quadratic equations lntaylor © 03/03/12. quadratic equations table of contents learning objectives...

Quadratic Equations

lntaylor © 03/03/12

Quadratic EquationsTable of Contents

Learning Objectives

Finding a, b and c

Finding the vertex and min/max values

Finding the discriminant

Finding the roots

Factoring

Completing the Square

Quadratic Formula

Quadratics with Function Tables

Graphing Quadratics


Learning Objectives

TOClntaylor © 03/03/12

Learning Objectives

LO 1

LO 2

Understand what a Quadratic Equation represents

Perform basic operations with Quadratic Equations

LO 3 Build a Quadratic equation using various techniques


Definitions

Definition 1 Quadratic Equations are in the form ax² + bx + c = 0

TOC

Definition 2 “a” determines the direction and Magnitude (width) of the curve

Definition 3 “-b/2a” determines whether the min/max has ± x value : note the OPPOSITE of b!

Definition 4 “c” determines whether the y intercept has a ± y value

Definition 5 The Discriminant determines the number of roots (b² - 4ac)

Definition 6 The Roots (x intercepts) are determined by factoring the equation


Previous knowledge

PK 1 Basic Operations and Properties


Rule 1

Rule 2

Know how to find a, b, -b, and c

Know how to find the vertex (-b/2a) and the discriminant (b² - 4ac)

Rule 3 Know how to find the roots by factoring, completing the square and the formula

Basic Rules of Quadratics

TOC

Rule 4 Always check your work


Quadratic Equations

• Quadratic Equations f(x) = ax² + bx + c

– You are given certain information in a function f(x)

• Width of the curve is determined by (a)

• Symmetry is determined by -b/2a (note the opposite of b)

• Y intercept of the curve is determined by (c)

• Remember all function tables are the same regardless of the equation

• Go up or down the same amount and look for a pattern


Finding a, b, -b and c


Q1


QuadraticsFinding a, b and c

In the quadratic equation below find a, b, -b and c

f(x) = 3x² + 5x + 2

a =

b =

-b =

c =


Q1 Answer



In the quadratic equation below find a, b and c

f(x) = 3x² + 5x + 2

a =

b =

-b =

c =

3 25

TOC

- 5


MathTV.com


http://www.mathtv.com/videos_by_topic

Now you try

f(x) = -2x² - 4x - 8




f(x) = -2x² -4x -8

a =

b =

-b =

c =

-2 -8-4

TOC

4


Now you try

f(x) = -¾ x² - ¼x + ½




f(x) = -¾ x² - ¼x + ½

a =

b =

-b =

c =

-¾ + ½- ¼¼

lntaylor © 03/03/12TOC

Changes in a


0,0

f(x) = 4x²f(x) = - 4x²

Step 1 – Compare Graphs

Graph 4x²

Graph - 4x²

Step 2 - Compare a values

a = 4

a = - 4

Step 3 – Conclusion

When a is + , curve goes up

When a is - , curve goes down


0,0

f(x) = 4x²f(x) = - 1/4x²


Graph 4x²

Graph - 1/4x²

Step 2 - Compare a values

a = 4

a = - 1/4


When a is large, curve narrows

When a is small, curve widens


Changes in b


0,0

f(x) = x² + 2xf(x) = x² - 4x


Graph x² + 2x

Graph x² - 4x

Step 2 - Compare b values

b =+2b= - 4

Step 3 – Question

If b is + , why is the curve on the negative side of x axis?

If b is -, why is the curve on the positive side of x axis?


0,0

f(x) = x² - 2xf(x) = x² + 4x


Previous Graph x² + 2x

Previous Graph x² - 4x

Step 2 - Compare b values

b = -2

b = + 4


If b is + , why is the curve on the negative side of x axis?

If b is -, why is the curve on the positive side of x axis?

Hint: use –b/2aTOC


Changes in c


0,0

f(x) = x² + 2xf(x) = x² - 2x - 1


Previous Graph x² + 2x

New Graph x² - 2x - 1

Step 2 - Compare c values

c = 0

c = - 1


C represents where the curve crosses the y axis

+ c means positive y intercept

No c means the origin

- c means negative y intercept


Q2


0,0

f(x) = x² + 2x - 3

Step 1 – Is this function:

Up or down?

Wide, Normal or Narrow?

Left or right of 0,0?

Positive or negative y intercept?

Step 2 - Answers:

Up

Normal

Left of 0,0

Y intercept - 3


Q 3


0,0

f(x) = ?

Step 1 – What are the values:

a is/is not a fraction

a <0 , a = 0 , a>0

b < 0, b = 0, b>0

C <0, c = 0, c>0

Step 2 - Answers:

a is a fraction

a<0 (meaning a is negative)

b = 0

c >0

The equation is:

f(x) = - 1/4x² + 5TOC


Finding the Vertex


Q4


0,0

f(x) = x² + 2x - 3

What is the vertex of the equation?


Q4 Answer


0,0

f(x) = x² + 2x - 3


TOC

-1, -4


0,0

f(x) = x² + 2x - 3

Step 1 – The vertex

The vertex is a coordinate value

Minimums go with + a values

Maximums go with – a values

a in this equation is +

Step 2 – The formula

Identify -b in the equation

Use the formula (-b/2a)

This is the x value (-1 )

Plug x into equation; find y

f(x) = x² + 2x – 3

f(x) = (-1)² + 2(-1) -3 = -4

This is the y value

TOC

min

- 2

-2 2(1)

-1,

-4 lntaylor © 03/03/12

Now you try

x² + 6x + 8


0,0

f(x) = x² + 6x + 8






Use the formula(-b/2a)



f(x) = x² + 6x + 8

f(x) = (-3)² + 6(-3) + 8 = -1

This is the y value TOC

min

- 6

-6 2(1)

-3,

-1


Now you try

- 2x² - 8x - 4


0,0

f(x) = - 2x² - 8x - 4


Maximums go with - a values

a in this equation is -






f(x) = - 2x² - 8x - 4

f(x) = -2(-2)² - 8(-2) - 4 = 4

This is the y value TOC

max

+ 8

8 2(-2)

-2,

4


Last one

x² - 4


0,0

f(x) = x² - 4







This is the x value (0 )


f(x) = x² - 4

f(x) = (0)² - 4 = - 4

This is the y value

No middle term means?

The vertex is the y intercept!

TOC

min

+ 0

0 2(1)

0,

- 4


Q5


0,0

f(x) = x² + 2x - 3


Is it a min or max point?

Answers:

Vertex = (-1, -4)

Minimum

TOC

min


Q6


0,0

f(x) = - 2x² - 8x - 4

TOC

maxWhat is the vertex of the equation?

Is it a min or max point?

Answers:

Vertex = (-2, 4)

Maximum


Finding the Discriminant


Q7


0,0

f(x) = x² + 2x - 3

What is the discriminant?

How many roots are there?


Q7 Answer


0,0

f(x) = x² + 2x - 3

TOC



Answers:

discriminant is +

There are two roots

-3, 0 1, 0


0,0

f(x) = - 2x² - 8x - 4

Step 1 – The discriminant

Discriminant give # of roots

+ means 2 roots

0 means 1 root

- means no roots

Step 2 – The formula b² - 4ac

Locate a, b, -b and c

a = -2b = -8-b = 8C = -4

Substitute into the formula

(-8)² - 4(-2)(- 4)

64 – 32 = + means two roots

Roots are x intercepts

Watch your signs!!!!


Now you try

x² - 4


0,0

f(x) = x² - 4

TOC



+ means 2 roots

0 means 1 root

- means no roots



a = 1b = 0-b = 0C = -4


(0)² - 4(1)(- 4)

0 + 16 = + means two roots




Now you try

x² + 4x + 4


0,0

f(x) = x² + 4x + 4

TOC



+ means 2 roots

0 means 1 root

- means no roots



a = 1b = 4-b = -4C = 4


(4)² - 4(1)(4)

16 - 16 = 0 means one root

One root touches the x axis

Did you recognize the equation?

This is a perfect square (x + 2)²lntaylor © 03/03/12

Last one

- x² + x - 1


0,0

f(x) = -x² + x - 1

TOC



+ means 2 roots

0 means 1 root

- means no roots



a = - 1b = 1-b = -1C = -1


(1)² - 4(-1)(-1)

1 - 4 = - means no roots

No roots means no x intercepts

Did you watch your signs?


Q8


0,0

f(x) = x² - 6x + 9



Answers:

discriminant is 0

One root

TOC

3, 0


Q9


0,0

f(x) = - 2x² - 8x - 4

TOC


How many roots?

Answers:

Discriminant is +

Two roots


Finding the Roots


Roots of a Quadratic Equation

Definition

Factor


Factoring is the quickest way to finding the roots

CTS Completing the Square is easy – once you get the hang of it

TOC

QF Quadratic Formula is the most versatile way of finding the roots


Factoring Quadratics


x2 + 6x + 8

Step 1 Check to see if last term is positive

+

Step 2 Divide middle term coefficient by 2

+ 6 2 = 3

Step 3 Square the answer and check last term

(3)(3) = 9

8

Step 4 If they are both the same you have your factors

No match

Step 5 If they are not the same subtract one and add one

(2)(4) = 8

Step 6 Continue until numbers match

Yes

Step 7 Add an x to each parenthesis

(x + 2)(x + 4)

Step 8 Set each () = 0 and solve

x + 2 = 0 and x + 4 = 0 x = - 2 and x = - 4


Now you try

x2 – 20x + 96


x2 – 20x + 96


+


– 20 2 = – 10

Step 3 Square the answer and check last term

(-10)(-10) = 100

96



(-11)(-9) = 99


Yes

(-12)(-8) = 96


(x – 12)(x – 8)


x - 12 = 0 and x - 8 = 0 x = 12 and x = 8


Now you try

x2 + 15x + 56


x2 + 15x + 56


+


+ 15 2 = 7.5

Step 3 Since the answer includes 0.5 round up and down

(8)(7) = 56

56




Yes


(x + 8)(x + 7)


x + 8 = 0 and x + 7 = 0 x = - 8 and x = - 7


Now you try

x2 – 19x + 84


x2 – 19x + 84


+


– 19 2 = – 9.5

Step 3 Since the answer includes 0.5 round up and down

(-10)(-9) = 90

84



(-11)(-8) = 88


Yes

(-12)(-7) = 84


(x – 12)(x – 7)


x - 12 = 0 and x - 7 = 0 x = 12 and x = 7


Factoring Negative Constants


x2 – 2x – 48

Step 1 Check to see if last term is negative

–

Step 2 Construct a table for the “difference” of factors for the constant

2 24 22

3 16 13

4 12 8

6 8 2

Step 3 Match the middle term coefficient to the last column

2

Step 4 Use the numbers in the 1st and 2nd columns; add a variable to each

(x 6)(x 8)

Step 5 Put the middle term sign next to the largest number

–

Step 6 Put the opposite sign next to the smallest number!!!!

+


(x + 6) = 0 and (x – 8) = 0 x = -6 and x = 8


Now you try

x2 – 11x – 12


x2 – 11x – 12

Step 1 Check to see if last term is negative

–

Step 2 Construct a table for the “difference” of factors for the constant

1 12 11

2 6 4

3 4 1

Step 3 Match the middle term coefficient to the last column

11

Step 4 Use the numbers in the 1st and 2nd columns; add a variable to each

(x 1)(x 12)

Step 5 Put the middle term sign next to the largest number

–

Step 6 Put the opposite sign next to the smallest number!!!!

+


(x + 1) = 0 and (x - 12) = 0 x = - 1 and x = 12


Special cases

x2 – 49


x2 – 49

Step 1 There are only two terms and the last term is negative

Step 2 Square root the last term

49½

= ± 7

Step 3 Write one factor with + and one factor with –

(x + 7)(x – 7)


(x + 7) = 0 and (x – 7) = 0 x = - 7 and x = 7


Now you try

x2 – 225


x2 – 225

Step 1 There are only two terms and the last term is negative

Step 2 Square root the last term

225½

= ± 15

Step 3 Write one factor with + and one factor with –

(x + 15)(x – 15)


(x + 15) = 0 and (x – 15) = 0 x = - 15 and x = 15



Step 1 Move the constant

Step 2 Take b/2; square the answer and add to both sides

Step 3 Factor out the perfect square; square root and solve


x² + 6x

Step 1 Set the equation = 0

Step 2 Move c to the opposite side

Step 3 Divide the middle term coefficient by 2

62

= + 3

x² + 6x

+ 5

(x+3)² = 4

x + 3 = 2 and x + 3 = -2

x = -1 and x = - 5


=

- 5

0

Step 4 Square the number and add to both sides

² = + 9

+ 5

= - 5 + 9

Step 5 Factor the perfect square

Step 6 Square root and solve

+ 9

TOC

Now you try!

f(x) = x² + 2x - 15


x² + 2x




22

= + 1

x² + 2x

- 15

(x+1)² = 16

x + 1 = 4 and x + 1 = - 4

x = 3 and x = - 5


=

15

0


² = + 1

- 15

= 15 + 1



+ 1

TOC

Last one!

f(x) = x² -10x + 18


x² - 10x




- 10 2

= - 5

x² - 10x

+ 18

(x - 5)² = 7

x - 5 = + √ 7 and x - 5 = - √ 7

x = 5 + √ 7 and x = 5 - √7


=

- 18

0


² = + 25

+ 18

= - 18 + 25



+ 25

TOC

Quadratics

The Quadratic Formula _______-b ± √b² - 4ac

2a


Quadratic Formula

Step 1 Train yourself to write down a, b, -b, and c

Step 2 Set up the equation slowly; make sure signs are correct

Step 3 Set up and Solve Slowly!!!! Make sure the signs are correct!!!!


x² + 2x - 15

Step 1 Write down a, b, -b and c

Step 2 Write down the formula

Step 3 Set up the equation to solve


a = 1b = 2-b = -2c = -15

______-b ± √b² - 4ac 2a

__________-2 ± √2² - 4(1)(-15) 2(1)

Step 4 Solve SLOWLY

Watch your signs!!!! (wait for the program)

_____-2 ± √4 + 60 2

__-2 ± √64 2

-2 ± 8 2

-2 + 8 2

-2 - 8 2

6 2

- 10 2

x = 3 x = - 5

The curve crosses the x axis at 3,0 and -5,0


Now you try

f(x) = x² + 6x + 9


x² + 6x + 9





a = 1b = 6-b = -6c = 9

______-b ± √b² - 4ac 2a

__________-6 ± √6² - 4(1)(9) 2(1)

Step 4 Solve SLOWLY


_____-6 ± √36 - 36 2

__-6± √ 0 2

-6 ± 0 2

-6 2

x = - 3

The curve touches the x axis at -3,0


Last one!

f(x) = -2x² + 3x - 15


-2x² - 3x - 15





a = -2b = -3-b = 3c = -15

______-b ± √b² - 4ac 2a

____________ 3 ± √(-3)² - 4(-2)(-15) 2(-2)

Step 4 Solve SLOWLY


_____ 3 ± √9 - 120 2

____-2 ± √ -111 2

Cannot square root a negative!

The curve does not cross the x axis


Build a Function Table


f(x) = x² - 5x - 6

Step 1 – Construct Table

Build 3 column Table

Build Headings

Step 2 – Choosing x values

Start with x = 0

Plug 0 into equation

Solve for y

Build x column

Build middle column

Build y column

Check your work!

Note that every x has one y

x and y together are called ordered pairs (coordinates and a single point on a graph)

TOC

x² - 5x - 6

x f(x) or y

0 0² -5(0) -6 -6

-3

-2

-1

0

1

2

3

(-3)² -5(-3) - 6

(-2)² -5(-2) - 6

(-1)² -5(-1) - 6

(0)² -5(0) - 6

(1)² -5(1) - 6

(2)² -5(2) - 6

(3)² -5(3) - 6

18

8

0

-6

-10

-12

-13


Now you try!

f(x) = x² + 2x + 1


f(x) = x² + 2x + 1



Build Headings


Start with x = 0


Solve for y

Build x column

Build middle column

Build y column

Check your work!



TOC

x² + 2x + 1

x f(x) or y

0 0² + 2(0) +1 1

-3

-2

-1

0

1

2

3

(-3)² + 2(-3) + 1

(-2)² + 2(-2) + 1

(-1)² + 2(-1) + 1

(0)² + 2(0) + 1

(1)² + 2(1) + 1

(2)² + 2(2) + 1

(3)² + 2(3) + 1

4

1

0

1

4

9

16


Now you try!

f(x) = x² - 4


f(x) = x² - 4



Build Headings


Start with x = 0


Solve for y

Build x column

Build middle column

Build y column

Check your work!



TOC

x² - 4

x f(x) or y

0 0² - 4 - 4

-3

-2

-1

0

1

2

3

(-3)² - 4

(-2)² - 4

(-1)² - 4

(0)² - 4

(1)² - 4

(2)² - 4

(3)² - 4

5

0

- 3

- 4

- 3

0

5


Graphing a Quadratic


0,0

f(x) = x² + 2x - 3

Step 1 – Graph

Write down:

a = 1

b = 2

-b = -2

c = -3

Step 2 – Key measures:

-b/2a = -2/2(1) = -1

Plug in x to find y = -4

Min = (-1, -4)

b² - 4ac = + two roots

Factor (x+3) (x-1)

Solve x = -3 and x = 1

Y intercept - 3 TOClntaylor © 03/03/12

Now you try!

f(x) = x² - 4


0,0

f(x) = x² - 4

Step 1 – Graph

Write down:

a = 1

b = 0

-b = 0

c = -4


-b/2a = 0/2(1) = 0

Plug in x to find y = -4

Min = (0, -4)

b² - 4ac = + two roots

Factor (x+2) (x-2)

Solve x = -2 and x = 2

Y intercept - 4 TOClntaylor © 03/03/12

Now you try!

f(x) = - x² + x - 1


0,0

f(x) = -x² + x - 1

TOC

Step 1 – Graph

Write down:

a = - 1

b = 1

-b = - 1

c = - 1


-b/2a = - 1/2(- 1) = 1/2

Plug in x to find y = - 3/4

Min = (1/2, - 3/4)

b² - 4ac = - no roots

Pick another x value and solve

x = 2; y = -3 (2, -3)

Y intercept - 1


Post Test


quadratic equations lntaylor © 03/03/12. quadratic equations table of contents learning objectives...

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