: Payroll Deductions Part 1 - CPP, EI, etc 453

Summary of Payroll Deductions

Government Payroll Deductions Year 2000

Canada Pension Plan (CPP) • 3.9% of gross annual income between $3500 and $37 600• maximum annual contribution is $1329.90

Employment Insurance (E.I. Premiums) • 2.4% of gross annual income up to $39 000• maximum annual contribution is $936.00.

Income Tax

Other Possible Payroll Deductions

Tax Exempt Not Tax Exempt

• Union or Professional Dues• RPP (Registered Pension Plan) • RRSP (Registered Retirement Savings Plan)• Child Care Expenses

• Dental Plans, Health Plans, • Life Insurance Premiums• others

Note The rates of CPP Contributions and EI Premiums change according to government policy. You may wish to use the current rates when doing the assignment questions. Current rates can be obtained by contacting the Government of Canada.

Government Payroll Deductions for the Year

Canada Pension Plan (CPP)

•

• maximum annual contribution is $

Employment Insurance (E.I. Premiums)

•

• maximum annual contribution is $



Assignment

1. Calculate the annual CPP contribution for the following employees:a) Mrs. Makowichuk who has b) Lauren, a junior accountant who

an annual salary of $27 750. works 25 weeks part time for $475 per week.

c) Mr. Marshall, a city manager, d) Mr. Foster, a zoning enforcement

who earns $2750 per month. officer, who earns $41, 500 per year.

2. Calculate the annual EI Premium for the following employees:a) Cheyenne, a teacher, b) Rebecca, a consultant, who earns a gross

who earns $38 350 per year. monthly salary of $2 850.

3. Siddhu, a web designer, has a weekly gross income $650. Find his weeklya) Canada Pension Plan contribution b) Employment Insurance Premium

4. Lee Huang, a Marketing Manager, worked 55 hours last week. He is paid $17.50 for the first forty hours and double time for overtime.a) Calculate his gross earnings for the week.

b) Assuming he works 55 hours per week for forty weeks and has 12 weeks vacation, calculate his annual CPP and EI deductions.

Personal Finance Lesson #2: Payroll Deductions Part 1 - CPP, EI, etc 455

5. Mr. McDaniel, a Project Manager, earns a gross monthly income of $5 250 per month. He pays union dues of $85 per month, contributes $750 per month to a Registered Retirement Savings Plan, and pays into a health plan $65 per month. Calculate his annual taxable income.

6. Luigi is a lawyer who has an annual salary of $83 000. He pays professional association fees of $120/month, contributes $1 150 per month to a RPP, and has life insurance premiums of $175 per month. His annual CPP contribution is $1 329.90 and his annual EI premiums are $936. Calculate his annual taxable income.

7. Marlyn is a loans officer at the local bank. She has a gross monthly income of $3 979 per Numerical Response month. She contributes $275 per month to a Registered Retirement Savings Plan, and has

weekly child care expenses of $62. Her monthly mortgage payment is $775. Her monthly taxable income, to the nearest dollar, is _____ . (Record your answer in the numerical response box from left to right)

Answer Key (based on the year 2000)

1.a) $945.75 b) $326.63 c) $1150.50 d) $1329.90 2.a) $920.40 b) $820.80 3.a) $22.73 b) $15.60 4.a) $1225 b) CPP $1329.90 EI $936.00

5 . $52 980 6. $67 760 7. 3 4 3 5


Personal Finance Lesson #3:Payroll Deductions Part Two - Income Tax

Income Tax

When calculating the federal and provincial income tax of an employee we need to know the following:

• the employee’s taxable income.• the employee’s tax credits. These include CPP contributions, EI premiums, medical

expenses, donations, and a basic personal amount. These tax credits reduce the amount of tax imposed on employees. Employees with a low income will therefore pay a small amount of tax. For the year 2000 the basic personal amount is $7131.

• the current rate of tax for each level of income.

Calculating an Employee’s Total Income Tax

The following procedure is a guide to determine an employee’s total income tax.

Step 1: Calculate the employee’s annual taxable incomeAnnual Taxable Income = Gross Income - Tax Exempt Deductions

Step 2: Calculate the employee’s tax creditsEmployee’s Tax Credits = CPP + EI Premium + Basic Personal Amount + others

Step 3: Calculate the employees Federal Tax Taxable

Income From T o Tax Rate

$0 $29590 17%$29590 $59180 25%$59180 29%

Federal Tax Rate Table for Year 2000

Province Provincial Tax RateAlberta 45.5% of Basic Federal TaxBritish Columbia 50.5% of Basic Federal TaxManitoba 52% of Basic Federal TaxSaskatchewan 50% of Basic Federal Tax

Provincial Tax Rate Table for Year 2000

using the Federal Tax Table for the current year. The Federal Tax is calculated on taxable income and not on gross income. Eg for a taxable income of $30 590 the federal tax would be 17% of $29 590 plus 25% of $1 000.

Step 4: Calculate the employee’s Basic Federal Tax

Basic Federal Tax = Federal Tax - 17% of Employee’s Tax Credits

Step 5: Use the basic federal tax to calculate the employee’s provincial tax using the appropriate provincial tax rate.

Step 6: Add the basic federal tax and the provincial tax to calculate the total income tax the employee must pay.

Net Annual Income - Take Home Pay

An employee’s net annual income is the income left over after all government deductions, income tax, and other voluntary deductions are subtracted from the gross income.

Net Annual Income = Gross Income - CPP - EI Premiums - Federal and Provincial Tax - other deductions

Class Ex. #1 Melissa works as a customs officer in Vancouver. In the year 2000 she earned a gross income of $67 400. She pays 4% of her gross income into an RRSP, and she has union dues of $50 per month. The union dues are a direct deduction from her salary, but the RRSP is purchased after receiving her take-home pay.a) Complete the steps below to determine her total income tax for the year 2000.

Step 1: Calculate Melissa’s annual taxable income.

Gross Income = Tax Exempt Deductions =

Annual Taxable Income = Gross Income - Tax Exempt Deductions

Step 2: Calculate Melissa’s tax credits.

CPP =

EI Premium =

Basic Personal Amount =

Melissa’s Tax Credits = CPP + EI Premium + Basic Personal Amount

Step 3: Calculate Melissa’s Federal Tax using the Federal Tax Table for the year 2000.

Step 4: Calculate Melissa’s Basic Federal Tax.

Melissa’s Basic Federal Tax = Federal Tax - 17% of Melissa’s Tax Credits

Step 5: Calculate Melissa’s provincial tax using the appropriate provincial tax rate.

Step 6: Calculate Melissa’s total income tax.

b) Calculate Melissa monthly take home pay if she has other deductions amounting to $32.40 per month.


458 Personal Finance Lesson #3: Payroll Deductions Part 2 - Income Tax

Assignment 1. Antonio, an animal control officer in Edmonton, had a gross annual income of $36 880 in

the year 2000. After receiving his take home pay he paid $170 per month into an RRSP, and had child care expenses of $150 per month, both of which are paid for after receiving his take-home pay.

a) Complete the steps below to determine his total income tax for the year 2000.

Step 1: Calculate Antonio’s annual taxable income.

Gross Income = Tax Exempt Deductions =

Annual Taxable Income = Gross Income - Tax Exempt Deductions

Step 2: Calculate Antonio’s tax credits.

CPP =

EI Premium =

Basic Personal Amount =

Antonio’s Tax Credits = CPP + EI Premium + Basic Personal Amount

Step 3: Calculate Antonio’s Federal Tax using the Federal Tax Table for the year 2000.

Step 4: Calculate Antonio’s Basic Federal Tax.

Antonio’s Basic Federal Tax = Federal Tax - 17% of Antonio’s Tax Credits

Step 5: Calculate Antonio’s provincial tax using the appropriate provincial tax rate.

Step 6: Calculate Antonio’s total income tax.

b) Calculate Antonio’s monthly take home pay if he has other deductions amounting to $27.50 per month.

Personal Finance Lesson #3: Payroll Deductions Part 2 - Income Tax 459

2. Dr. Fuller is a physician in Winnipeg and had a gross monthly salary of $8 650 in the year 2000. Her basic personal tax credit was $7131. Her monthly deductions included: union dues $98, RRSP $650, health insurance plan $75, dental plan $45. Her annual deduction for CPP was $1 329.90 and for EI was $936. She had an annual tax credit of $1 230 for allowable medical expenses. Calculate her total annual income tax.

3. John worked 52 hours last week. He is paid $6.50 per hour for the first 40 hours and time and a half for overtime.a) Calculate John’s gross earnings for the week.

b) Calculate his total deductions for the week using the following rates

CPP at 3.5% of gross earnings EI at 2.1% of gross earnings Income tax at 11% of gross earnings after CPP and EI have been deducted

c) Calculate John’s net pay for the week.


4. Scott is a computer analyst who is earning a monthly income of $6 800. Deductions from his salary are CPP, EI, and an RRSP contribution of 8% of his gross income. Calculate his net annual income if he has a total of $97.89 per month of other non tax-exempt deductions. Use the following data in your calculations.

Provincial Tax Rate Taxable

Income From T o Tax Rate$0 $31800 15%

$31800 $54600 22%$54600 $73000 29%$73000 35%

Federal Tax Rate Table= 10% of taxable income

Basic Personal Amount = $8240

CPP = 4.2% of gross annual income

between $4100 and $39 100 EI

= 1.9% of gross annual income up to $40 000

Basic Federal Tax = Federal Tax - 15% of Tax Credits

Personal Finance Lesson #3: Payroll Deductions Part 2 - Income Tax 461

Answer Key

1 . a) $6269.23 b) $2341.15 2 . $31 856.14

3 . a) $377.00 b) $60.27 c ) $316.73 4 . $49 883.42


Personal Finance Lesson #4:Bank Statements

Bank Statements

When you open a chequing account at a bank, you receive a cheque book with which to write cheques and/or a personal debit card to pay for purchases. At the end of each month you receive a financial statement listing the debits (money taken out of your account),credits (money paid into your account), and the balance (the total money in your account).

Class Ex. #1 Below is an example of Guido’s monthly statement sent by his bank.

Date Activity

Code

Description

of Activity

Wtihdrawn

from Your

Account

( D e b i t )

Deposited to

Your Account

( C r e d i t )

Balance

Apr 1 Opening Balance $2471.00Apr 1 DD Property Tax $143.50 $2327.50Apr 1 DD Cable T.V. $39.85 $2287.65Apr 1 DD Mortgage Payment $935.86 $1351.79Apr 5 CHQ Cheque #131 $653.25 $698.54Apr 5 DD Gas Company $68.00 $630.54Apr 7 DEP Deposit $389.99 $1020.53Apr 10 ATM Cash Withdrawal $200.00 $820.53Apr 13 CHQ Cheque #132 $25.00 $795.53Apr 15 PAY Salary $1422.75 $2218.28Apr 17 DD Util it ies $96.50 $2121.78Apr 25 DD Insurance $115.26 $2006.52Apr 26 CHQ Cheque #135 $1333.87 $672.65Apr 26 ATM Cash Withdrawal $200.00 $472.65Apr 29 CHQ Cheque #134 $36.50 $436.15Apr 30 PAY Salary $2735.36 $3171.51Apr 30 DD Car Payment $375.00 $2796.51Apr 30 INT Interest $0.37 $2796.88Apr 30 SC Service Charge $8.50 $2788.38

Apr 30 Closing Balance $2788.38

DD - Direct Debit CHQ - Cheque DEP - Deposit ATM - Teller MachinePAY - Salary Payment INT - Interest SC - Service Charge

DD - use of your bank card to pay for purchases in a store, to pay a bill at a bank machine, or a preauthorized bill payment taken from your accountCHQ - a cheque written by you resulting in a withdrawal from your accountDEP - money put into your account eg. a cheque made payable to youATM - a cash withdrawal from a bank machinePAY - a preauthorized salary deposit put directly into your accountINT - money deposited in your account depending on your balanceSC - a fee charged by the bank to manage your account

a) Explain why cheque #135 appears before cheque #134 on Guido’s bank statement.

b) Assuming that Guido wrote cheque #133, why do you think it does it not appear on his April bank statement?

Reconciling a Bank Statement

Reconciling a bank statement is the process of checking your own personal records against the statement from the bank. If the two balances are the same then they are reconciled.

Steps in Reconciling a Bank Statement

• Take the statement balance and subtract any outstanding debits and add any outstanding credits.

• Take your own personal record and subtract any service charges and add any interest payment.

• The balance on your personal record should be the same as the adjusted bank statement balance. If there are any discrepancies check your personal records for errors, and if necessary contact your bank for clarification or correction.

Class Ex. #2 Guido keeps a record of his banking activities.

Date Cheque

Number

Description

of Activity

Debi t C r e d i t Balance

Opening Balance $2471.00DD Property Tax $143.50 $2327.50DD Cable T.V. $39.85 $2287.65DD Mortgage Payment $935.86 $1351.79DD Gas Company $68.00 $1283.79DD Util it ies $96.50 $1187.29DD Insurance $115.26 $1072.03DD Car Payment $375.00 $697.03

Apr 2 # 1 3 1 Auto Repair $653.25 $43.78Apr 7 Insurance Coverage Rebate $389.99 $433.77Apr 10 Cash $200.00 $233.77Apr 11 # 1 3 2 Cancer Donation $25.00 $208.77Apr 15 Salary $1422.75 $1631.52Apr 23 # 1 3 3 Birthday (Trevor) $50.00 $1581.52Apr 23 # 1 3 4 Phone Bill $36.50 $1545.02Apr 23 # 1 3 5 Credit Card Payment $1333.87 $211.15Apr 26 Cash $200.00 $11.15Apr 30 Salary $2735.36 $2746.51

a) Guido noticed that the closing balance on his personal record is not the same as the closing balance on his bank statement. Should he contact his bank immediately?

b) Compare Guido’s personal record with the bank statement by using a check mark on both. Are there any transactions missing?

c) Guido checked there were no calculation errors on either statement. Reconcile his account for the month of April.

464 Personal Finance Lesson #4: Bank Statements

Class Ex. #3 Guido is currently paying a flat fee of $8.50 per month for service charges. The bank offers an alternative monthly plan of

• $0.70 per cheque.

• $0.50 per ATM transaction

• $1.10 maintenance fee (covering all direct debits and deposits)

If April is a typical month for Guido’s bank transactions, should he consider switching plans?


Assignment

1. a) Reconcile Logan’s personal record with his accompanying bank statement.

Personal RecordDate Cheque

Number

Description

of Activity

Debi t C r e d i t Balance

Opening Balance $875.06

DD Cable T.V. $53.45

DD Utilities $87.80

May 1 #28 Phone Bill $35.45

May 3 Cash $200.00

May 5 #29 Groceries $275.24

May 6 Cash $100.00

May 7 Cheque from Dad $50.00

May 7 #30 VCR $99.99

May 16 Salary $1727.00

May 20 Cash $200.00

May 21 #31 Magazine Subscription $65.89

May 21 #32 Rent $750.00

Bank StatementDate Code Transaction Type Debi t C r e d i t Balance

May 1 Opening Balance $875.06May 3 CHQ Cheque #28 $35.45 $839.61May 3 ATM Cash Withdrawal $200.00 $639.61May 4 DD Cable TV $53.45 $586.16May 6 ATM Cash Withdrawal $100.00 $486.16May 7 DEP Deposit $50.00 $536.16May 9 CHQ Cheque #30 $99.99 $436.17May 14 CHQ Cheque #29 $275.24 $160.93May 16 PAY Deposit $1727.00 $1887.93May 20 ATM Cash Withdrawal $200.00 $1687.93May 25 CHQ Cheque #32 $750.00 $937.93May 29 DD Util it ies $87.80 $850.13May 31 INT Interest $0.17 $850.30May 31 SC Service Charge $6.50 $843.80

May 31 Closing Balance $843.80

Personal Finance Lesson #4: Bank Statements465

b) Logan is currently paying a flat fee of $6.50 per month for service charges. The bank offers an alternative monthly plan of• $0.65 per cheque.

• $0.40 per ATM withdrawal

• $1.00 per month maintenance fee

Would Logan have been better off with this alternative plan for the month of May?

c) Describe circumstances when this alternative plan would not be the better alternative.

Answer Key

1 . b) Yes, by $1.05

c ) If he wrote more than 8 cheques or had more than 13 ATM withdrawals or some combination of both that would exceed $5.50.

466 Personal Finance Lesson #4: Bank Statements

Personal Finance Lesson #5:Investing Money - Simple Interest and Compound Interest

Warm-Up #1

If you deposit money in a bank account, you are in effect lending money to a bank. In exchange the bank pays you interest. There are two types of interest:

Simple Interest and Compound Interest.

Simple Interest

Simple Interest is usually applicable to short term investments of one year or less or to longer term investments where the annual interest is paid to the investor and not reinvested.

Warm-Up #2 Exploring Simple Interest

a) If you invest $500 which earns interest at a rate of 6% per year, how much interest would you earn in:

i ) one year i i ) half a year iii) one month?

b) If r is the annual interest rate (expressed as a decimal) and $P is the initial investment, calculate how much interest would be earned in:

i ) one year i i ) half a year iii) t years.

The formula to calculate simple interest is:

I ==== Prtwhere

I represents the amount of interestP represents the principal (the initial investment)r represents the annual rate of interest - expressed as a decimalt represents the time in years for which the money is invested

Class Ex. #1 Millie invests in $2350.00 at 7% per year for 6 months. Calculate:

a) the simple interest b) the amount of the investment at the end of six months.

Class Ex. #2 Wes invested $3000 for 4 months and received $66 interest. What was the annual rate of interest?

Compound Interest

In simple interest the principal at the beginning of the second year is the same as the principal at the beginning of the first year.

In compound interest the interest earned during the first year is added to the original principal to form a new principal.

To understand the comparison between simple interest and compound interest do the following Warm-Up.

Warm-Up #3 Exploring Compound Interest

A bank offers two types of savings bonds:

• Regular Savings Bond which pays simple interest at 6% per year

• Compound Savings Bond which pays interest at 6% per year compounded annually.

a) Complete the following tables to find the value of each bond at the end of 8 years.

Year Principal ($) Interest ($) Value ($)

1 5000 5000(0.06)(1) = 300 5300

2 5000 5000(0.06)(1) = 300 5600

3 5000 5000(0.06)(1) = 300

4

5

6

7

8

Simple

Interest

Table

468 Personal Finance Lesson #5: Investing Money - Simple and Compound Interest

Year Principal ($) Interest ($) Value ($)

1 5000 5000(0.06)(1) = 300 5300

2 5300 5300(0.06)(1) = 318 5618

3 5618 5618(0.06)(1) = 337.084

5

6

7

8

Compound

Interest

Table

b) Plot the data from both tables on the grid below

5000

6000

7000

8000

9000

10 000

11 000

1 5 10 122 3 4 6 7 8 9 11 13 14 15 16 17 18

Growth of Savings Bonds

Time (Years)c) Use the graph to answer the following:

i ) Extrapolate to determine the approximate value of each bond after 10 years.

i i ) Estimate the time needed for the simple interest bond to double in value.

i i i) Estimate the time needed for the compound interest bond to double in value.

Note The graph shows that the simple interest bond is growing in a linear pattern and the compound interest bond is growing more quickly, or exponentially.

Personal Finance Lesson #5: Investing Money - Simple and Compound Interest 469

Using this method to calculate compound interest is tedious. In the next exploration we will develop a formula for compound interest.

Warm-Up #4 Developing a Formula for Compound Interest

a) Let A = the final amount after the compound interest has been calculated. To calculate the final amount, we will need to add the principal and compound interest. Complete:

End of Year 1: Value of Bond = Principal + Interest

= 5000 + 5000(0.06)

= 5000[1 + 0.06] factor out 5000

= 5000(1.06)


= 5000(1.06) + 5000(1.06)(0.06)

= 5000(1.06)[1 + 0.06] factor out 5000(1.06)

= 5000(1.06)(1.06)

= 5000(1.06)2


=

= factor out

=

= b) To derive a general formula for compound interest, complete the following:

End of Year 1 Value of bond = 5000(1.06)End of Year 2 Value of bond = 5000(1.06)2

End of Year 3 Value of bond = 5000(1.06)3

End of Year 4 Value of bond =

End of Year n Value of bond =

c) Complete the following:If A = the final amount (or the value of the bond), P = the initial principal, i = the annual interest rate (expressed as a decimal), and n = the number of years, we have the formula:

A=


Note In the previous explorations, interest is compounded on an annual basis. But in practice, compounding can take place over any period of time, eg semi-annually, monthly, daily, continuously, etc.

Compound Interest Formula

The formula which can be used to calculate compound interest is

A ==== P(1 ++++ i)n

where,

A represents the final amountP represents the initial principali represents the interest rate per compounding periodn represents the number of compounding periods.

Note • The compounding period is also known as the conversion period.• Note that i does NOT always represent the annual interest rate.• Note that n does NOT always represent the number of years.

Class Ex. #3 $1000 is invested for 5 years at an annual interest rate of 6%. Complete the table to calculate the final value of the investment if interest is compounded according to the period of time given in the table.

Compounding Period

Number of Compounding

Periods Per Year

Total Number of

Compounding Periods

Interest Rate per

Compounding Period

Formula Amount

Annually

Semi-Annually

Quarterly

Monthly

Class Ex. #4 $7000 is invested in a 6 year GIC compounded quarterly at a rate of 5% per annumDetermine the value of the investment at the end of the term.


Class Ex. #5 Superbank and Bank for More offer the following investment opportunities for an initial investment of $10 000:

• Superbank pays interest at an annual rate of 7.3% compounded annually. • Bank for More pays interest at an annual rate of 7.2% compounded monthly.

a) Which bank provides the greater interest at the end of 1 year?

b) Which bank provides the greater interest at the end of 10 years?

c) How much interest did each investment earn at the end of thirty years?

Class Ex. #6 Christine invested $2500 for 4 years compounded semi-annually and received $843.26 interest. What was the annual rate of interest to the nearest tenth of a percent?

Class Ex. #7 Barbara wants to invest some money so that she will have $8000 in 5 years. The bank offers an annual rate of 7% compounded quarterly. How much should her initial investment be?



Assignment

1. Calculate the simple interest in each case:

a) $740 is invested at 6% per annum for six months.

b) $1500 is invested at 8%/a for 3 months.

2. Determine the annual rate of interest in each case:

a) Joan invested $2000 for six months and received $45 simple interest.

b) Jason invested $500 for eight months and received $25 simple interest.

3. Determine the length of the investment in each case:

a) Orlando invested $4000 at 6% per annum and received $40 simple interest.

b) Mariko invested $800 at 5% p.a. and received $20 simple interest.

4. If the annual rate of interest is 9% per annum, state the interest rate per compounding period and the total number of compounding periods in each case:

a) compounded semi-annually for 4 years

b) compounded quarterly for 3 years

c) compounded monthly for 41 2 years

d) compounded annually for 6 years

5. $4000 is invested for 4 years at an annual interest rate of 7.2%. Complete the table to calculate the final value of the investment if interest is compounded according to the period of time given in the table.

Compounding Period


Periods Per Year

Total Number of

Compounding Periods

Interest Rate per

Compounding Period

Formula Amount

Annually

Semi-Annually

Quarterly

Monthly


6. $8000 is invested in an RRSP for 7 years compounded quarterly at a rate of 9.2% per year. Determine the value of the investment at the end of the term.

7. Mrs. Sim wanted to invest some money so that her daughter will have $20 000 for college education in ten years time. The bank offers an annual rate of 6.8% compounded semi-annually. How much should her initial investment be? Answer to the nearest dollar.

8. Levi invested $3400 for 5 years and received interest compounded quarterly. If, at the end Numerical Response of the 5 years, the investment was worth $4670.39, the annual rate of interest, to the nearest

tenth of a percent, is _____ .(Record your answer in the numerical response box from left to right)

9. Patrick invested $1500 for 3 years and received interest compounded monthly. If he received $295.02 in interest at the end of the term, then the annual rate of interest, to the nearest tenth of a percent, is _____ .(Record your answer in the numerical response box from left to right)

Answer Key 1 . a) $22.20 b) $30 2 . a) 4.5% b) 7.5% 3 . a) 2 months b) 6 months4 . a) 4.5%, 8 b) 2.25%, 12 c ) 0.75%, 54 d) 9%, 65 .

Compounding Period


Periods Per Year

Total Number of

Compounding Periods

Interest Rate per

Compounding Period

Formula Amount

Annually 1 4 7.2% $5282.50

Semi-Annually 2 8 3.6% $5308.09

Quarterly 4 16 1.8% $5321.38

Monthly 12 48 0.6% $5330.44

A = 4000(1.072)4

A = 4000(1.036)8

A = 4000( 1.018)16

A = 4000( 1.006)48

6 . $15 121.95 7 . $10 248 8. 6 . 4 9. 6 . 0


Personal Finance Lesson #6:Using TVM (on a Calculator) to Solve Investment Problems

Some graphing calculators like the TI-83 Plus have capabilities to analyze financial calculations such as annuities, investments, savings, loans, mortgages and leases.

The TVM Solver (time-value-of-money solver) function on the TI-83 Plus Graphing calculator is one such function which can carry out financial calculations.In this lesson we will refer to some of the Class Examples in Lesson #5.

Consider Class Ex. #4 Lesson #5:$7000 is invested in a 6 year GIC compounded quarterly at a rate of 5% per annum

Determine the value of the investment at the end of the term.Step 1: Access the “Finance” functions of the calculator by pressing APPS .

Step 2: Access the TVM Solver by pressing 1 or Enter . A screen similar to the one below will appear. N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGINEND

N represents the total number of payment periodsI% represents the annual interest ratePV represents the present value, or inital valuePMT represents the payment amountFV represents the future valueP/Y represents the number of payment periods per yearC/Y represents the number of compounding periods per yearPMT: END BEGIN represents the timing of the payments

Step 3: Enter the given values using the Enter key or arrow keys after each entry.N = 6x4 (After pressing the Enter key, this will change to 24)I% = 5PV = –7000 (See the note below as to why this is a negative value)PMT = 0 (Since no additional payments are made after the initial investment)FV = 0 (See Step 4 below)P/Y = 4 (If PMT = 0, set P/Y = C/Y)C/Y = 4PMT: END BEGIN (Since no additional payments use calculator default END)

Note For the TI-83 Financial functions, you must enter:

• cash received as positive values (money entering your “pocket”)

• cash paid as negative values (money leaving your “pocket”)The amount $7000 is entered as a negative value because it is the initial principal that has been paid from you to the financial institution.

Step 4: The final value of the bond is represented by FV. Although FV is not known at this time, the calculator will not proceed unless a value is entered - use 0. To find the final value of the bond place the cursor on the value of FV and then press Alpha Enter to solve for the final amount. An indicator square on the left of FV indicates the future value has been determined.

Final value: _______________ Compare the answer with Class Ex. #4

In the compound interest formula A = P(1 + i)n

A is represented by FV on the calculatorP is represented by PV on the calculatorn is represented by N on the calculator

Solving for Different Variables using TVM Solver

Step 1: Access the TVM Solver under the Finance Menu

Step 2: Enter in:• the given values• a zero for the variable to be solved• the timing of the payment (END or BEGIN)

Step 3: Place the cursor on the TVM variable to be solved and press Alpha Enter .

An indicator square on the left of the variable indicates the variable solved.

Class Ex. #1 Verify the answers from the following Class Examples using N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

the TVM Solver.

a) Class Ex. #6 Lesson #5

Christine invested $2500 for 4 years compounded semi-annually and received $843.26 interest. What was the annual rate of interest to the nearest tenth of a percent?

b) Class Ex. #7 Lesson #5 N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

Barbara wants to invest some money so that she will have $8000 in 5 years. The bank offers an annual rate of 7% compounded quarterly. How much should her initial investment be?

Class Ex. #2 Mr. Rate decided to invest $10 000 in a RRSP which pays 6% N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

compounded quarterly. He wants to double his investment . How many years (to the nearest quarter of a year) will it take Mr. Rate to achieve his goal?

476 Personal Finance Lesson #6: Using TVM to Solve Investment Problems

Effective Annual Interest Rate

If you invest $100 at 8% per annum compounded annually, the interest received would be $8, which is 8% of $100.

If you invest $100 at 8% per annum compounded semi-annually, the interest received would be $8.16, so in effect the interest received is 8.16% of $100.

This rate is called the effective annual interest rate.

A nominal interest rate of 8% per annum compounded semi-annually is equivalent to an effective annual interest rate of 8.16%.

Class Ex. #3 Calculate the effective annual interest rate for a nominal interest rate of 5% compounded semi-annually, using:

a) the compound interest formula A = P(1 + i)n.

N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

b) TVM Solver

c) the effective annual interest rate key, found under APPS , Finance, scroll to C, ENTER

Eff (rate per annum, number of compounding periods)


Personal Finance Lesson #6: Using TVM to Solve Investment Problems 477

Assignment Use TVM on a calculator to answer the following questions

1. Find the amount of each investment, to the nearest cent, given the initial investment, the annual rate of interest, the compounding period, and the length of the investment.

a) An investment of $1 000 b) An investment of $7 500 at 8% per annum at 4.8% per annumcompounded annually compounded monthlyfor 5 years. for six years.



2. How much money, to the nearest dollar, would you need to invest today, given the following annual interest rates, compounding periods, and length of the investment, for the investment to accumulate at least $10 000?

a) 5.4% per annum, compounded b) 6.2% per annum, compounded quarterly, for four years. semi-annually, for three years.



3. Determine the annual rate of interest, to the nearest tenth of a percent, given the initial investment, the final amount, the compounding period, and the length of the investment.

a) An investment of $2 000 b) An investment of $3 500 compounded semi-annually compounded monthlyfor 5 years to produce for 2 years to producea final amount of $2 850 a final amount of $4 000




4. Determine the annual rate of interest, to the nearest tenth of a percent, given the initial investment, the interest, the compounding period, and the length of the investment.

a) An investment of $6 000 b) An investment of $75 000 compounded quarterly compounded semi-annuallyfor 3 years to produce for 4 years to produce$985 interest. $25 000 interest.



5. Determine the length of the investment, given the initial investment, the final amount, the annual interest rate, and the compounding period.

a) An investment of $9 000 b) An investment of $450 compounded quarterly compounded dailyat 7% /a to produce at 7.3% per year to producea final amount of $12 300 a final amount of $475(Answer to the nearest (Answer to the nearest day)

quarter of a year)



6. $8 250 is invested in an RRSP for 7 years compounded quarterly at a N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

rate of 7.8% per year. Determine the value of the investment at the end of the term.

7. Jill invested $2 575 for three years and received interest compounded N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

monthly. If, at the end of the three years the interest accumulated was $447.95, determine the annual rate of interest to the nearest hundredth of a percent.


8. Mr. Forsythe wanted to invest some money so that his grandson will N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

have at least $30 000 for college education in eighteen years time. The bank offers an annual rate of 4.92% compounded semi-annually. How much should his initial investment be? Answer to the nearest dollar.


It has been determined that the time taken for money to double in value can beapproximated by the “The Rule of 72”. This rule states that for money compoundedannually, the annual interest rate, multiplied by the number of years is approximately 72.

9. Investigate using an initial investment of $100.

“Rule of 72” TVM SolverAnnual Interest Rate Time (years)

9%

12

4%

Annual Interest Rate Time (years)

9%

12

4%

10. a) Use the “Rule of 72” to estimate the number of years it would take for an investment of $1000 to amount to $8000 if the annual interest rate was 7.2%.


b) Check the accuracy of the answer in a) using TVM Solver.


11. Determine the effective annual interest, to the nearest thousandth of a percent, in each of the following investments:

a) a nominal interest rate of 5.2% per annum compounded semi-annually

b) a nominal interest rate of 6.8% per annum compounded quarterly

c) a nominal interest rate of 6% per annum compounded monthly.

12. An investment compounded quarterly has an effective annual interest rate of 7.1859%. Numerical Response Determine the nominal interest rate using the “Nom(” feature found under the

FINANCE menu. The nominal interest rate, to the nearest tenth of a percent, is _____ .(Record your answer in the numerical response box from left to right)

Answer Key 1 . a) $1 469.33 b) $9 997.43 2 . a) $8 070 b) $8 327

3 . a) 7.2% b) 6.7% 4 . a) 5.1% b) 7.3%

5 . a) 18 quarters or 4 12 years b) 270 days 6 . $14 167.60 7 . 5.36%

8 . $12 508 9 . 8 years, 8.04 years 6%, 5.946% 18 years, 17.673 years

1 0 . a) 30 years b) 29.9 years 1 1 . a) 5.268% b) 6.975% c ) 6.168%

1 2 . 7 . 0


Personal Finance Lesson #7:Annuities

Annuities

An annuity is a series of equal payments made at regular time intervals.

There are different forms of annuities. For example;

• A regular savings plan can be considered as an investment annuity. • Paying off a loan with regular payments is a form of an annuity. • Retirement income is often received in the form of an annuity.

Class Ex. #1 Two grade 11 students, Mark and Robin, were asked to solve the following problem.

Susan invests $800 in a savings account at the beginning of each quarter. If the account pays interest at 6% per year compounded quarterly, what is the value of the investment at the end of one year?

Mark choose to use the compound interest formula A = P(1 + i)n to determine the amount of each quarterly deposit. Susan choose to use the TVM Solver on her calculator. Complete each of their solutions.

Mark Investment

DateP i n A

Jan 1 $800 0.06 ÷ 4 = 0.015 4 $800(1.015) =

April 1 $800 0.015 3

July 1

October 1

Total

4

Robin

PMT = –800 since money is leaving her “pocket”PV = 0 since there is no initial payment over and above the regular payments



Class Ex. #2 A university graduate at the age of 25 has $1 000 in a savings fund. N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

He has recently found employment in his profession and decides to invest $250 at the beginning of each month into the savings fund paying 5.5% interest per annum compounded quarterly.

a) At this rate of interest, how much will he have saved by the time he retires at the age of 55?

b) How much of this saving is interest?

Class Ex. #3 Banker Bill wanted to set up a fund for his granddaughters Becky and Bonnie. The girls had to choose between the following two options:

Option A: Two thousand dollars will be invested on each birthday between age 27 and age 64 inclusive. The money will stay in your account until age 65.

Option B: Two thousand dollars will be invested on each birthday between age 19 and age 26 inclusive. The money will stay in your account until age 65.

Both options will receive interest at the rate of 10% per year compounded annually.

Becky recognized that more money will be invested under Option A and chose this option.

Bonnie chose Option B.

a) How many payments were made on Becky’s behalf? How much money was invested

for her?

b) How many payments were made on Bonnie’s behalf? How much money was invested for her?

c) At age 65 what was:

i ) Becky’s investment worth? i i ) Bonnie’s investment worth?



N=I%=PV=PMT=FV=P/Y=C/Y=PMT: BEGIN END

d) What conclusion can you draw from this class example?


484 Personal Finance Lesson #7: Annuities

Assignment

1. Assunta deposited $300 into her savings account at the beginning of N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

each month for a period of two years. If the account pays interest at 6.2% per annum compounded monthly, how much interest has she earned at the end of the two years?

2. On February 1, Nicole had $3 500 in an RRSP account. She decided N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

to make monthly payments of $100 into that account starting on that date. If the rate of interest is 5.8% per annum compounded quarterly,determine:

a) the value of the RRSP after three years of payments have been made

b) the interest earned.

3. Gabriel invested $6 500 in a GIC for a three year term at 6.6% compounded semi-annually. At the end of the term, he transferred the money into a savings account paying 5.7% per annum compounded monthly. He also made monthly payments of $250 into the savings account starting on the day he opened the account. What was the value of his investment four years after the savings account was opened?



Personal Finance Lesson #7: Annuities 485

4. Greg and Rose decided to invest $1000 into an education fund for their son Luke on his second birthday and on every birthday thereafter until age 18 inclusive. The fund earns interest at 7.1% per annum compounded semi-annually.

a) How much money will Luke have on his 19th birthday to pay for N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

his post secondary studies?

b) If Luke decides to leave the money in the fund until his 21st N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

birthday, how much money will be in the fund?

c) If Luke leaves the money in the fund until his 60th birthday, what N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

would be the value of his parents’ investment for him? How does this compare with the amount of money initially invested?

Answer Key

1 . $483.95 2 . a) $8099.09 b) $999.09 3 . $23 421.524 . a) $33 746.36 b) $38 799.61 c ) $589 570.03 compared to an initial investment of $17 000.

486 Personal Finance Lesson #7: Annuities

Personal Finance Lesson #8:Loans and Consumer Credit

Loans

When you want to purchase an item or pursue an idea which requires money you do not have, then you may apply for a loan. A loan is an agreement for a specified period between an institution which lends money (usually a bank) and a borrower.

To borrow the money, the consumer will be charged interest by the lending institution.

Using TVM on a Calculator to Solve Loan Problems

In the previous lessons we have used the TI-83 Plus to analyze financial calculations such as investments and savings. We can also use TVM Solver of the TI-83 Plus to analyze loans and mortgages. The steps are outlined in the example below.

Consider the Following ExampleMr. Cruiser is borrowing $5700 from a bank to buy a used car. The bank which is issuing him the loan is charging 10.5% interest per year compounded semi-annually. If he makes $150 monthly payments at the end of each month for three years, how much of the loan does he have left to repay?

Step 1: Access the TMV Solver N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGINEND


Step 2: Enter the given values.N = 3x12 (Since the borrower is making monthly payments for three years. The value will

change to 36 after you press Enter)I% = 10.5PV = 5700 (This is a positive value since the borrower is receiving cash from the bank)PMT = –150 (This is a negative value since the borrower is paying cash to the bank)FV = 0 (The amount the borrower still owes the lending institution - see step 3)P/Y = 12 (The number of payments made in one year)C/Y = 2 (The number of compounding periods per year)PMT: END BEGIN (END)

Step 3: • The final value of the bond is represented by FV. Although FV is not known at this time, the calculator will not allow you to proceed unless a value is entered - use 0. To find the final value of the loan place the cursor on the value of

FV and then press Alpha Enter to solve for the final amount. • A negative FV value indicates that the money has still to leave his “pocket” and

is therefore the amount left to be repaid.

Amount left on the Loan: _______________

Class Ex. #1 Darma Asphalt applied for a loan to have her basement developed. The N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

contractors she hired estimated that it would cost her approximately $13 700 to develop the basement. Darma had already saved $7 000 towards home improvements. She received a loan for the remaining amount and paid it off in four years with payments of $175 made at the end of each month. What was the annual rate of interest (to the nearest tenth of a percent) if the interest was compounded quarterly?

Class Ex. #2 Rod is borrowing $7280 from a bank to purchase a boat. The bank N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

which is issuing him the loan is charging 9.8% interest per year compounded semi-annually. He is making payments of $250 at the beginning of each month.

a) How many years and months will it take him to pay off the loan?

b) How much did Rod’s boat really cost him?

c) How much interest will Rod have paid for his loan?


488 Personal Finance Lesson #8: Loans and Consumer Credit

Consumer Credit

Consumer credit occurs when an item is purchased and payment for it is made at a later date. There are often interest charges which result in the consumer paying more for the item than the original purchase price.

The amount of interest paid is called the finance charge.

Class Ex. #3 Ginger buys a drum set advertised at $2950.00. She arranges to pay for it in 24 monthly installments of $170.00.

a) How much did Ginger pay for the drum set?

b) Calculate the finance charge.

Class Ex. #4 Chad sees the following T.V. advertisement:

Home Theatre System:

Option A: $5000 now

Option B: No payment for 12 months then 36 monthly payments of $205

Option C: 12.5% down plus 48 payments of $130

Chad only has $1000 available. Which option is less costly?

Personal Finance Lesson #8: Loans and Consumer Credit 489

Credit Cards

Most people have a credit card which allows them to access consumer credit. In effect, using a credit card to pay for purchases is similar to borrowing money. Although similar to loans and mortgages, there are major differences in using a credit card.

• If you choose to pay off the balance on the card by an agreed date then there is no interest charge.

• If you choose not to pay for the item by the agreed date, then interest is charged and the item(s) you purchased ends up costing more because of the interest charges.

• The interest rate on purchases made by credit card is a great deal more than the interest rate on a small loan or mortgage.

• Unlike a small loan or mortgage, the consumer must make a minimum payment each month and then has the option of paying off as much of the remainder of the credit card balance as desired.

• You can use a credit card to borrow money in the form of cash, but interest is charged immediately at a much higher rate than a bank loan.

Credit Card Charges

Well-known Canadian credit card companies are similar in the way they apply credit card charges.

An example of monthly credit charges is given below. This will vary from credit card company to credit card company.

• No interest is charged if the entire balance is paid within 25 days of the billing date (except for cash advances).

• Interest is charged at a rate of 18% per annum on the entire previous monthly balance if the entire balance is not paid by the due date.

• Interest on balance owing is calculated from the date the transaction was posted until, but not including, the current statement date. To stop interest accumulation, contact the credit card company for your payout balance for that day and make the full payment.

• Interest on cash advances is calculated from the day the money was withdrawn.

• The minimum monthly payment is 3% of the statement balance or $10, whichever is greater.

Note The class examples and assignment questions use the above credit card charges.


Class Ex. #5 Matt received the following credit card statement.

Date of

T r a n s

Date of

Post

Transact ion Debits/

Credits (-)

Previous Balance $1728.441 1 / 2 3 1 1 / 2 5 Electronics Extraordinaire $858.561 1 / 2 4 1 1 / 2 6 Dairy King $14.251 1 / 2 8 1 1 / 3 0 Hurlies Gas $32.711 2 / 0 7 1 2 / 1 0 Shoppers Paradise Dept Store $31.261 2 / 1 4 1 2 / 1 4 Payment Thank-you $1728.44-1 2 / 2 0 1 2 / 2 1 Groceries for Less $325.79

Statement

Date

Payment

Due Date

Minimum

Payment

Overdue

Payment

Total Minimum

Payment Due

Dec. 23 Jan 16 $37.88 $0.00 $37.88

New Balance

a) Explain the entry $1728.44 –.

b) Calculate the new balance.

c) Explain how the entry $37.88 was determined.

d) Why may the date of posting be different from the date of transaction?


Assignment

1. Jody has saved $2200 to put towards the cost of a new car. The car she plans to buy costs $19 757 and she needs to take out a loan to pay the balance. The loan must be paid off in four years and interest is charged at 8.2% per annum compounded semi-annually.


a) Determine the payment she requires to make at the end of each month.

b) How much did the car cost her?


2. Lee had to borrow $7500 to build a new deck. He could only afford payments of $250 at the beginning of each month. The bank charges interest at 7.8% per annum compounded annually.

a) How much will he owe after 2 years? b) In which month will he finally pay off the loan?



c) If Lee was required to make a full payment in the last month, how much interest did he pay for the loan?

3. Home Audio has a special on Kyoto home theatre systems. The system can be purchased for $8788 or with a down payment of 20% followed by 36 monthly payments of $249. The Carter family wants to buy one of these systems, but can only afford the down payment. Before agreeing to the Home Audio deferred payment plan, Mr. Carter goes to his local bank to inquire about a loan for the remaining amount. The bank offers Mr. Carter a 3 year loan at 8.5% per annum compounded quarterly. If the bank payments are made at the beginning of each month, which of the two methods is more economical?


4. Jane sees the following ad in

$2499.99 or 15% down + 12 payments of $199.99

a local newspaper.

a) How much is the down payment?

b) What is the total cost of the computer using the instalment method?

c) Calculate the difference between the two payment options.

5. Isabel received the following credit card statement.

Date of

T r a n s

Date of

Post

Transact ion Debits/

Credits (-)

Previous Balance $0.000 6 / 3 0 0 7 / 0 2 Theatre Tickets $107.000 7 / 0 4 0 7 / 0 4 Robert's Diner $27.890 7 / 0 9 0 7 / 1 0 Quik Gas $33.710 7 / 1 6 0 7 / 1 7 Riverfront Grocery Store $178.950 7 / 1 6 0 7 / 1 6 Big Al's Towing and Repair $493.260 7 / 1 8 0 7 / 2 0 Simpson's Electronics $38.71

Statement

Date

Payment

Due Date

Minimum

Payment

Overdue

Payment

Total Minimum

Payment Due

New Balance

July 26 A B $0.00 C

a) Calculate the new balance.

b) Determine the entries for positions A, B, C.

6. Pierre has a balance owing of $37.58 on his credit card. What is his minimum payment?


7. A certain type of projection television can be purchased for the advertised price of $3500 or by a down payment of 20% of the advertised price plus 25 monthly instalments of $135. Find the difference in cost between the two methods and express the difference as a percentage, to the nearest 1%, of the advertised price.

Questions #8 and #9 refer to the following information. Hale buys a set of golf clubs advertised at $1 050. He makes a $200 down payment and pays 36 monthly installments of $28.50

8. The amount, to the nearest dollar, that Hale paid for the golf clubs, is _____ .Numerical Response


9. The finance charge expressed as a percentage of the advertised price, to one decimal place,is _____ .(Record your answer in the numerical response box from left to right)

Answer Key 1 . a) $429.14 b) $22 798.72

2 . a) $2 221.27 b) 34 months c ) $1000

3 . The bank by $28.81 per month 4 . a) $375.00 b) $2 774.88 c ) $274.89

5 . a) $879.52 b) A - Aug. 19, B - $26.39, C - $26.39 6 . $10

7 . $575, 16% 8 . 1 2 2 6 9 . 1 6 . 8


Personal Finance Lesson #9:Mortgages and Property Tax

Mortgages

A mortgage is a special type of loan which is used to purchase property. The property itself is used as collateral - which the lending institution uses to recover its loan in the event that the borrower does not pay back the loan. The person borrowing the money is called the mortgagor and the institution lending the money is called the mortgagee.

Amortization

To amortize a mortgage is to repay a mortgage in equal periodic payments over a given period of time. This period of time is called the amortization period.

In practice most mortgage agreements are for a much shorter term - at the end of which the mortgage agreement is renegotiated with the lending institution.

The following amortization table gives the monthly payment for each $1000 of loan. The interest rate is the annual rate with interest being compounded semi-annually. Payments are made at the end of each month.

Note • By Canadian law, mortgage interest must be calculated annually or semi-annually. • In this lesson we will consider payments to be made at the end of the month unless

otherwise stated

Monthly Payments per $1000 of Mortgage

Interest Rate

5 Years 10 Years 15 Years 20 Years 25 Years

6.00% $19.30 $11.07 $8.40 $7.12 $6.406.25% $19.41 $11.19 $8.53 $7.26 $6.556.50% $19.53 $11.31 $8.66 $7.41 $6.706.75% $19.64 $11.43 $8.80 $7.55 $6.857.00% $19.75 $11.56 $8.93 $7.69 $7.007.25% $19.87 $11.68 $9.07 $7.84 $7.167.50% $19.98 $11.81 $9.21 $7.99 $7.327.75% $20.10 $11.94 $9.34 $8.13 $7.478.00% $20.21 $12.06 $9.48 $8.28 $7.638.25% $20.33 $12.19 $9.62 $8.43 $7.798.50% $20.45 $12.32 $9.76 $8.59 $7.95

Amortization Period

Class Ex. #1 Use the amortization table to calculate the monthly mortgage payment for each of the following mortgages.a) $120 000 at 6.75% per annum b) $240 850 at 7.5% per annum

amortized over 15 years. amortized over 25 years.

Class Ex. #2 Solve Class Ex. #1 using TVM Solver.

a) $120,000 at 6.75% per annum b) $240,850 at 7.5% per annumamortized over 15 years. amortized over 25 years.



b) Do the answers agree with the solution to Class Ex. #1?

c) Can you suggest why there might be discrepancies?

Class Ex. #3 Jimmy and Sue bought a house in Scenic View for $225,000. They had a 30% down payment and negotiated a three year mortgage at 8 1

4 % per annum compounded semi-annually. The mortgage is amortized over 20 years. Determine;

a) the amortization period b) the term of the mortgage

c) the amount of the down payment. d) the amount to be financed

e) the monthly mortgage payment N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

f) what their house would really cost if the interest rate remained the same throughout the amortization period.

496 Personal Finance Lesson #9: Mortgages and Property Taxes

Property Taxes

Owning property involves paying property taxes to the local municipality to help pay for services such as police, fire fighters, schools, roads, garbage collection etc.

The amount of property tax payable by an owner depends on the assessed value of the property and the mill rate set by the municipality. The mill rate is determined each year by the municipality when it sets its budget. Some municipalities list the mill rate as a tax rate.

If the municipality lists its taxes using a mill rate, then property tax is determined by:

Property Tax = assessed value1000 ¥¥¥¥ mill rate

If the municipality lists its taxes using a tax rate, then property tax is determined by:

Property Tax = assessed value ¥¥¥¥ tax rate ËÊÁÁTax rate =

mill rate1000

ˆ˜

Class Ex. #4 Calculate the annual property tax for Darcy’s ranch if:

a) in 2001 the assessed value was $384 325 and the mill rate was 11.230

b) in 2002 the assessed value was $391 250 and the tax rate was 0.010296.

Class Ex. #5 The mortgage term on the mall owned by XY Holdings Inc. has come up for renewal. The company negotiates a three year, $385 290 mortgage at 6.5% compounded semi-annually and amortized over twenty years. The mall has an assessed value of $1,945,525 and the mill rate is 17.5695. If the company pays its property tax and mortgage at the beginning of every month, calculate the monthly payment.


Personal Finance Lesson #9: Mortgages and Property Taxes 497

Assignment

1. Use the amortization table to calculate the monthly mortgage payment and the total cost for each of the following mortgages.

a) $215 000 at 7.25% per annum b) $179 578 at 7.75% per annumamortized over 10 years. amortized over 20 years.

2. Solve assignment question #1 using TVM Solver.

a) $215 000 at 7.25% per annum b) $179 578 at 7.75% per annumamortized over 10 years. amortized over 20 years.



3. Giselle bought a condominium for $102 500. She had a 15% down payment and negotiated a four year mortgage at 7.75% per annum compounded semi-annually. The mortgage is amortized over 25 years. Determine:

a) the term of the mortgage b) the amortization period

c) the amount of the down payment.

d) the monthly mortgage payment N=I%=PV=PMT=FV=P/Y=C/Y=PMT: END BEGIN

e) the total amount paid for the condominium if the interest rate remained at 7.75% throughout the amortization period.


4. A financial institution offers a $125 000 mortgage amortized over 20 years at 7% per year compounded semi-annually.

a) Determine the monthly payments.

b) Calculate the total amount to be paid.

c) Calculate the amount of interest to be paid.

d) What would happen to the monthly payments and the total to be paid if the mortgage was amortized over 25 years? (No calculations are required)

5. Aziz needs a $95 000 mortgage. He visited two banks and was given the following options:• Mega Bank: a two year term at 7.5% amortized over 10 years.• Bank for More: a five year term at 7.25% amortized over 15 years.

a) Which option has the lower monthly payment?Mega Bank Bank For More



b) Assuming the interest rate remains the same over the amortization period of each loan, determine the total amount of interest paid for each option.

c) Even though the monthly payments are higher, why might Aziz choose Mega Bank?

d) Even though the total amount of interest paid is higher, why might Aziz choose Bank for More?


6. In 1998 Mike and Barb moved into a new home and assumed a $135 000 mortgage amortized in the year 2013 at 6.5% per annum compounded semi-annually for a three year term. At the end of the three year term, the mortgage rate had increased by 1.5%. They renegotiated their mortgage for a further three year term at this new rate. Calculate the monthly payment for their renegotiated mortgage if the mortgage was still to be amortized in the year 2013.

7. Calculate the annual property tax of the following:

a) The assessed value of Henry’s home is $235 000 and the mill rate is 12.750 .

b) The tax rate of the building owned by Adi Corp is 0.019405 and the assessed valueis $1 759 000.

c) The assessed value of a hotel is $3 235 775 and the mill rate is 15.463 .

d) The tax rate of a home is 0.009894 and the assessed value is $159 000.


8. What is the tax rate for a house which is assessed at $145 500 and where the monthly property tax bill is $141.67?

9. Craig and Jennifer have arranged a four year $110 000 mortgage at 6.8% per annum compounded semi-annually and amortized over 18 years. Their house has an assessed value of $185 000 and the residential tax rate is 0.015718. If Craig and Jennifer pay their property tax and mortgage at the end of every month, calculate their monthly housing costs (mortgage payment plus property tax).

Answer Key

1 . a) $2,511.20 b) $1,459.97

2 . a) $2,512.19 b) $1,460.73

3 . a) 4 years b) 25 years c ) $15,375 d) $651.10 e ) $210,705

4 . a) $961.64 b) $230,793.60 c ) $105,793.60d) The monthly payments would decrease but the total to be paid would increase

5 . a) Mega Bank - $1 121.99 Bank for More - $861.49b) Mega Bank - $39 638.80 Bank for More - $60 068.20c ) The amortization period is less, so the total amount of interest to be paid is less.d) The monthly payments are less and he may not be able to afford a higher monthly

payment for Mega Bank

6 . Original monthly payment - $1 169.60, amount owed at the end of three years - $117 265.58,renegotiated monthly payment - $1,260.99

7 . a) $2,996.25 b) $34,133.40 c ) $50,034.79 d) $1,573.15

8 . 0.011684 9 . $878.24 + $242.32 = $1,120.56


Personal Finance Lesson #10:Foreign Exchange

Exchange Rates

The rate of exchange from one currency to another varies from day to day. Current rates can be obtained from the financial section of a newspaper, from a bank, from the internet, etc. In this lesson we will use the following rates of exchange.

Country Cur rency Canadian $

per unit

(buy C $)

Canadian $

per unit

(sell C $)

Australia Dollar 0.8196 0.8794

Chile Peso 0.00244 0.00257

Europe Euro 1.3392 1.4174

India Rupee 0.0275 0.02789

Japan Yen 0.009574 0.0102

Mexico Peso 0.1142 0.115

Russia Rouble 0.0404 0.0417

Switzerland Franc 0.8597 0.9097

United Kingdom Pound 1.9608 2.0384

United States Dollar 1.143 1.185

In the table above, the financial institution pays less if you are selling Canadian dollars than they charge if you are buying Canadian dollars. This is how the financial institution makes money. They may also charge a transaction fee.

In all the examples and assignment questions in this lesson, we will assume that the transaction fee is built into the exchange rate.

Note For example, the rate of exchange for converting Canadian dollars into US dollars (i.e. selling Canadian dollars) is:

• 1.1850 Canadian dollars per US dollar, or

• 11.1850 = 0.8439 US dollars per Canadian dollar.

Class Ex. #1 Complete the following:The rate of exchange for converting: a) Canadian dollars into Euros is _____________ Canadian dollars per Euro

or _____________ Euros per Canadian dollar.

b) Swiss francs into Canadian dollars is _____________ Canadian dollars per franc or _____________ francs per Canadian dollar.

c) Canadian dollars into Japanese yen is _____________ Canadian dollars per yen or _____________ yen per Canadian dollar.

Class Ex. #2 Determine the equivalent amount for converting $1000 Canadian into thefollowing currencies:

a) US dollars b) Mexican pesos c) India rupees

Class Ex. #3 The following items were bought by Canadian tourists. Calculate the equivalent cost in Canadian dollars. Assume that the Canadian tourist had to change Canadian dollars into foreign currency to purchase the item.

a) A soft drink in Sydney, Australia costing 4 dollars.

b) A Scottish kilt costing 37.50 pounds.

c) A meal costing 25 Euros.

Class Ex. #4 The Johnston family booked an all inclusive holiday in Cancun, Mexico. For spending money they converted $1800 Canadian to pesos. While in Cancun, they spent 9200 pesos and upon returning home to Vancouver, they changed their remaining pesos to Canadian dollars. How much did they receive?


504 Personal Finance Lesson #10: Foriegn Exchange

Assignment 1. Complete the following:

The rate of exchange for converting: a) Canadian dollars into Russian roubles is _____________ Canadian dollars per rouble

or _____________ roubles per Canadian dollar.

b) Chilean pesos into Canadian dollars is _____________ Canadian dollars per peso or _____________ pesos per Canadian dollar.

c) Australian dollars into Canadian dollars is _____________ Canadian dollars per Australian dollar or _____________ Australian dollars per Canadian dollar.

2. Determine the equivalent amount for converting the following currencies:a) $200 Cdn to $US b) 450 Swiss francs to $Cdn c) $650 Cdn to Japanese yen

d) $3000 Cdn to euros e) 1700 rupees to $Cdn f) $775 Cdn to Chilean pesos

3. The following items were bought by Canadian tourists. Calculate the equivalent cost in Canadian dollars. Assume that the Canadian tourist had to change Canadian dollars into foreign currency to purchase the item.

13.99 pounds 7560 Chilean pesos

$270 US 23 000 yen

a) b)

c) d)

Personal Finance Lesson #10: Foriegn Exchange 505

4. Sabine, an exchange student from Geneva, Switzerland, changed 3 000 Swiss francs into Canadian dollars. While in Alberta, she spent $1 580 and changed the remainder of her money into US dollars. She spent $820 US and upon returning home converted the remaining US dollars into francs at the rate of 1.3296 francs per US dollar. How many francs did she receive?

5. A wooden carving in Helsinki costs 38 markka. The cost in Canadian dollars, to the nearest Numerical Response cent, if the exchange rate is $1 Cdn = 4.52 markka is _____ .


6. A student from Hong Kong buys a Canadian souvenir for $75. If the exchange rate is Numerical Response $0.1551 Cdn = 1 Hong Kong dollar, then the cost in Hong Kong dollars, to the nearest

dollar, is _____ .


Answer Key

1 . a) 0.0147, 23.9808 b) 0.8196, 1.2201 c ) 0.8794, 1.1371

2 . a) $168.78 US b) $386.87 c ) 63 725 yend) 2116.55 euros e ) $46.75 f ) 301 556 pesos

3 . a) $28.52 b) $19.43 c ) $319.95 d) $234.60

4 . 30.74 francs 5 . 8 . 4 1 6 . 4 8 4

506 Personal Finance Lesson #10: Foriegn Exchange

Personal Finance Lesson #11:Budgeting

Budget

A budget is an itemized plan over a period of time for predicting expenses and deciding how to pay for these expenses. To create a budget you need to be aware of money earned (income) and money spent (expenditures).

Expenditures

When creating a budget be aware of the following three types of expenses;

• regular fixed expenses

• regular variable expenses

• irregular expenses

• unexpected expenses

Class Ex. #1 Classify the following expenses according to the criteria given above.

a) mortgage b) food c) holiday travel

d) clothing e) internet f) cell phone

g) car repair h) savings i ) home renovation

Class Ex. #2 Provide two more examples of each of the four types of expenditure

• regular fixed expenses

• regular variable expenses

• irregular expenses

• unexpected expenses

Balanced Budget

A balanced budget is a budget in which the total expenditure equals the total income.

If the expenditure is less than the income over the time period of the budget, then the budget is said to have a surplus.

If the expenditure is greater than the income over the time period of the budget, then the budget is said to have a deficit.

It is important when preparing a budget that the figures you use are realistic and are subject to review due to changing circumstances.

Class Ex. #3 Before the start of the girls soccer season, Mr. Elbel, the head coach, prepared the following budget.

Expenditures Income

• league entry fee Æ $120 • player user fee Æ to be announced• replacement uniforms Æ $150 • donations Æ $300• equipment Æ $190 • fund raising Æ $180• transportation Æ $400• officials Æ $200• trophies Æ $60• ground maintenance Æ $150• miscellaneaous Æ $100

If 18 girls made the final soccer roster, how much should Mr. Elbel charge each player for a user fee in order to have a balanced budget.

508 Personal Finance Lesson #11: Budgeting

Analyzing a Budget

Class Ex. #4 The Addams’ family have a monthly net income of $4250. They prepared a budget for their predicted monthly expenses. The breakdown of expenses is shown in the circle graph.

25%

9 %

5 %

20%

10%

6 %

10%

15%

mortgage

uti l i t ies

cable/phone

food

clothing

transportation

savings

others

a) How much money did they budget for:

i ) food? i i ) transportation?

b) Calculate the amount of money budgeted for entertainment if 15% of the “others” category was budgeted for entertainment.

c) Mr. Addams created the circle graph above with compasses and protractor. Calculate the measure of the central angle for the “food” sector.


Personal Finance Lesson #11: Budgeting 509

Assignment

1. The circle graph shows the way in which a company’s projected manufacturing costs of 1.2 million dollars for a year were budgeted.

35%

30%

18%

8 %5 % 4 %

Wages & Salaries

Materials Used

Sales Expenses

Packaging

Util it ies

Others

How much money was budgeted for:

a) wages and salaries? b) packaging? c) sales expenses?

2. Paying the fine for a speeding ticket would be classified as a(n) MultipleChoice

A. regular fixed expenseB. regular variable expenseC. irregular expenseD. unexpected expense

3. Sara’s mortgage payment accounts for 24% of her monthly budget. Food accounts for Numerical Response 18% and transportation accounts for 10%. If the total expenditure for these three items is

$1320, then her monthly budget amount, to the nearest dollar is _____ .(Record your answer in the numerical response box from left to right)


4. Pierre’s net monthly income is $1750. He prepares a budget for next month’s expenditure. Rent $750, food $400, cable $50, clothing $100, transportation 100, savings $200.The bar graph shows the budget expenditure.

9007506004503001500Monthly Expenditure in Dollars

Others

Savings

Transportation

Clothing

Cable

Food

Rent Series 1

a) Calculate the amount in the “others” category.

b) At the start of the month, the price of oil increased dramatically and his transportation costs were increased by 50%. Pierre revised his budget by finding the extra money from the “savings” category. Draw a circle graph which represents the revised budget.

Personal Finance Lesson #11: Budgeting 511

5. Helen and Jenny are about to leave high school. Helen has enrolled in an engineering course in a local university and is planning to live in student housing. Jenny has been offered employment as a payroll officer at a local lumber yard for which she will earn $1500 net per month. Jenny is planning to rent a small basement suite.

For each girl prepare a detailed balanced monthly budget outlining income and expenditure. Explain some of the choices you have made in each budget. Provide a graphical representation of each budget.

If possible, use a computer spreadsheet program or the table function of a graphing calculator to help develop the budgets.

Answer Key

1 . a) $420 000 b) $96 000 c ) $216 000

2 . D 3 . 2 5 3 8

4 . a) $150b)

8.6%

11.4%

5.7%

5.7%2.9%

22.9%

42.9%

Others 8.6%Savings 11.4%

Transportation 5.7%Clothing 5.7%

Cable 2.9%Food 22.9%

Rent 42.9%


Notes Page 1

Notes Page 2

Notes Page 3

Notes Page 4

Notes Page 5

Notes Page 6

Notes Page 7

Notes Page 8

Notes Page 9

Notes Page 10

Notes Page 11

Notes Page 12

Notes Page 13

Notes Page 14

Notes Page 15

Notes Page 16

Notes Page 17

Notes Page 18

Notes Page 19

Notes Page 20

Notes Page 21

Notes Page 22

Notes Page 23

Notes Page 24

Notes Page 25

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