review of essential skills and knowledge -...

Review of Essential Skills andKnowledge

R–1 Exponent Laws...................................................................................540

R–2 Expanding and Simplifying Polynomial Expressions...........................542

R–3 Factoring Polynomial Expressions.......................................................544

R–4 Working with Rational Expressions ....................................................545

R–5 Slope and Rate of Change of a Linear Function..................................549

R–6 The Zeros of Linear and Quadratic Functions ....................................551

R–7 Exponential Functions....................................................................... 553

R–8 Transformations of Functions .............................................................554

R–9 Families of Functions..........................................................................557

R–10 Trigonometric Ratios and Special Angles ............................................559

R–11 Graphing and ......................................................562

R–12 Transformations of Trigonometric Functions ......................................564

R–13 Solving Trigonometric Equations in Degrees ......................................567

R–14 Proving Trigonometric Identities ........................................................570

y � cos xy � sin x

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539Review of Essential Skills and Knowledge

08-036_25_AFTR_APPR.qxd 9/15/08 12:25 PM Page 539

540 Advanced Functions: Review of Essential Skills and Knowledge

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R–1 Exponent Laws

Rule Word Description Algebraic Description Example

Multiplication If the bases are the same,add the exponents.

am � an � am�n 107 � 105 � 1012

Division If the bases are the same,subtract the exponents.

a � 0am

an � am�n, 10100 � 1095 � 105

Power of aPower

Keep the base, andmultiply the exponents.

(am)n � amn (2x)2 � 22x

Power of aProduct

Raise each factor to theexponent.

(ab)n � anbn

� 8x15(2x5)3 � 23(x5)3

Power of aQuotient

Raise the numerator andthe denominator to theexponent separately.

b � 0aabb n

�an

bn, a x3b 2

�x2

9

Zero Exponent A power with zero as the exponent equals 1,except when zero is alsothe base.

if Exception: is undefined.00

a � 0a0 � 1, 170 � 1

NegativeExponents

A power with a negativeexponent equals thepower with the reciprocalbase and a positiveexponent.

,

a, b � 0aabb�n

� abab n

,

a � 0a�n � a1ab n

�1

an

x � 0a x2b�4

� a2x b

4

�16

x4,

10�4 �1

104�

1

10 000

RationalExponentswithNumerator 1

The denominatordetermines the root.

a1n � �n

a

� 15 � 10 � 5

100013 � 25

12 � �3

1000 � �25

RationalExponentswithNumeratornot 1

The denominatordetermines the root andthe numerator indicatesthe exponent applied tothe root.

or

amn � �n

am

amn � Q�n

aRm

� 64 � 26

3265 � Q�5

32R6

EXAMPLE

Simplify. Express your answers using positive exponents.

a) b) c)Ax2y3B2Axy2B�4

64�23A23x�yB A2�x�4yB


R-1 Exponent Laws: Review of Essential Skills and Knowledge 541

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Solution

a) b) c)

� x8y14

�x4y6

x�4y�8

Ax2y3B2Axy2B�4

�1

16

�1

42

�1

Q�364R2

�1

6423

64�23

� 22x�5y� 23x�y�x�4y

A23x�yB A2�x�4yB

Practising

1. Simplify. Express your answers using positive exponents.

a) e) i) m)

b) f ) j) n)

c) g) k) o)

d) h) l)

2. Evaluate. Express your answers in fraction form.a) b) c) d) 64

13 � 16

32�1000.5225

3227 �1

3

a4

3b�2A20 � 2B0a9

a3

( y � 1)3( y � 2)4

( y � 1)5( y � 2)a 1

10b�1

80(x � 2)4(x � 2)

a x2y�5

x�2y�3 b�4

�4�5Ax2yB2Axy3B4A y3B5

Aa3bc0B�24�1y4

y7x2y7


R–2 Expanding and Simplifying PolynomialExpressions

To convert a polynomial expression from factored form to expanded form, use thedistributive property:

Some patterns occur frequently and are worth memorizing.

a(b � c) � ab � ac


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Square of a Sum Square of a Difference Difference of Squares

� a2 � 2ab � b2� (a � b) (a � b)(a � b)2

� a2 � 2ab � b2� (a � b) (a � b)(a � b)2

� a2 � b2� a2 � ab � ab � b2(a � b) (a � b)

EXAMPLE 1

Expand and simplify

Solution

(x � 5)(3x2 � 4x � 5).

� 3x3 � 11x2 � 15x � 25

� 20x � 25� 3x3 � 4x2 � 5x � 15x2

(x � 5)(3x2 � 4x � 5)

� 2x3 � 4x2 � 10x � 12� 16x � 12� 2x3 � 8x2 � 6x � 4x2

� (2x � 4)(x2 � 4x � 3)

� (2x � 4)(x2 � 3x � x � 3)

� 32(x � 2) 4 � 3 (x � 1)(x � 3) 42(x � 2)(x � 1)(x � 3)

EXAMPLE 2

Expand and simplify .

Solution

2(x � 2)(x � 1)(x � 3)

Use the distributive property tomultiply each term in the binomialby each term in the trinomial.

There are terms in theexpanded form, before it issimplified.

Collect like terms to simplify theexpanded form.

2 � 3 � 6

Since multiplication isassociative, you can multiply theexpressions in any order you like.

Use the distributive property tomultiply. Drawing arrows willhelp you to keep track of themultiplications.


R–2 Expanding and Simplifying Polynomial Expressions: Review of Essential Skills and Knowledge 543

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EXAMPLE 3

Expand and simplify

Solution

(2x � 3)2 � (3x � 2)(3x � 2).

Use the patterns for and (a � b) (a � b).(a � b)2

� �5x2 � 12x � 13� 4x2 � 12x � 9 � 9x2 � 4

� 3 (3x)2 � (2)2 4� (2x)2 � 2(2x) (3) � (3)2(2x � 3)2 � (3x � 2)(3x � 2)

Practising

1. Expand and simplify.a)

b)c)

d)

e)f )

2. Write in simplified expanded form.a)b)c)d)e)f ) (x � 3)4

(3x � 4)2(2x � 3)(x � 5)(x � 2)(x � 5)(x � 2)4(x � 2)32(x2 � x � 3)(x � 7)5(x � 1)(x � 1)(x � 2)

32x(x � 1) 4 25x(2x � 4)2

6ax �1

3b ax �

1

2b

3(x � 4)2 � (2 � x) (2 � x)(2x � 7)2

3x A5x2 � 3x � 4B

08-036_25_AFTR_APPR.qxd 9/15/08 12:25 PM 43

R–3 Factoring Polynomial Expressions


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Type Example Comment

Common Factoring

Factor out the largest commonfactor of each term.

ab � ac � a(b � c) � 2x3(5x � 4 � 3x2)10x4 � 8x3 � 6x5 Each term has a common

factor of 2x3.

Factoring Trinomialswhen

Write the trinomial as theproduct of two binomials.Determine two numberswhose sum is b and whoseproduct is c.

a � 1ax2 � bx � c, � (x � 7)(x � 3)x2 � 4x � 21 and

4 � 7 � (�3)(�21) � 7(�3)

Factoring Trinomialswhen

Look for a common factor. Ifnone exists, use decompositionand write the trinomial as theproduct of two binomials.Check by expanding andsimplifying.

a � 1ax2 � bx � c,

Check:

� 3x2 � 4x � 4� 3x2 � 6x � 2x � 4

� (�2)(2)(3x) (x) � (3x) (2) � (�2)(x)

� (3x � 2)(x � 2)� x(3x � 2) � 2(3x � 2)� (3x2 � 2x) � (6x � 4)� 3x2 � 2x � 6x � 4

3x2 � 4x � 4 Multiply Find two numbers whoseproduct is andwhose sum is 4. In thiscase, the numbers are 6and Using thesenumbers, decompose thex-term. Group the terms,and factor out thecommon factors.

�2.

�12

3(�4) � �12.

Factoring a Difference ofSquares

This is a special case offactoring trinomials, whenb 5 0.

x2 � y2 � (x � y) (x � y) � 2(x � 3)(x � 3)� 2(x2 � 9)

2x2 � 18 Common factor first,when possible.

Practising

1. Factor.a)b)c)d)e)f )

2. Factor.a)b)c)d)

3. Expand to show that is the factored form of x3 � y3.

(x � y) (x2 � xy � y2)

12x2 � 18x � 125a2 � 7a � 612x2 � x � 16y2 � y � 2

x4 � 81x2 � 643y2 � 18y � 24x2 � x � 628x � 14xy6x2 � 5x


R–4 Working with Rational Expressions: Review of Essential Skills and Knowledge 545

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A rational expression is an algebraic expression that can be written as the quotientof two polynomials. A rational expression is undefined if the denominator is zero,so we write restrictions on the variables to avoid this.

Simplifying Rational Expressions

A rational expression can be simplified by factoring the numerator and thedenominator, and then dividing out the common factors.

EXAMPLE 1

Simplify and state restrictions.

a) b) c)

Solution

4x � 6

9 � 6xx2 � 1

x2 � 4x � 3

21m3n2

6mn4

R–4 Working with Rational Expressions

a) b)Factor the numerator and the denominator to find the largestpossible common factorto divide out.

Write restrictions on the variables to preventthe denominator fromequalling zero.

x � � 1,�3

�x � 1

x � 3

�(x � 1)

1(x � 1)

(x � 1)1

(x � 3)

�(x � 1)(x � 1)

(x � 1)(x � 3)

x2 � 1

x2 � 4x � 3

n � 0m � 0,

�7m2

2n2

�3mn2(7m2)

1

3mn2

1(2n2)

21m3n2

6mn4

c)

x �3

2

� �2

3

��2(3 � 2x) 1

3(3 � 2x) 1

�2(2x � 3)

3(3 � 2x)

4x � 6

9 � 6x

When factors are opposites, factor out from one of the factors to make the

factors identical.�1

opposites


Multiplying and Dividing Rational Expressions

To multiply or divide rational expressions, factor the numerators and the denominators(where possible), and then look for common factors that can be divided out.

EXAMPLE 2

Simplify x2 � 92x � 6

�4x � 20

x2 � 6x � 9.


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Solution

x � �3, 3

�2(x � 5)

x � 3

�(x � 3)

1(x � 3)

1

21(x � 3)

1

�42(x � 5)

(x � 3)1

(x � 3)

�(x � 3)(x � 3)

2(x � 3)�

4(x � 5)

(x � 3)(x � 3)

x2 � 9

2x � 6�

4x � 20

x2 � 6x � 9Factor each polynomial.

Divide out the commonfactors to reduce theexpression to lowest terms.

Write restrictions toprevent the denominatorfrom equalling zero, whichwould result in undefinedvalues.

Change the division intomultiplication by thereciprocal.

Factor.

Divide out the identicalfactors.

Write restrictions to avoidundefined values.

EXAMPLE 3

Simplify

Solution

4x2 � 4x � 1

x � 3�

2x2 � 9x � 5

x � 5.

12

�5,x � �3,

�2x � 1

x � 3

�(2x � 1)(2x � 1)

1

x � 3�

x � 51

(2x � 1)1

(x � 5)1

�(2x � 1)(2x � 1)

x � 3�

x � 5

(2x � 1)(x � 5)

�4x2 � 4x � 1

x � 3�

x � 5

2x2 � 9x � 5

4x2 � 4x � 1

x � 3�

2x2 � 9x � 5

x � 5


R–4 Working with Rational Expressions: Review of Essential Skills and Knowledge 547

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Adding and Subtracting Rational Expressions

To add or subtract rational expressions, you must have a common denominator. To ensure that you will use the lowest common denominator, factor the numeratorsand the denominators first. This will keep the expressions as simple as possible.

EXAMPLE 4

Simplify

Solution

3x � 6

x2 � 4�

x2 � 8x � 15

7x � 21.

2, 3x � �2,

��x2 � 7x � 11

7(x � 2)

�21 � x2 � 7x � 10

7(x � 2)

�21 � (x2 � 7x � 10)

7(x � 2)

�21

7(x � 2)�

x2 � 7x � 107(x � 2)

� a 3x � 2

b a7

7b � ax � 5

7b ax � 2

x � 2b

�3

x � 2�

x � 5

7

�3(x � 2)

1

(x � 2)1

(x � 2)�

(x � 3)1

(x � 5)

7(x � 3)1

�3(x � 2)

(x � 2)(x � 2)�

(x � 3)(x � 5)

7(x � 3)

3x � 6

x2 � 4�

x2 � 8x � 15

7x � 21Factor the numerators andthe denominators.

If possible, divide out likefactors (but only withineach rational expression).

The lowest commondenominator is

Multiply the numeratorand denominator of eachrational expression to createan equivalent expressionwith the desired commondenominator.

Simplify the numerator.

Write restrictions to avoidundefined values.

7(x � 2).



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Practising

1. State the restrictions (if any) on each rational expression.

a) b) c) d)

2. Simplify, and state restrictions. Write your answers using positive exponents.

a) c) e)

b) d) f )

3. Simplify, and state restrictions.

a)

b)

c)

d)

e)

f )

4. Simplify, and state restrictions.

a)

b)

c)

d)

e)

f )

5. Show that x � 3.4x � 1 �2

x � 3 �4x 2 � 11x � 1

x � 3 ,

x � 1

x2 � 2x � 3�

x � 2

x2 � 4x � 5

x � 3

x � 4�

x � 1x � 2

6x

x2 � 5x � 6�

3x

x2 � x � 12

2x � 1

3�

5

x � 2

5

x � 1�

2

x � 1

4

5x�

2

3x

�9m2 � 4

4m2 � 4m � 1

3m2 � 7m � 6

6m2 � 3m�

9m � 6

2m2 � 5m � 3

x2 � 5x � 6

x2 � 1�

x2 � 4x � 5

x2 � 4�

x � 5

x2 � 3x � 2

x � 7

10�

2x � 14

25

(x � 1)2

x2 � 2x � 3�

x2 � 2x � 1

x2 � 4x � 3

2ab5bc

�6ac10b

6x8y

�2y2

3x

t3 � t2

t � t3

b3 � a2b

b2 � 2ab � a2

x2 � 5x

x2 � 4x � 5

x2 � 3x � 2

x2 � 5x � 6

3h2 � 6h

h2 � 4h � 4

2ab4b

3x

x2 � 36

17x � 2

x5

5x


R–5 Slope and Rate of Change of a Linear Function: Review of Essential Skills and Knowledge 549

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R–5 Slope and Rate of Change of a Linear Function

The slope of a line is a ratio that compares the change in the dependent variable, y,with the change in the independent variable, x.

The equation of a linear relation can be written in the form where m isthe slope and b is the y-intercept.

Values of the Slope• The slope of a line that rises to the right is positive.• The slope of a line that drops to the right is negative.• The slope of a horizontal line is zero. The equation of the line can be written in

the form • The slope of a vertical line is undefined. The equation of the line can be written in

the form

Equations of Straight Lines• point-slope equation of a line: • general form of the equation of a line: • slope-intercept equation of a line:

Parallel and Perpendicular LinesTwo lines, with slopes and are• parallel if and only if • perpendicular if and only if that is, if their slopes are negative

reciprocals:

EXAMPLE

Find the slope and equation of a line that passes through points (5, 6) and (15, 2).Explain how the slope is a rate of change.

Solution

The slope is

Substituting and into

The slope of the line is and the equation is y � �25 x � 8.�2

5,

y � �2

5x � 8

y � �2

5x � 2 � 6

y � 6 � �25

(x � 5)

y � y1 � m(x � x1),(x1, y1) � (5, 6)m � �25

m �y2 � y1

x2 � x1�

2 � 615 � 5 � �

410 � �

25.

m2 � �1m1

m1 m2 � �1;m1 � m2

m2,m1

y � mx � bAx � By � C � 0

y � y1 � m(x � x1)

x � a.

y � b.

y � mx � b,

Slope � m �riserun �

change in y

change in x�

�y

�x�

y2 � y1

x2 � x1



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Practising

1. Determine the slope of a line that passes througheach pair of points.a) and b) and (7, 4)c) and d) and

2. Describe the graph ofa) b)

3. Suppose that you buy a plant. The height of theplant t weeks after you buy it is where h is the height in centimetres. What is theslope of the height function, and what does theslope mean in the context of this situation?

4. Determine the slope and y-intercept of each line.a) b) Ax � By � C � 03x � 5y � 10 � 0

h(t) � 26 � 1.2t,

y � 6x � �3

(�2, 9)(�3, 5)(5,�4)(5, �2)

(�1, 4)(�4, �9)(1, �5)

The slope is a rate of change because y will decrease by 2 units for each 5 unitincrease in x.

2

4

6

8

10

y

x0 5 10

run = 5

rise = –2


R–6 The Zeros of Linear and Quadratic Functions: Review of Essential Skills and Knowledge 551

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R–6 The Zeros of Linear and Quadratic Functions

The Zero of a Linear Function

A linear function of the form has one zero (x-intercept), unless theline is horizontal. (A horizontal line has no x-intercepts, unless it lies on the x-axis.Then every point on the line is an x-intercept.)Factoring out the slope will give the x-intercept.

EXAMPLE 1

What is the x-intercept of

Solution

y � 2x � 6?

y � mx � b

Solving for the zero, let

Factor out the slope.

The x-intercept is 3, since substitutingresults in a y value of zero.

(Hence the name “zero of the function”for an x-intercept.)

–6 60

2

–4–2

–6–8

y

x

2 4–4 –2

x � 33 � x0 � 2(x � 3)

y � 0.

y � 2(x � 3)y � 2x � 6

The Zeros of a Quadratic Function

A quadratic function can have two zeros, one zero, or no zeros.The zeros of a quadratic function are also found by factoring the equation. Whenfactoring is not possible, the quadratic formula can be used. The discriminant canbe used to determine the number of zeros.

Quadratic Formula Discriminant

The zeros of the functionare

x ��b � �b2 � 4ac

2a .

y � ax2 � bx � cIf there are two zeros.If there is one zero.If there are no zeros.b2 � 4ac 0,

b2 � 4ac � 0,b2 � 4ac 0,



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EXAMPLE 2

Determine the zeros of each function.a) b)

Solution

g(x) � x2 � x � 2f (x) � 2x2 � x � 6

a) Factoring gives

Solving for the zeros, let

or

or

b) To solve for the zeros, let

Using the quadratic formula gives

Since the square root of is not a realnumber, the function g has no zeros.

�7

x �1 � ��7

2

x ��(�1) � �(�1)2 � 4(1)(2)

2(1)

x ��b � �b2 � 4ac

2a

0 � x2 � x � 2g(x) � 0.

x � 2x � �3

2

0 � x � 20 � 2x � 30 � (2x � 3)(x � 2)

f (x) � 0.

f (x) � (2x � 3)(x � 2)f (x) � 2x2 � x � 6

The graph of g is entirely abovethe x-axis. Therefore, the functionhas no zeros.

–6 60

246

–4–2

–6

y

x

2 4–4 –2

–6 60

2468

–4–2

y

x

2 4–4 –2

Practising

1. What are the zeros of each function?a)b)c)d)

2. Determine the zero of each linear function.

a)

b) line with y-intercept 3 and slope 12

y �1

3x � 2

y � 2x2 � x � 15y � 4(x � 9)2y � �2(x � 3)(x � 7)y � �3(x � 4)

3. Determine the zeros of each quadratic function.a) c)b) d)

4. The zeros of a quadratic function are and 2, andthe y-intercept is 8. Write the equation of the function.

5. Use the discriminant to determine the number ofzeros for each quadratic function.a)b) y � 16x2 � 40x � 23

y � 16x2 � 40x � 25

�2

y � 3x2 � 5x � 4y � 16x2 � 8x � 1g(x) � x2 � 3x � 1f (x) � x2 � x � 42


R–7 Exponential Functions: Review of Essential Skills and Knowledge 553

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x

23

22

21

0 1

1 2

2 4

3 8

12

1

4

1

8

y � 2x

R–7 Exponential Functions

The exponential function has the following characteristics:

• The base is restricted to or • The domain is and the range is • The x-axis is a horizontal asymptote.• The y-intercept is 1.• If the graph increases (is a growth function).• If the graph decreases (is a decay function).

EXAMPLE

Sketch the graph of each exponential function.

a) b)

Solution

f (x) � a13b x

y � 2x

0 b 1,b 1,

{ y�R � y 0} .{x�R } ,b 1.0 b 1

f (x) � bx

–3 30

2468

10

–2

y

x

1 2–2 –1 –3 30

2468

10

–2

y

x

1 2–2 –1

x

22 921 3

0 1

1

21

9

13

y � a13b xa) b)

Note that is a growth curve and is a decay curve.y � A13Bxy � 2x

Practising

1. Sketch the graph of each exponential function.

a) c)

b) d)

2. Compare the graphs of and How arethey related?

3. For the function state the domain, range,intercepts, and equation of the asymptote.

4. The function models thetemperature, in °C, of a cup of coffee t minutes after it is poured.a) What is the initial temperature of the coffee?b) What is the temperature after 10 min?c) What is the temperature after 60 min?d) Determine the equation of the horizontal

asymptote. What does it represent?e) What is the significance of the number 76

in the equation?

T � 20 � 76(0.92)t

f (x) � 4x,

y � 3x.y � A13Bxf (x) � (1.5)xf (x) � 10x

f (x) � a1

2b x

f (x) � 3x


Advanced Functions: Review of Essential Skills and Knowledge

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EXAMPLE 1

What transformations to the parent function would you perform tocreate the graph of What happens to the coordinates of each point on the parent function?

Solution

Comparing the transformed function with the general formwe have and c � �5.d � 4,k � 3,a � �2,y � af (k(x � d)) � c,

y � �2f (3(x � 4)) � 5?y � f (x)

R–8 Transformations of Functions

You can graph functions of the form by applying theappropriate transformations to key points on the parent function Stretches/compressions and reflections (based on a and k) must be applied beforetranslations (based on c and d ).

The value of a determines whether there is a vertical stretch or compression andwhether there is a reflection in the x-axis. The y-coordinate of each point ismultiplied by a.

• If the graph of is stretched vertically by the factor • If the graph is compressed vertically by the factor • If a is negative, the graph is also reflected in the x-axis.

The value of k determines whether there is a horizontal stretch or compression andwhether there is a reflection in the y-axis. The x-coordinate of each point is

multiplied by

• If the graph of is compressed horizontally by the factor

• If the graph is stretched horizontally by the factor • If k is negative, the graph is also reflected in the y-axis.

The value of c determines the vertical translation. This value is added to the y-coordinate of each point.

• If the graph is translated c units up.• If the graph is translated c units down.

The value of d determines the horizontal translation. This value is added to the x-coordinate of each point.

• If the graph is translated d units to the right.• If the graph is translated d units to the left.d 0,

d 0,

c 0,c 0,

1�k �.0 �k � 1,

1�k �.y � f (x)�k � 1,

1k.

�a�.0 �a � 1,Za Z.y � f (x)�a � 1,

y � f (x).y � af (k(x � d )) � c

554


R–8 Transformations of Functions: Review of Essential Skills and Knowledge 555

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Table of values for Graph of We start with points on the parentfunction y � 2x.

y � 2xy � 2x

• Since there is a vertical stretch by a factor of 2 and also a reflection inthe x-axis. The y-coordinate of each point is multiplied by

• Since there is a horizontal compression by a factor of The

x-coordinate of each point is multiplied by

• Since there is a vertical translation 5 units down. The value isadded to the y-coordinate of each point.

• Since there is a horizontal translation 4 units to the right. The value 4 isadded to the x-coordinate of each point.

EXAMPLE 2

Graph the function by applying of the appropriatetransformations to the parent function

Solution

y � 2x.y � 3

2 � 2�12(x�3) � 1

d � 4,

�5c � �5,

13.

13.k � 3,

�2.a � �2,

60

246

–2

y

x

2 4–4 –2

Table of values for Graph of Apply any stretches/compressionsand reflections next.

Since there is a verticalstretch. Each y-coordinate ismultiplied by

Since there is a horizontalstretch and also a reflection in the y-axis. Each x-coordinate ismultiplied by �2.

k � �12

32.

a � 32,

y � 32 � 2�1

2xy � 32 � 2�1

2x

60

246

–2

y

x

2 4–4 –2

x y

438

234

03

2�2 3

�4 6

x y

2214

2112

0 1

1 2

2 4



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Table of values for Graph of Apply any translations last.Since there is a translationup. The value 1 is added to each y-coordinate.Since there is a translation tothe right. The value 3 is added toeach x-coordinate.

Notice that the horizontal asymptoteis shifted up to y � 1.

d � 3,

c � 1,y �

32 � 2�1

2(x�3) � 1y �

32 � 2�1

2(x�3) � 1

60

2

6

–2

y

2 4–4 –2

4

Practising

1. Describe the transformations that you would applyto the graph of to graph each of thefollowing functions.a)

b)

c)

d)

e)

f )

2. The point (2, 5) is on the graph of Statethe coordinates of the image of this point under eachof the following transformations.a) c)b) d)

3. Given the function state the equation ofthe transformed function under a vertical stretch offactor 3, a reflection in the x-axis, a horizontaltranslation 3 units to the right, and a verticaltranslation 2 units up.

4. Consider the function a) Make a table of values for f using

b) Describe the transformations to f that result in

the function c) Determine the five points on the graph of g that

are the images of the five points in your table ofvalues for f in part a).

5. Consider the functions and What transformations to result in Y2?Y1

Y2 � �4 � x.Y1 � �x

g(x) � 12(x � 4)3 � 5.

x � {�2, �1, 0, 1, 2} .

f (x) � x3.

f (x) � x2,

y � f (x) � 7y � �2f (x)y � f (x � 4)y � f (3x)

y � f (x).

y � �1

5f (�x) � 3

y � �f (�x) � 4

y � �3f (2(x � 1)) � 2

y � f (2x) � 7

y � f a12

(x � 3)by � 3f (x) � 2

y � f (x)

x y

7 13

8

5 13

4

3 21

21 4

�1 7


R–9 Families of Functions: Review of Essential Skills and Knowledge 557

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R–9 Families of Functions

Families of Linear Functions

Consider the equation It represents a straight line with y-intercept 3 and slope m. Different values of the parameter m will result in lines with differentslopes. Together, these lines make up a family of lines with the same y-intercept.

Families of Quadratic Functions

The equation represents a family of quadratic functions.Each member of the family has zeros at 1 and 3. Their differences are determinedby the value of the vertical stretch factor a.

y � a(x � 1)(x � 3)

y � mx � 3.

0

2

6

–2

y

2 4–4 –2

4

60

42

–4–2

–6

6y

x

2 4–2

EXAMPLE 1

What member of the family of quadratic functions with the vertex passesthrough point (5, 5)?

Solution

(3, �1)

The graph shows several members of the family ofquadratic functions with the vertex

This family has an equation of the form

Substitute point (5, 5) into the equation, and solve for a.

The equation is f (x) � 32(x � 3)2 � 1.

a �6

4�

3

2

6 � 4a5 � a(2)2 � 15 � a(5 � 3)2 � 1

f (x) � a(x � 3)2 � 1.

(3, �1).

60

42

–4–2

–6

6y

x

2 4–2


Advanced Functions: Review of Essential Skills and Knowledge

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Practising

1. a) Determine the general equation of the family of straight lines with slope 3, but varying y-intercepts.

b) Find the equation of the member of this familythat passes through point (4, 7).

2. a) Determine the equation of the family ofquadratic functions with zeros at 2 and 4.

b) What is the equation of the member of thisfamily with y-intercept

3. A family of exponential functions has equation

a) At what point do all the members of this familymeet?

b) Why does the parameter k vary in the graphs of this family?

c) Show that results in a curve that passesthrough point (4, 8).

4. Determine the equation of the quadratic functionthat has vertex and passes through (1, 8).

5. Determine the equation of the quadratic functionthat has x-intercepts 5 and and passes through

6. Determine the equation of the quadratic functionif f (2) � 3.f (x) � ax2 � 6x � 7

(7, �40).�1,

(�2, 5)

k � 34

y � 2(kx).

�4?

558


R–10 Trigonometric Ratios and Special Angles: Review of Essential Skills and Knowledge 559

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The CAST rule is an easy way to remember which trigonometric ratios are positive in each quadrant. Since r is always positive, the sign of each ratio depends on the signs of the coordinates of the point.• In quadrant 1, all (A) ratios are positive because both x and y are positive.• In quadrant 2, only sine (S) and its reciprocal cosecant are positive, since x is

negative and y is positive.• In quadrant 3, only tangent (T) and its reciprocal cotangent are positive,

because both x and y are negative.• In quadrant 4, only cosine (C) and its reciprocal secant are positive, since x is

positive and y is negative.

R–10 Trigonometric Ratios and Special Angles

Right-Triangle Definitions of Trigonometric Ratios

The trigonometric ratios for an acute angle can be defined using a right triangle, asshown below.

The Pythagorean theorem is often useful for solving problems that involve righttriangles:

The right-triangle definitions given above cannot be used for an angle that is notacute, so we need to broaden the definitions.

Definitions of Trigonometric Ratios for Any Angle

An angle in standard position has its vertex at the origin and rotates counterclockwisefrom the positive x-axis to its terminal arm. If point (x, y) is on the terminal arm ofangle at a distance r units from the origin, we define the trigonometric ratios of

as follows:

In the diagram above, is the acute angle related to The related acute anglealways has one arm on the x-axis. The trigonometric ratios for angle are equal inmagnitude to those for angle but they are always positive.u,

b

u.b

cot u �xy tan u �

yx

sec u �rx cos u �

xr

csc u �ry sin u �

yr

u

u,

(adjacent)2 � (opposite)2 � (hypotenuse)2

cot u �adjacent

opposite tan u �

opposite

adjacent

sec u �hypotenuse

opposite cos u �

adjacent

hypotenuse

csc u �hypotenuse

opposite sin u �

opposite

hypotenuse

u

opposite

adjacent

hypotenuse

y

xbx

y

(x, y)

ru

1

4

2

3

0

y

xS A

T C


The exact values of the primary trigonometric ratios for 30°, 45°, and 60° anglescan be found by using an isosceles right triangle and half of an equilateral triangle,as shown below. These triangles are often referred to as special triangles.

EXAMPLE 1

Determine the exact value of

Solution

The following diagram shows that a 240° angle is related to a 60° acute angle.

sin 240�.


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1

1

45˚

45˚

2

1

2 3

60˚

30˚

u usin ucos utan

30°12

� 0.5 �32

� 0.86601

�3�

�3

3� 0.5774

45° �2

2� 0.7071

�2

2� 0.7071 1

60° �3

2� 0.8660

1

2� 0.5 �3 � 1.7321

y

x240˚

60˚


R–10 Trigonometric Ratios and Special Angles: Review of Essential Skills and Knowledge 561

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EXAMPLE 2

If tan , find an exact value for . Confirm the value with a calculator.

Solution

There are two possible terminal arms, as shown in the following diagram.

Using gives

Therefore, .sin u �yr �

�2

�5

r � �5.r2 � x2 � y2

sin uu � �2

sin 240� � ��32

� sin 240� � � sin 60� ��32

To determine the sign of sin 240°, consider that sineis the ratio

Since 240° is in quadrant III,y is negative. Therefore, sineis negative.

yx.

Since we knowthat any point on theterminal arm of the angle

must satisfy

A point on the terminal armin quadrant II is A point on the terminal armin quadrant IV is (1, �2).

(�1, 2).

yx � �2.

tan u � �2,

–2 20

1

–1

–2

2y

x

1–1

(1, –2)

(–1, 2)

5

u2

5

u1

Practising

1. Using exact values, show that for each angle.a)b)

2. Determine the acute angle that each line makes withthe x-axis.a)b)

3. Determine the angle(s) between 0° and 360° ifa)b)

4. Determine an exact value fora)b)c)d) csc 300�

tan 225�

cos 210�

sin 135�

sin u � �1cos u � sin u

u

y 5 23xy 5 x

u � 45�

u � 60�

sin2 u � cos2 u � 1


R–11 Graphing y = sin x and y = cos x

The functions and are periodic functions since their graphsconsist of a regularly repeating shape. The period of both of these functions is 360°.

The minimum value of these functions is and the maximum value is 1.

The amplitude of a function is defined as, so the amplitude of these functions is

The axis (or midline) of a function is the horizontal line halfway between themaximum value and the minimum value. For both of these functions, the equation

of the axis is

EXAMPLE

At what values of x, in the interval to 360°, does

Solution

Draw the sine function from to 360° along with the line

We can read the intersection points directly from the graph. The values of x in theinterval are and 150�.30�,�210�,�330�,3�360�, 360� 4

y � 12.�360�

sin x � 12?�360�

y �maximum � minimum

2 � 1 � (�1)2 � 0.

1 � (�1)2 � 2

2 � 1.

maximum � minimum2

,

�1,



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360°0

1

–1

y

x

–180° 180°

y = sin x y = cos x

360°0

1

–1

y

x

180°–180°–360°


R–11 Graphing y = sin x and y = cos x: Review of Essential Skills and Knowledge 563

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Practising

1. At what values of x does the function havea maximum value?

2. At what values of x do the functions andmeet?

3. Consider the graph of from 0° to 360°.

a) Determine the values of x that correspond to anangle in

i) quadrant Iii) quadrant IIiii) quadrant IIIiv) quadrant IV

b) Describe the behaviour of the sin x function ineach quadrant.

y � sin x


y � sin x

180° 270° 360°0

2

–2–1

y

x

90°

1

EXAMPLE

At what values of x does the function have a minimum value?

Solution

Examine the graph.

We can see that the minimum values occur at 180°, 540°, and so on.There are an infinite number of values of x, so we cannot list all of them. (Thiswould take forever!) Notice, however, that they occur at regular intervals due tothe periodic nature of the function. Each minimum value is a multiple of 360°that is either less than or more than 180°. Therefore, we can write all the values ofx as follows:

where k�Ix � �180� � k(360�),

x � �180�,

y � cos x

360° 540° 720°0

21

–2–1

y

x

180°–180°–360°



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R–12 Transformations of Trigonometric Functions

We can transform the sine and cosine functions in the same way that we transformother functions.

The general forms of the transformed functions are and

EXAMPLE

Describe the transformations that have been applied to a parent function to geteach of the following transformed functions. State the amplitude, period, and axisof the transformed function, and determine whether there is a horizontal shift.Then sketch the graph of the transformed function.

a) b)

Solution

a) To obtain the transformed function the parent functionundergoes a vertical stretch of factor 2, a reflection in the x-axis, a

horizontal compression of factor and a vertical translation 1 unit down.

The amplitude of the transformed function is 2, the period is and theaxis is There is no horizontal shift.y � �1.

360�2 � 180�,

12,

y � cos xy � �2 cos 2x � 1,

g(x) �2

3 sina1

2x � 15�by � �2 cos 2x � 1

y � a cos(k(x � d )) � c.y � a sin(k(x � d )) � c

Transformation Characteristics of the Transformed Function

The value of a determines whether there is a verticalstretch/compression and whether there is a reflection inthe x-axis.

Since the vertical stretch/compression factor is theamplitude of the transformed function is �a �.

�a �,

The value of k determines whether there is horizontalstretch/compression and whether there is a reflection inthe y-axis.

Since the horizontal stretch/compression factor is

the period of the transformed function is 360��k � .

1�k �,

The value of c determines the vertical translation. The axis of the transformed function is y � c.

The value of d determines the horizontal translation. The horizontal shift of the transformed function is d.

180° 270° 360° 450°0

2

–2–3

–1

y

x

90°

1

–90°

Start with the parent function The amplitude is 1.The period is 360°.

y � cos x.


R–12 Transformations of Trigonometric Functions: Review of Essential Skills and Knowledge 565

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b) The transformed function is not in the general

form we work with, so we must factor the argument of the function.

The parent function undergoes a vertical compression of factor a

horizontal stretch of factor and a horizontal translation 30° to the right.

The amplitude of the curve is the period is , and the axis of the

curve is There is a horizontal shift 30° to the right.y � 0.

360�12

� 720�23,

112

� 2,

23,f (x) � sin x

g(x) �2

3 sin a1

2(x � 30�)b

g(x) �2

3 sin a1

2x � 15�b

g(x) � 23 sin A12 x � 15�B

Apply the stretch, compression, and reflection.

The function is reflected in the x-axis.The amplitude is 2.The period is 180°.

Apply the translation.

The function is reflected in x-axis.The amplitude is 2.The period is 180°.The axis is There is no horizontal shift.

y � �1.

y � �2 cos 2x � 1

y � �2 cos 2x

180° 270° 360° 450°0

2

–2–3

–1

y

x

90°

1

–90°

180° 270° 360° 450°0

2

–2–3

–1

y

x

90°

1

–90°

The argument of a function is theinput to the function. For the function

the argumentis A12x � 15�B.g(x) � 2

3 sin A12 x � 15�B,

180° 360 540° 720°0

2

–2–1

y

x1

–180°

Start with the parent function

The amplitude is 1.The period is 360°.

f (x) � sin x.



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Apply the compression and stretch.

There is no reflection.

The amplitude is

The period is 360°.

Apply the translation.

The amplitude is

The period is 360°.The axis is There is a horizontal shift of 30°to the right.

y � 0.

23.

g(x) �2

3 sin a1

2(x � 30�)b

23.

y �2

3 sin a1

2x b180° 360 540° 720°0

2

–2–1

y

x1

–180°

180° 360 540° 720°0

2

–2–1

y

x1

–180°

Practising

1. For each of the following transformed functions, identify the parent function. Describe the transformationsthat have been applied to create the transformed function. State the amplitude, period, and axis, anddetermine whether there is a horizontal shift. Then sketch a graph of the transformed function.

a)

b)

c) y �1

2 sin c�2

3(x � 60�)d

f (x) � �cos(3x � 90�) � 2

f (x) � 4 sin(2x) � 4


R–13 Solving Trigonometric Equations in Degrees: Review of Essential Skills and Knowledge 567

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R–13 Solving Trigonometric Equations in Degrees

Trigonometric functions can have many solutions, due to their periodic nature.The number of solutions for a function depends on the domain of the function.The solutions can be found using various approaches, as shown in the followingexamples.

EXAMPLE 1 USING SPECIAL TRIANGLES AND THE CAST RULE

Solve where

Solution

u� 30�, 360� 4 . sin u � 0.5

Since we can see

that is a solution. Theterminal arm for is in quadrant I.

The angle in quadrant II is

There are no other values of in thedomain Therefore, the solutions are or

.

The value of 0.5 is recognizable as aspecial value from the special

triangle.

The CAST rule states that sine valuesare positive for first quadrant andsecond quadrant angles. The terminalarms of these angles are mirror imagesin the y-axis.

30��60��90�

150�

u � 30�

30�, 360� 4 .u

180� � 30� � 150�.

y

x

S A

CT

150˚30̊

12 2

1

– 3 3

( 3, 1)(– 3, 1)

30�

u � 30�

sin u �yr �

oppositehypotenuse,



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EXAMPLE 3 USING THE x-y-r DEFINITIONS AND A CALCULATOR

Solve for , to the nearest degree, where

Solution

u� 30�, 360� 4u tan u � �2

EXAMPLE 2 USING A CALCULATOR AND THE CAST RULE

Solve for , to the nearest degree, where

Solution

u� 30�, 360� 4 .ucos u � �0.8

The related acute angle is about

The angle in quadrant II is

The angle in quadrant III is

Therefore, the two solutions in therequired domain are and

The cosine ratio is negative inquadrants II and III. Since is notfrom a special triangle, we use acalculator to determine the relatedacute angle.

20.8

217°.143�

180� � 37� � 217�.u

180� � 37� � 143�.u

y

x

S A

CT

143˚

37�.

so we can use

and as thepoints on the terminal arm.

By definition, tan u �yx.

(�1, 2)(x, y) � (1,�2)

yx � �2 � �2

1 � 2�1,


R–13 Solving Trigonometric Equations in Degrees: Review of Essential Skills and Knowledge 569

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Practising

1. Solve each trigonometric function, to the nearest degree, where

a) c) d)

b) d) f )

2. Solve, where a) c)

b) d) 3 cos u � 2 � 13 tan u � ��3

�4 sin u � 12 cos u � 1 � 00 � u � 3600.

cos u 5 1.5 sin u � �1 tan u �2

5

sin u � �0.554 tan u � 1 cos u ��3

2

0 � u � 3600.

The related acute angle is about The diagram shows that the angles inthe required domain are

andTherefore, the two solutions in therequired domain are and

Determine the inverse cosine of 2 tofind the related acute angle.

297�.117�

u � 180� � 63� � 117�

u � 360� � 63� � 297�

63�.

y

x

(–1, 2)

(1, –2)

S A

CT

117˚

297˚ –63˚



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R–14 Proving Trigonometric Identities

An identity is an equation that is true for all possible values of its variable.

To disprove an identity (in other words, to prove that an equation is not anidentity), we need to find only one value of the variable that does not satisfy theequation.

To prove that an equation is an identity, we need to show that the two sides of theequation are equal for all possible values of the variable. To do this, we need torewrite one or both sides of the equation by substituting known identities and/orusing algebraic techniques.

EXAMPLE 1

Show that is not an identity.

Solution

To prove that this equation is not an identity, we need to find a value of thatdoes not satisfy the equation.

Try

For the equation is not satisfied. Therefore, the equationis not an identity.cos u � sin u � 1

u � 45�,

� �2

�2

�2

�1

�2�

1

�2

Right side � 1Left side � cos 45� � sin 45�

u � 45�.

u

cos u � sin u � 1

AlgebraicTechniques

ReciprocalIdentities

Quotient Identities

PythagoreanIdentities

• factoring

sin u � 0

csc u �1

sin u,

cos u � 0

tan u � sin u

cos u,

cos 2u � sin

2u � 1

• expanding andsimplifying

cos u � 0

sec u �1

cos u,

sin u � 0

cot u � cos u

sin u,

1 � tan 2u � sec

2u

• adding orsubtractingrationalexpressionsusing acommondenominator

tan u � 0

cot u �1

tan u, cot2 u � 1 � sec

2 u


A–14 Proving Trigonometric Identities: Review of Essential Skills and Knowledge 571

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d.

EXAMPLE 2

Prove each identity.

a) b)

Solution

a) To prove the identity, we must work with each side independently.

Since we have shown that the left side and right side are equal, the equation is anidentity—every possible value of will satisfy the equation. There are values of thatwill result in each side being undefined, however, so a restriction is needed:

b) Work with each side independently.

Since the left side and right side are equal, the equation is an identity.Again, restrictions are needed to avoid undefined values:

Practising

1. Prove each of the following identities.

a) c)

b) d)

2. Prove that where

3. Prove that (1 � cos2 x) (1 � cos2 x) � 2 sin2 x � sin4 x.

sin x � 0. sin2 x a1 �1

tan2 xb � 1,

tan2 u

1 � tan2 u� sin2 u sin

4u 2 cos 4u 5 sin

3u 2 cos 2u

1 � cos2 u � sin u cos u tan u1

cos u� tan u �

1 � sin u

cos u

cos u � 0.sin u � 0,

�1

sin u cos u

� sin

2u

sin u cos u�

cos 2u

sin u cos u

� sin u

cos u�

cos u

sin u

RS �1

sin u cos uLS � tan u �

1

tan u

cos u 2 0.uu

� sin u

cos 2 u

� sin u

cos 2u

�

sin u

cos u

cos u

RS � sin u

1 � sin 2 u

LS � tan u

cos u

tan u �1

tan u�

1 sin u cos u

tan u

cos u�

sin u

1 � sin 2 u


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