P1: FXS/ABE P2: FXS CHAPTER 3...P1: FXS/ABE P2: FXS 0521842344c03.xml CUAT030-EVANS August 27, 2008 1:39 Chapter3—Families of functions 83 In the diagram below, the graphs of f (x)

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0521842344c03.xml CUAT030-EVANS August 27, 2008 1:39

C H A P T E R

3Families of functions

ObjectivesTo consider functions with equation y = A f(n(x + c)) + b, where A, n, c and b ∈ R,

for

f (x) = x n, where n is a non-zero rational number

and f (x) = |x|To use dilations, reflections and translations to sketch graphs of such functions.

To determine the rule for the function of such graphs.

To use addition of ordinates to sketch such graphs.

To find and graph the inverse relations of such functions.

To use matrices to describe transformations.

3.1 Functions with rule f(x) = xn

In this section functions of the form f (x) = xn , where n is a rational number, are considered.

These functions are called power functions. We need to use calculus to study all aspects of

these functions, but at this stage we can consider some members of this family as an important

addition to the functions already introduced.

f (x) = xn where n is a non-zero rational numberWhen n = 1, f (x) = x , i.e. the basic linear function is formed.

When n = 2 and n = 3, f (x) = x2 and x3 respectively. These functions are part of the

family of functions of the form f (x) = xn where n is a positive integer. It is appropriate to

delay the introduction of this family until Chapter 4.

f(x) = xn where n is a negative integerWhen n = −1, f (x) = x−1

= 1

x

The maximal domain of this function is R\{0}.

The graph of the function is as shown.

x0

y

81Cambridge University Press • Uncorrected Sample Pages • 2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

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82 Essential Mathematical Methods 3 & 4 CAS

Asymptotes:

There is a horizontal asymptote with equation y = 0

As x → ∞,1

x→ 0+, i.e. from the positive side.

As x → −∞,1

x→ 0−, i.e. from the negative direction.

There is a vertical asymptote with equation x = 0

As x → 0+, i.e. from the positive direction,1

x→ ∞

As x → 0−, i.e. from the negative side,1

x→ −∞

f (x) = 1

xis an odd function, since f (−x) = − f (x)

When n = −2, f (x) = x−2 = 1

x2. The maximal

domain of this function is R\{0}. The graph of

the function is as shown on the right. x0

y

Asymptotes:

There is a horizontal asymptote with equation y = 0

As x2 → ∞,1

x2→ 0+, i.e. from the positive side.

There is a vertical asymptote with equation x = 0

As x → 0+, i.e. from the positive direction,1

x2→ ∞

As x → 0−, i.e. from the negative side,1

x2→ ∞

f (x) = 1

x2is an even function, since f (−x) = f (x)

In the diagram on the right, the graphs of f (x) = 1

x2and

f (x) = 1

x4are shown on the one set of axes.

The graphs intersect at the points with

coordinates (1, 1) and (−1, 1).

Note that1

x2>

1

x4for x > 1 and x < −1,

and1

x2<

1

x4for 0 < x < 1 and

−1 < x < 0

x0

(1, 1)(–1, 1)f (x) =

x21

f(x) =x41

y

In general, if n and m are positive even integers with n < m:

The graphs of f (x) = 1

xnand f (x) = 1

xmintersect at the points with coordinates

(1, 1) and (−1, 1).1

xn>

1

xmfor x > 1 and x < −1, and

1

xn<

1

xmfor 0 < x < 1 and −1 < x < 0

When n is a positive even integer, f (x) = 1

xnis an even function, since f (−x) = f (x)

Cambridge University Press • Uncorrected Sample Pages • 2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

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Chapter 3 — Families of functions 83

In the diagram below, the graphs of f (x) = 1

xand f (x) = 1

x3are shown on the one set of

axes.

The graphs intersect at the points with coordinates (1, 1) and (−1, −1).

Note that1

x>

1

x3for x > 1 and −1 < x < 0,

and1

x<

1

x3for 0 < x < 1 and x < −1

x0

(1, 1)

(–1, –1)

f (x) = x1

f (x) =x3

1

y

In general, if n and m are positive odd integers with n < m:

The graphs of f (x) = 1

xnand f (x) = 1

xmintersect at the points with coordinates

(1, 1) and (−1, −1).1

xn>

1

xmfor x > 1 and −1 < x < 0, and

1

xn<

1

xmfor 0 < x < 1 and x < −1

When n is a positive odd integer, f (x) = 1

xnis an odd function, since

f (−x) = − f (x)

f (x) = x1n where n is a positive integer

First recall that x1n = n√x . In particular x

12 = √

x . When n is even, the maximal domain is

[0, ∞) and when n is odd, the maximal domain is R. The graphs of f (x) = √x = x

12 and

f (x) = 3√x = x13 are as shown.

x0

y

x0

y

When n is a positive odd integer, f (x) = x1n is an odd function, since f (−x) = − f (x).

x0

(1, 1)

f (x) =x1

f(x) =1

√x

y

f(x) = x− 1n where n is a positive integer

The expression x− 1n = 1

x1n

= 1n√x

. The maximal

domain of the function f with rule f (x) = 1n√x

is (0, ∞) if n is even, and R\{0} if n is odd.

The graphs of y = 1√x

and y = 1

xare shown for

x ∈ (0, ∞) on the one set of axes.


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The graphs intersect at the point with coordinates (1, 1).

Note that1√x

>1

xfor x > 1, and

1√x

<1

xfor 0 < x < 1

Asymptotes:

For each of the graphs there is a horizontal asymptote with equation y = 0 and a vertical

asymptote with equation x = 0.

The graphs of y = 13√x

and y = 1

xare shown for x ∈ R\{0} on the one set of axes.

The graphs intersect at the points with

coordinates (1, 1) and (−1, −1).1

3√x>

1

xfor x > 1 and −1 < x < 0, and

13√x

<1

xfor 0 < x < 1 and x < −1

When n is a positive odd integer, f (x) = x− 1n

is an odd function, since f (−x) = − f (x)

x0

(1, 1)

(–1, –1)

f(x) =1

√x3

f (x) = x1

y

f (x) = xpq where p and q are non-zero integers and

the highest common factor (p, q) = 1The special cases where p = ±1 have been considered above.

xpq is defined as

(x

1q)p

.

It is necessary to have the fractionp

qin simplest form. For example, (−1)

26 �= (−1)

13 since

(−1)26 = ((−1)2)

16 = 1

16 = 1, but (−1)

13 = −1.

An investigation of these graphs with your calculator is worthwhile. Every case will not be

listed here.

Example 1

For each of the following, use your calculator to help sketch the graph of y = f (x) for the

maximal domain. State this maximal domain and the range corresponding to this domain. Also

state the equations of asymptotes if appropriate, and whether the function is odd, even or

neither.

a f (x) = x23 b f (x) = x− 2

3 c f (x) = x32

d f (x) = x− 32 e f (x) = x

35 f f (x) = x− 3

5

Solution

a The maximal domain of f (x) = x23 is R.

The range is [0, ∞).

f (x) = x23 is an even function, since

f (−x) = f (x)


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b The maximal domain of f (x) = x− 23 is R\{0}.

The range is (0, ∞).

The asymptotes have equations x = 0 and y = 0

Note that x− 23 = 1

x23

As x23 → 0,

1

x23

→ ∞ and as x23 → ∞,

1

x23

→ 0

f (x) = x− 23 is an even function, since f (−x) = f (x)

c The maximal domain of f (x) = x32 is [0, ∞).

The range is [0, ∞).

f (x) = x32 is neither odd nor even.

d The maximal domain of f (x) = x− 32 is (0, ∞).

The range is (0, ∞). The asymptotes have

equations x = 0 and y = 0. Note that

x− 32 = 1

x32

As x32 → 0+,

1

x32

→ ∞ and as x32 → ∞,

1

x32

→ 0

f (x) = x− 32 is neither odd nor even.

e The maximal domain of f (x) = x35 is R.

The range is R.

f (x) = x35 is an odd function, since

f (−x) = − f (x)

f The maximal domain of f (x) = x− 35 is R\{0}.

The range is R\{0}.

The asymptotes have equations x = 0 and y = 0

Note that x− 35 = 1

x35

.

As x35 → 0+,

1

x35

→ ∞ and as x35 → ∞,

1

x35

→ 0+

f (x) = x− 35 is an odd function, since f (−x) = − f (x)

Exercise 3A

1 For the function with equation f (x) = 1

x4:

a state the maximal domain, the corresponding range and the equations of any asymptotes

b sketch the graph without using your calculator.


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2 For each of the following rules of functions:

i state the maximal domain of the function, the corresponding range and the equations of

any asymptotes

ii sketch the graph using your calculator to assist you.

a f (x) = 1√x

b f (x) = x35 c f (x) = x− 3

5

d f (x) = 13√x

e f (x) = 5√x

3 a Find {x : x32 > x2} b Find {x : x− 3

2 < x−2}

4 For each of the following, state whether the function is odd, even or neither:

a f (x) = 1

xb f (x) = 1

x2c f (x) = 3√x

d f (x) = 13√x

e f (x) = x23 f f (x) = x

57

Introducing transformations of functionsMany graphs of functions can be described as transformations of graphs of other functions, or

‘movements’ of graphs about the cartesian plane. For example, the graph of the function

y = −x2 can be considered as a reflection, in the x-axis, of the graph of the function y = x2.

x (mirror line)0

y = x2

y

x0

y = –x2

y

Formally, a transformation is a one-to-one function (or mapping) from R2 to R2. A good

understanding of transformations, combined with knowledge of the ‘simplest’ function and its

graph in each family, provides an important tool with which to sketch graphs and identify rules

of more complicated functions.

There are three basic types of transformations that are considered in this course: dilations

from the coordinate axes, reflections in the coordinate axes and translations. A graph of a

function may be transformed to the graph of another function by a dilation from the x- or

y-axis, a reflection in either the x- or y-axis, a translation in the positive or negative direction of

the x- or y-axis, or a combination of these. The following three sections consider dilations,

reflections and translations separately.

3.2 DilationsA transformation which, for example, dilates each point in the plane by a factor of 2 from the

x-axis can be described as ‘multiplying the y-coordinate of each point in the plane by 2’ and

can be written as (x, y) → (x, 2y). This is read as ‘the ordered pair (x, y) is mapped onto the


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ordered pair (x, 2y)’. The dilation is a one-to-one mapping from R2 to R2 and uniquely links

any ordered pair (a, b) to the ordered pair (a, 2b).

Similarly, a transformation which dilates each point in the plane by a factor of 3 from the

y-axis can be described as ‘multiplying the x-coordinate of each point in the plane by 3’ and

can be written as (x, y) → (3x, y). This is read as ‘the ordered pair (x, y) is mapped onto the

ordered pair (3x, y)’. The dilation is a one-to-one mapping from R2 to R2 and uniquely links

any ordered pair (a, b) to the ordered pair (3a, b).

A dilation of a graph of a function can be thought of as the

graph ‘stretching away from’ or ‘shrinking towards’ an axis.

Consider dilating the graph of, say, a circle in various ways,

and observe the effect on a general point with coordinates

(x, y) on a circle.

x

(x, y)

0

y

1 A dilation of factor 2 from the x-axis

x

(x, y)

(x, 2y)

0

y

The graph is ‘stretched’ to twice the

height. The point (x, y) is mapped

onto (x, 2y),

i.e. (x, y) → (x, 2y)

2 A dilation of factor1

2from the x-axis

0

(x, y)

x

y

x, y12

The graph is ‘shrunk’ to half the height.

The point (x, y) is mapped onto

(x,

1

2y

),

i.e. (x, y) →(

x,1

2y

)

3 A dilation of factor 2 from the y-axis

x

(2x, y)(x, y)

0

y

The graph is ‘stretched’ to twice the

width. The point (x, y) is mapped onto

(2x, y),

i.e. (x, y) → (2x, y)

4 A dilation of factor1

2from the y-axis

x0

(x, y)x, y

12

y

The graph is ‘shrunk’ to half the width.

The point (x, y) is mapped onto

(1

2x, y

),

i.e. (x, y) →(

1

2x, y

)


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Example 2

Determine the rule of the image when the graph of y = 1

x2is dilated by a factor of 4:

a from the x-axis b from the y-axis

Solution

a (x, y) → (x, 4y)

Let (x ′, y ′) be the coordinates of

the image of (x, y), so x ′ = x, y ′ = 4y

Rearranging gives x = x ′, y = y ′

4

Therefore, y = 1

x2becomes

y ′

4= 1

(x ′)2

So the rule of the transformed function is y = 4

x2

x0

(1, 4)

(1, 1)

y

b (x, y) → (4x, y)

Let (x ′, y ′) be the coordinates of the

image of (x, y), so x ′ = 4x, y ′ = y

Rearranging gives x = x ′

4, y = y ′

Therefore, y = 1

x2becomes y ′ = 1(

x ′4

)2

So the rule of the transformed function

is y = 16

x2

0x

(4, 1)

(1, 1)

y

In general, a dilation of factor a, where a > 0, from the x-axis is a transformation that

maps (x, y) → (x, ay). To find the equation of the image of y = f (x), under the

dilation (x, y) → (x, ay), replace y withy

a, i.e. the image of the graph with equation

y = f (x) under the dilation (x, y) → (x, ay) is the graph with equationy

a= f (x),

which is more commonly written as y = af (x)

In general, a dilation of factor b, where b > 0, from the y-axis is a transformation

that maps (x, y) → (bx, y). To find the equation of the image of y = f (x), under the

dilation (x, y) → (bx, y), replace x withx

b, i.e. the image of the graph with equation

y = f (x) under the dilation (x, y) → (bx, y) is the graph with equation y = f(x

b

)

Example 3

Determine the factor of dilation when the graph of y = √3x is obtained by dilating the graph

of y = √x :

a from the y-axis b from the x-axis


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Solution

a Note that a dilation from the y-axis ‘changes’ the x-values. Write the transformed

function as y ′ = √3x ′, where (x ′, y ′) are the coordinates of the image of (x, y).

Therefore, x = 3x ′ (‘changed’ x) and y = y ′

Rearranging gives x ′ = x

3and y ′ = y

So the mapping is given by (x, y) →( x

3, y

)and

the graph of y = √x is dilated by a factor

of1

3from the y-axis to produce the graph of y = √

3x

x0

(1, 1)

, 11

3

y

b Note that a dilation from the x-axis ‘changes’ the y-values. Write the transformed

function asy ′√

3=

√x ′, where (x ′, y ′) are the coordinates of the image of (x, y).

Therefore x = x ′ and y = y ′√

3(‘changed’ y)

Rearranging gives x ′ = x and y ′ = √3y

So the mapping is given by (x, y) → (x,√

3y)

and the graph of y = √x is dilated by a factor

of√

3 from the x-axis to produce the graph of y = √3x

x0

(1, 1)

(1, ) √3

y

Exercise 3B

1 Sketch the graph of each of the following:

a y = 4

xb y = 1

2xc y = √

3x d y = 2

x2

2 For y = f (x) = 1

x2, sketch the graph of each of the following:

a y = f (2x) b y = 2 f (x) c y = f( x

2

)d y = 3 f (x) e y = f (5x) f y = f

( x

4

)3 Sketch the graphs of each of the following on the one set of axes:

a y = 1

xb y = 3

xc y = 3

2x

4 Sketch the graph of the function f : R+ → R, f (x) = 3√

x

5 State a transformation which maps the graphs of y = f (x) to y = f1(x) for each of the

following:

a i f (x) = 1

x2ii f1(x) = 5

x2

b i f (x) = √x ii f1(x) = 4

√x

c i f (x) = √x ii f1(x) = √

5x


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6 Write down the equation of the rule when the graph of each of the functions below is

transformed by:

i a dilation of factor 4 from the x-axis ii a dilation of factor2

3from the x-axis

iii a dilation of factor1

2from the y-axis iv a dilation of factor 5 from the y-axis

a y = |x | b y = 3√x c y = 1

x3d y = 1

x4

e y = 13√x

f y = x23 g y = 1

x34

3.3 ReflectionsA transformation which, for example, reflects each point in the plane in the x-axis can be

described as ‘multiplying the y-coordinate of each point in the plane by −1’ and can be written

as (x, y) → (x, −y). This is read as ‘the ordered pair (x, y) is mapped onto the ordered pair

(x, −y)’. The reflection is a one-to-one mapping from R2 to R2 and uniquely links any ordered

pair (a, b) to the ordered pair (a, −b).

Similarly, a transformation which reflects each point in the plane in the y-axis can be

described as ‘multiplying the x-coordinate of each point in the plane by −1’ and can be written

as (x, y) → (−x, y). This is read as ‘the ordered pair (x, y) is mapped onto the ordered pair

(−x, y)’. The reflection is a one-to-one mapping from R2 to R2 and uniquely links any ordered

pair (a, b) to the ordered pair (−a, b).

This course of study considers reflections in the x- or y-axis

only. (Note: The special case where the graph of a function

is reflected in the line y = x produces the graph of the

inverse relation and is discussed separately in Section 3.10.)

Consider reflecting the graph of the function shown here

in each axis, and observe the effect on a general point (x, y)

on the graph.

x

(x, y)

0

y

1 A reflection in the x-axis:

x0

(x, y)

(x, –y)

y

The x-axisacts as a ‘mirror’ line.The point (x, y) is mapped onto (x, –y),i.e. (x, y) (x, –y)

2 A reflection in the y-axis:

x0

(x, y)(–x, y)

y

The y-axisacts as a ‘mirror’ line.The point (x, y) is mapped onto (–x, y),i.e. (x, y) (–x, y)


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Example 4

Find the rule of the function obtained from the graph of the function with equation

y = √x by a reflection:

a in the x-axis b in the y-axis

Solution

a Note that a reflection in the x-axis changes the

y-values, so (x, y) → (x, −y)

Let (x ′, y ′) be the coordinates of the image of

(x, y), so x ′ = x, y ′ = −y

Rearranging gives x = x ′, y = −y ′

So y = √x becomes (−y ′) = √

x ′

The rule of the transformed function is y = −√x

(1, 1)

0

(1, –1)

x

y

b Note that a reflection in the y-axis changes

the x-values, so (x, y) → (−x, y)


image of (x, y), so x ′ = −x, y ′ = y

Rearranging gives x = −x ′, y = y ′

So y = √x becomes y ′ = √

(−x ′)The rule of the transformed function is y = √−x

0

(–1, 1) (1, 1)

x

y

In general, a reflection in the x-axis is the transformation that maps (x, y) → (x, −y).

To find the equation of the image of y = f (x), under the reflection (x, y) → (x, −y),

replace y with −y; i.e. the image of the graph with equation y = f (x) under the

reflection (x, y) → (x, −y) is the graph with equation −y = f (x), which is more

commonly written as y = − f (x)

In general, a reflection in the y-axis is the transformation that maps (x, y) → (−x, y).

To find the equation of the image of y = f (x), under the reflection (x, y) → (−x, y),

replace x with −x; i.e. the image of the graph with equation y = f (x) under the

reflection (x, y) → (−x, y) is the graph with equation y = f (−x)

Exercise 3C

1 Sketch the graphs, and state the domain, of:

a y = −√x b y = √−x

2 State a transformation which maps the graph of y = f (x) to y = f1(x), where f (x) = √x

and f1(x) = √−x


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3 Find the equation of the rule when the graph of each of the functions below is transformed

by:

i a reflection in the x-axis

ii a reflection in the y-axis

a y = |x | b y = 3√x c y = 1

x3d y = 1

x4

e y = 13√x

f y = x23 g y = 1

x34

4 Identify whether the functions given in Question 3 are odd, even or neither.

3.4 TranslationsA transformation which, for example, translates each point in the plane 2 units in the positive

direction of the y-axis can be described as ‘adding 2 units to the y-coordinate of each point in

the plane’ and can be written as (x, y) → (x, y + 2). This is read as ‘the ordered pair (x, y) is

mapped onto the ordered pair (x, y + 2)’. The translation is a one-to-one mapping from R2 to

R2 and uniquely links any ordered pair (a, b) to the ordered pair (a, b + 2).

Similarly, a transformation which translates each point in the plane 3 units in the positive

direction of the x-axis can be described as ‘adding 3 units to the x-coordinate of each point in

the plane’ and can be written as (x, y) → (x + 3, y). This is read as ‘the ordered pair (x, y) is

mapped onto the ordered pair (x + 3, y)’. The translation is a one-to-one mapping from R2 to

R2 and uniquely links any ordered pair (a, b) to the ordered pair (a + 3, b).

A translation moves each point on the graph the same

distance in the same direction. Consider translating the

graph of the function shown here in various ways, and

observe the effect on a general point (x, y) on the graph. x(x, y)

0

y

a A translation of 1 unit in the positive

direction of the x-axis:

x(x + 1, y)

(x, y)

01 unit

‘to the right’

y

The point (x, y) is mapped onto (x + 1, y),

i.e. (x, y) → (x + 1, y)

b A translation of 1 unit in the negative

direction of the x-axis:

x(x, y)

01 unit

‘to the left’

(x – 1, y)

y

The point (x, y) is mapped onto (x − 1, y),

i.e. (x, y) → (x − 1, y)


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c A translation of 1 unit in the positive

direction of the y-axis:

x

(x, y + 1)(x, y)

0

1 unit

‘up’

y

The point (x, y) is mapped onto (x, y + 1),

i.e. (x, y) → (x, y + 1)

d A translation of 1 unit in the negative

direction of the y-axis:

x(x, y – 1)0

1 unit‘down’

y

(x, y)

The point (x, y) is mapped onto (x, y − 1),

i.e. (x, y) → (x, y − 1)

In general, the translation of h units (h > 0) in the positive direction of the x-axis

and k units (k > 0) in the positive direction of the y-axis is the transformation that

maps (x, y) → (x + h, y + k). To find the equation of the image of y = f (x), under the

translation (x, y) → (x + h, y + k), replace x with x − h and y with y − k, i.e. the

image of the graph with equation y = f (x) under the translation (x, y) → (x + h, y + k)

is the graph with equation y − k = f (x − h), which is more commonly written as

y = f (x − h) + k

Example 5

Find the equation of the image when the graph of y = |x| is transformed by the following

sequence of transformations:

a translation of 4 units in the positive direction of the x-axis, and

a translation of 3 units in the negative direction of the y-axis.

Solution

(x, y) → (x + 4, y − 3)


image of (x, y), so x ′ = x + 4, y ′ = y − 3

Rearranging gives x = x ′ − 4, y = y ′ + 3

So y = |x| becomes y ′ + 3 = |x ′ − 4|The rule of the transformed function

is y = |x − 4| − 3

x

(4, –3)

0

(4, 4) (8, 4)

(8, 1)

y


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Exercise 3D

1 Sketch the graphs of each of the following. Label asymptotes and axis intercepts, and state

the domain and range.

a y = 1

x+ 3 b y = 1

x2− 3 c y = 1

(x + 2)2

d y = √x − 2 e y = 1

x − 1f y = 1

x− 4

g y = 1

x + 2h y = 1

x − 3i f (x) = 1

(x − 3)2

j f (x) = 1

(x + 4)2k f (x) = 1

x − 1+ 1 l f (x) = 1

x − 2+ 2

m y = 1

x2− 4

2 For y = f (x) = 1

x, sketch the graph of each of the following. Label asymptotes and axis

intercepts.

a y = f (x − 1) b y = f (x) + 1 c y = f (x + 3)

d y = f (x) − 3 e y = f (x + 1) f y = f (x) − 1

3 State a transformation which maps the graphs of y = f (x) to y = f1(x) for each of the

following:

a i f (x) = x2 ii f1(x) = (x + 5)2

b i f (x) = 1

xii f1(x) = 1

x+ 2

c i f (x) = 1

x2ii f1(x) = 1

x2+ 4

4 Write down the equation of the rule when the graph of each of the functions below is

transformed by:

i a translation of 7 units in the positive direction of the x-axis, and 1 unit in the positive

direction of the y-axis

ii a translation of 2 units in the negative direction of the x-axis, and 6 units in the negative


iii a translation of 2 units in the positive direction of the x-axis, and 3 units in the negative


iv a translation of 1 unit in the negative direction of the x-axis, and 4 units in the positive


a y = |x | b y = 3√x c y = 1

x3d y = 1

x4

e y = 13√x

f y = x23 g y = 1

x34


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3.5 Combinations of transformationsIn the previous three sections each of the three types of transformations was considered

separately. In the remainder of this chapter we look at situations where a graph may have been

transformed by any combination of dilations, reflections and translations.

Example 6

Find the rule of the image when the graph of the function with rule y = √x is translated

6 units in the negative direction of the x-axis, reflected in the y-axis and dilated by a factor

of 2 from the x-axis.

Solution

The translation of 6 units in the negative direction of the x-axis maps

(x, y) → (x − 6, y). The reflection in the y-axis maps (x − 6, y) → (−(x − 6), y).

The dilation by a factor of 2 from the x-axis maps (−(x − 6), y) → (−(x − 6), 2y). In

summary, (x, y) → (−(x − 6), 2y)

Let (x ′, y ′) be the coordinates of the image of (x, y), so x ′ = −(x − 6) and y ′ = 2y

Rearranging gives x = −x ′ + 6 and y = y ′

2

Therefore, y = √x becomes

y ′

2= √−x ′ + 6

The rule of the transformed function is y = 2√

6 − x

Example 7

Sketch the graph of the image of the graph shown under the following sequence of

transformations:

a reflection in the x-axis

a dilation of factor 3 from the x-axis

a translation of 2 units in the positive direction

of the x-axis and 1 unit in the positive direction

of the y-axis.0 x

(0, 0)

1

1

31, –

y

Solution

Consider each transformation separately and

sketch the graph at each stage. A reflection in

the x-axis produces the following graph:0

x

(0, 0)

–1

y

1

31,


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Next consider the dilation of factor 3 from

the x-axis:

0x

(0, 0)

(1, 1)

–3

y

Finally, apply the translation of 2 units in the

positive direction of the x-axis and 1 unit in

the positive direction of the y-axis.

0x

(2, 1)(3, 2)

–2

y

Example 8

For the graph of y = x2:

a Sketch the graph of the image under the sequence of transformations:

a translation of 1 unit in the positive direction of the x-axis and 2 units in the positive


a dilation of factor 2 from the y-axis

a reflection in the x-axis.

b State the rule of the image.

Solution

a Consider each transformation separately and

sketch the graph at each stage. The translation

produces the following graph:

x

3 (2, 3)

(1, 2)

0

y

Next consider the dilation of factor 2 from

the y-axis:

x

3

0

(2, 2)

(4, 3)

y


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Finally, apply the reflection in the x-axis:

x

–3

(2, –2)

(4, –3)

y

0

b The mapping representing the transformations is:

(x, y) → (x + 1, y + 2) → (2(x + 1), y + 2) → (2(x + 1), −(y + 2))

Let (x′, y′) be the coordinates of the image of (x, y), so x′ = 2(x + 1) and

y′ = −(y + 2)

Rearranging gives x = 1

2(x′ − 2) and y = −y′ − 2

Therefore, y = x2 becomes −y′ − 2 =(

1

2(x′ − 2)

)2

The rule of the transformed function is y = −1

4(x − 2)2 − 2

Using the TI-NspireDefine f (x) = x2.

The rule for the new function is

− f

(1

2(x − 2)

)− 2.

The calculator gives the equation of the

image of the graph under this sequence of

transformations.

Using the Casio ClassPadDefine f (x) = x2.

The rule for the new function is

− f

(1

2(x − 2)

)− 2.


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Exercise 3E

1 Find the rule of the image when the graph of each of the functions listed below undergoes

the following sequences of transformations:

i a dilation of factor 2 from the x-axis, followed by a reflection in the x-axis, followed by

a translation 3 units in the positive direction of the x-axis and 4 units in the negative


ii a dilation of factor 2 from the x-axis, followed by a translation 3 units in the positive

direction of the x-axis and 4 units in the negative direction of the y-axis, followed by a

reflection in the x-axis

iii a reflection in the x-axis, followed by a dilation of factor 2 from the x-axis, followed by

a translation 3 units in the positive direction of the x-axis and 4 units in the negative


iv a reflection in the x-axis, followed by a translation 3 units in the positive direction of

the x-axis and 4 units in the negative direction of the y-axis, followed by a dilation of

factor 2 from the x-axis

v a translation 3 units in the positive direction of the x-axis and 4 units in the negative

direction of the y-axis, followed by a dilation of factor 2 from the x-axis, followed by a


vi a translation 3 units in the positive direction of the x-axis and 4 units in the negative

direction of the y-axis, followed by a reflection in the x-axis, followed by a dilation of

factor 2 from the x-axis

a y = |x | b y = 3√x c y = 1

x3d y = 1

x4

e y = 13√x

f y = x23 g y = 1

x34

2 Sketch the graph of the image of the graph shown

under the following sequence of transformations:

a reflection in the x-axis


a translation of 3 units in the positive direction of the

x-axis and 4 units in the positive direction of the y-axisx

0

(5, 3)

2

y

3 Sketch the graph of the image of the graph shown

under the following sequence of transformations:

a reflection in the y-axis

a translation of 2 units in the negative direction of the

x-axis and 3 units in the negative direction of the y-axis

a dilation of factor 2 from the y-axisx

0

(2, 3)

4–2

y


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4 For the graph of y = |x |:a Sketch the graph of the image under the sequence of transformations:


a translation of 2 units in the negative direction of the x-axis and 1 unit in the

negative direction of the y-axis

a reflection in the x-axis.


5 For the graph of y = x13 :

a Sketch the graph of the image under the sequence of transformations:

a reflection in the y-axis

a translation of 1 unit in the positive direction of the x-axis and 2 units in the

negative direction of the y-axis

a dilation of factor1

2from the y-axis.


3.6 Determining transformations to sketch graphsBy considering a rule for a graph as a combination of transformations of a more ‘simple’ rule,

we are able to readily sketch graphs of many apparently ‘complicated’ functions.

Example 9

Identify a sequence of transformations that maps the graph of the function y = 1

xonto the

graph of the function y = 4

x + 5− 3, and use this to sketch the graph of y = 4

x + 5− 3,

stating the equations of asymptotes and the coordinates of axes intercepts.

Solution

Rearrange the rule of the function of the transformed graph into the formy′ + 3

4= 1

x ′ + 5(the ‘shape’ of y = 1

x), where (x ′, y′) are the coordinates of the

image of (x, y).

Therefore x = x ′ + 5 and y = y′ + 3

4. Rearranging gives x ′ = x − 5 and

y′ = 4y − 3.

So the mapping is given by (x, y) → (x − 5, 4y − 3) which identifies the sequence

of transformations as:

a dilation of factor 4 from the x-axis, followed by a translation of 3 units in the

negative direction of the y-axis, and


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a translation of 5 units in the negative direction of the x-axis.

Axes intercepts are found in the usual way, as below.

When x = 0, y = 4

5− 3

= −21

5

When y = 0,4

x + 5− 3 = 0

∴ 4 = 3x + 15

∴ 3x = −11

and x = −11

3

Transformations are shown below:

Original y = 1

x1 Dilation from

x-axis:

2 Translation in the

negative direction

of the x-axis:

3 Translation in the

negative direction

of the y-axis:

0

(–1, –1)

(1, 1)

x

y

0

(–1, –4)

(1, 4)

y

x

0

(–6, –4) x = –5

(–4, 4)

x

y

(–6, –7)

y = –3

0

(–4, 1)x

y

The result, with intercepts marked, is:

y = –3

x = –5

0x

, 0–11

3 0, –21

5

y

Once you have done a few of these types of exercises, you can identify the transformations

more quickly by carefully observing the rule of the transformed graph and relating it to the

‘shape’ of the simplest function in its family. Consider the following examples.


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Example 10

Sketch the graph of y = √x − 4 + 5

Solution

The graph is obtained from the graph of y = √x

through two translations:

4 units in the positive direction of the x-axis

5 units in the positive direction of the y-axis.x

(4, 5)

0

(8, 7)

y

Example 11

Sketch the graph of y = −√x − 4 − 5

Solution

The graph is obtained from the graph of y = √x by:


of the x-axis, and

a reflection in the x-axis, followed by a translation

of 5 units in the negative direction of the y-axis.

x

(4, –5)

0

y

Example 12

Sketch of graph of y = 3

(x − 2)2+ 5

Solution

This is obtained from the graph of y = 1

x2by:

a dilation of factor 3 from the x-axis, followed

by a translation of 5 units in the positive

direction of the y-axis, and


of the x-axis.

x

0, 534

y = 5

x = 2

0

y


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Example 13

Sketch the graph of y = 3√−x + 4

Solution

This is obtained from the graph of y = √x by:

a reflection in the y-axis, and

a dilation of factor 3 from the x-axis, followed by a translation of 4 units in the

positive direction of the y-axis.

x(1, 1)

0

y1 The original:

2

(–1, 1)x

0

y

4 Translated:

x0

(–1, 7)

4

y

3

x

(–1, 3)

0

y

Dilated:Reflected in the y-axis:

In general, the function given by the equation y = A f (n(x + c)) + b, where b, c ∈ R+

and A, n ∈ R, represents a transformation of the graph of y = f (x) by:

a dilation of factor |A| from (and if A < 0 a reflection in) the x-axis, followed by a

translation of b units in the positive direction of the y-axis, and

a dilation of factor

∣∣∣∣1

n

∣∣∣∣ from (and if n < 0 a reflection in) the y-axis, followed by a

translation of c units in the negative direction of the x-axis.

Exercise 3F

1 In each case below, state the sequence of transformations required to transform the graph

of the first equation into the graph of the second equation:

a y = 1

x, y = 2

x − 1+ 3 b y = 1

x2, y = −3

(x + 4)2− 7

c y = 1

x3, y = 4

(1 − x)3− 5 d y = 3√x , y = 2 − 3√x + 1

e y = 1√x

, y = 2√−x+ 3 f y = 2

3 − x+ 4, y = 1

x


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2 Sketch the graph of each of the following without using your calculator. State the equations

of asymptotes and axes intercepts. State the range of each function.

a f (x) = 3

x − 1b g(x) = 2

x + 1− 1 c h(x) = 3

(x − 2)2

d f (x) = 2

(x − 1)2− 1 e h(x) = −1

x − 3f f (x) = −1

x + 2+ 3

g f (x) = 2

(x − 3)3+ 4

3 Sketch the graph of each of the following without using your calculator. State the range of

each.

a y = −√x − 3 b y = −√

x − 3 + 2 c y = √2(x + 3)

d y = 1

2x − 3e y = 5

√x + 2 f y = −5

√x + 2 − 2

g y = −3

x − 2h y = −2

(x + 2)2− 4 i y = 3

2x− 5

j y = 5

2x+ 5 k y = 2|x − 3| + 5

4 Use your calculator to help you sketch the graph of each of the following. State the range

of each.

a y = − 3√x + 2 + 7 b y = 43√x − 1

+ 2 c y = −(x + 1)34 − 6

5 a Show that3x + 2

x + 1= 3 − 1

x + 1and hence, without using your calculator, sketch the graph of:

f : R\{−1} → R, f (x) = 3x + 2

x + 1

b Show that4x − 5

2x + 1= 2 − 7

2x + 1and hence, without using your calculator, sketch the graph of:

f : R\{−1

2

}→ R, f (x) = 4x − 5

2x + 1

Note: f (x) = 2 − 7

2(x + 1

2

)6 Sketch the graph of each of the following without using your calculator. State the range of

each.

a y = 2

x − 3+ 4 b y = 4

3 − x+ 4 c y = 2

(x − 1)2+ 1

d y = 2√

x − 1 + 2 e y = −3√

x − 4 + 1 f y = 5√

2x + 4 + 1


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3.7 Using matrices for transformationsA summary of some of the transformations and their rules which were introduced earlier in

this chapter is presented here. Suppose (x ′, y′) is the image of (x, y) under the mapping in the

first column of the table below.

Mapping Rule

Reflection in the x-axis x ′ = x = x + 0y

y′ = −y = 0x + −y

Reflection in the y-axis x ′ = −x = −x + 0y

y′ = y = 0x + y

Dilation by factor k from the y-axis x ′ = kx = kx + 0y

y′ = y = 0x + y

Dilation by factor k from the x-axis x ′ = x = x + 0y

y′ = ky = 0x + ky

Reflection in the line y = x x ′ = y = 0x + y

y′ = x = x + 0y

Translation defined by a vector

[a

b

]x ′ = x + a

y′ = y + b

The first five mappings are special cases of a general kind of mapping defined by

x ′ = ax + by

y′ = cx + dy

where a, b, c, d are real numbers.

These equations can be rewritten as

x ′ = a11x + a12 y

y′ = a21x + a22 y

which yields the equivalent matrix equation[x ′

y′

]=

[a11 a12

a21 a22

] [x

y

]

A transformation of the form

(x, y) → (a11x + a12 y, a21x + a22 y)

is called a linear transformation.


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The first five transformations above can be defined by a 2 × 2 matrix. This is shown in the

table below.

Mapping Matrix

Reflection in the x-axis

[1 0

0 −1

]

Reflection in the y-axis

[−1 0

0 1

]

Dilation by factor k from the y-axis

[k 0

0 1

]

Dilation of factor k from the x-axis

[1 0

0 k

]

Reflection in the line y = x

[0 1

1 0

]

Example 14

Find the image of the point (2, 3) under

a a reflection in the x-axis b a dilation of factor k from the y-axis

Solution

a

[1 0

0 −1

] [2

3

]=

[2

−3

]. Therefore (2, 3) → (2, −3)

b

[k 0

0 1

] [2

3

]=

[2k

3

]. Therefore (2, 3) → (2k, 3)

Example 15

Consider a linear transformation such that (1, 0) → (3, −1) and (0, 1) → (2, −4) . Find the

image of (−3, 5).

Solution[a11 a12

a21 a22

] [1

0

]=

[3

−1

]and

[a11 a12

a21 a22

] [0

1

]=

[2

−4

]

∴ a11 = 3, a21 = −1 and a12 = 2, a22 = −4


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i.e. the transformation can be defined by the 2 × 2 matrix

[3 −2

−1 4

]

Let (−3, 5) → (x ′, y′)

∴[

x ′

y′

]=

[3 −2

−1 4

] [−3

5

]=

[3 × −3 + −2 × 5

−1 × −3 + 4 × 5

]=

[−19

23

]

∴ (−3, 5) → (−19, 23)

The image of (−3, 5) is (−19, 23).

Note that non-linear mappings cannot be represented by a matrix in the way indicated above.

Thus for the translation defined by (0, 0) → (a, b)

x ′ = x + a

y′ = y + b

While this cannot be represented by a square matrix, the defining equations

suggest

[x ′

y′

]=

[x

y

]+

[a

b

]

where the ‘sum’ has the definition:

for each x, y, a, b in R,

[x

y

]+

[a

b

]=

[x + a

y + b

]

Composition of mappings

Consider a linear transformation defined by the matrix A =[

a11 a12

a21 a22

]composed with a

linear transformation defined by the matrix B =[

b11 b12

b21 b22

]

The composition consists of the transformation of A being applied first and then the

transformation of B.

The matrix of the resulting composition is the product

BA =[

b11a11 + b12a21 b11a12 + b12a22

b21a11 + b22a21 b21a12 + b22a22

]

Example 16

Find the image of the point (2, −3) under a reflection in the x-axis followed by a dilation of

factor k from the y-axis.


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Solution

Matrix multiplication gives the matrix. Let A be the transformation reflection in the

x-axis and B the transformation dilation of factor k from the y-axis. Then the required

transformation is defined by the product

BA =[

k 0

0 1

] [1 0

0 −1

]=

[k 0

0 −1

]and BA

[2

−3

]=

[2k

3

]

Example 17

Express the composition of the transformations, dilation of factor k from the y-axis followed

by a translation defined by the matrix C =[

a

b

], mapping a point (x, y) to a point (x ′, y′) as a

matrix equation. Hence find x and y in terms of x ′ and y′ respectively.

Solution

Let A be the dilation transformation, X =[

x

y

], and X′ =

[x ′

y′

]

The equation is AX + C = X′

Then AX = X′ − C and hence X = A−1(X′ − C)

Now A =[

k 0

0 1

]

det(A) = k and therefore A−1 = 1

k

[1 0

0 k

]=

1

k0

0 1

X = 1

k0

0 1

([

x ′

y′

]−

[a

b

])=

1

k0

0 1

[

x ′ − a

y′ − b

]=

1

k(x ′ − a)

y′ − b

Hence x = 1

k(x ′ − a) and y′ = y − b

Transforming graphsThe notation is now applied to transforming graphs. The notation is consistent with the

notation introduced earlier in this chapter.

Example 18

A transformation is defined by the matrix

[1 0

0 2

]. Find the equation of the image of the

graph of the quadratic equation y = x2 + 2x + 3 under this transformation.


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Solution

As before the transformation maps (x, y) → (x ′, y′).

Using matrix notation,

[1 0

0 2

] [x

y

]=

[x ′

y′

]

It can be written as the matrix equation TX = X′

Now multiply both sides of the equation by T−1.

Therefore T−1TX = T−1X′

and X = T−1X′

Therefore

[x

y

]=

1 0

01

2

[x ′

y′

]

[x

y

]=

x ′

1

2y′

x = x ′ and y = y′

2The curve with equation y = x2 + 2x + 3 is mapped to the curve with equation

y′

2= (x ′)2 + 2x ′ + 3

This makes quite hard work of an easy problem, but it demonstrates a procedure

that can be used for any transformation defined by a 2 × 2 non-singular matrix.

Example 19

A transformation is described through the equation T(X + B) = X′ where T =[

0 −3

2 0

]and

B =[

1

2

]. Find the image of the straight line with equation y = 2x + 5 under the

transformation.

Solution

First solve the matrix equation for X.

T−1T(X + B) = T−1X′

X + B = T−1X′

and X = T−1X′ − B

Therefore

[x

y

]=

0

1

2−1

30

[x ′

y′

]−

[1

2

]=

y′

2− 1

− x ′

3− 2

Therefore x = y′

2− 1 and y = − x ′

3− 2

The straight line with equation y = 2x + 5 is transformed to the straight line with

equation − x ′

3− 2 = 2

(y′

2− 1

)+ 5

Rearranging gives y′ = − x ′

3− 5


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Exercise 3G

1 Using matrix methods find the image of the point (1, −2) under each of the following

transformations:

a dilation of factor 3 from the x-axis b dilation of factor 2 from the y-axis

c reflection in the x-axis d reflection in the y-axis

e reflection in the line y = x

2 Find the matrix that determines the composition of transformations, in the given order:


dilation of factor 2 from the x-axis

3 Write down the matrix of each of the following transformations:

a reflection in the line x = 0

b reflection in the line y = x

c reflection in the line y = −x

d dilation of factor 2 from the x-axis

e dilation of factor 3 from the y-axis

4 Express the composition of the transformations, dilation of factor 3 from the x-axis

followed by a translation defined by the matrix C =[

2

1

], mapping a point (x, y) to a

point (x ′, y′) as a matrix equation. Hence find x and y in terms of x ′ and y′ respectively.

5 A transformation is defined by the matrix

[−4 0

0 2

]. Find the equation of the image of

the graph of the quadratic equation y = x2 + x + 2 under this transformation.


[1 0

0 −2


the graph of the cubic equation y = x3 + 2x under this transformation.


[0 3

−2 0


the graph of the straight line with equation y = 2x + 3 under this transformation.


[0 2

−3 0


the graph of the straight line with equation y = −2x + 4 under this transformation.

9 A transformation is described through the equation T(X + B) = X′ where

T =[

0 1

−3 0

]and B =

[−1

2

]. Find the image of the straight line with equation

y = −2x + 6 under the transformation.


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10 A transformation is described through the equation TX + B = X′

where T =[

0 −3

1 0

]and B =

[−3

2

]. Find the image of the straight line with equation

y = −2x + 6 under the transformation.


where T =[

4 0

0 2

]and B =

[−1

4

]. Find the image of the curve with equation

y = −2x3 + 6x under the transformation.


where T =[

−3 0

0 −2

]and B =

[−1

2

]. Find the image of the curve with equation

y = −2x3 + 6x2 + 2 under the transformation.

3.8 Determining the rule for a function of a graphGiven sufficient information about a curve, a rule for the function of the graph may be

determined. For example, if the coordinates of two points on a hyperbola of the form

y = a

x+ b

are known, the rule for the hyperbola may be found, i.e. the values of a and b may be found.

Sometimes a more specific rule is known. For example, the curve may be a dilation

of y = √x . It is then known to be of the y = a

√x family, and the coordinates of one point

(with the exception of the origin) will be enough to determine the value for a.

Example 20

It is known that the points (1, 5) and (4, 2) lie on a curve with the equation y = a

x+ b. Find

the values of a and b.

Solution

When x = 1, y = 5, therefore 5 = a + b (1)

and when x = 4, y = 2, therefore 2 = a

4+ b (2)

Subtract (2) from (1): 3 = 3a

4∴ a = 4

Substitute in (1) to find b: 5 = 4 + b

Therefore b = 1

and y = 4

x+ 1

Example 21

It is known that the points (2, 1) and (10, 6) lie on a curve with equation y = a√

x − 1 + b.

Find the values of a and b.


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Solution

When x = 2, y = 1.

Therefore 1 = a√

1 + b, i.e. 1 = a + b (1)

When x = 10, y = 6.

Therefore 6 = a√

9 + 6, i.e. 6 = 3a + b (2)

Subtract (1) from (2): 5 = 2a

∴ a = 5

2

Substitute in (1) to find b: 1 = 5

2+ b

Therefore b = −3

2

and y = 5

2

√x − 1 − 3

2

Exercise 3H

1 The graph shown has the rule:

y = A

x + b+ B

Find the values of A, b and B.x

0

(0, 1)y = 2

x = 1

y

2 The points with coordinates (1, 5) and (16, 11) lie on a curve which has a rule of the form

y = A√

x + B. Find A and B.

3 The points with coordinates (1, 1) and (0.5, 7) lie on a curve which has a rule of the form

y = A

x2+ B. Find the values of A and B.

4 The graph shown has the rule:

y = A

(x + b)2+ B

Find the values of A, b and B.x0

(0, –1)

y = –3

x = –2

y

5 The points with coordinates (1, −1) and

(2,

3

4

)lie on a curve which has a rule of the form

y = a

x3+ b. Find the values of a and b.

6 The points with coordinates (−1, −8) and (1, −2) lie on a curve which has a rule of the

form y = a3√x

+ b. Find the values of a and b.


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3.9 Addition of ordinatesIn Chapter 1 it was established that for functions f and g a new function f + g can be

defined by:( f + g) (x) = f (x) + g(x)

dom ( f + g) = dom ( f ) ∩ dom (g)

In this section graphing by addition of ordinates is considered.

Example 22

Sketch the graphs of f (x) = x + 1 and g(x) = 3 − 2x and hence the graph of ( f + g)(x).

Solution

If f and g are functions defined by:

f (x) = x + 1

and g(x) = 3 − 2x

then ( f + g)(x) = f (x) + g(x)

= 4 − x

We note ( f + g)(2) = f (2) + g(2) = 3 + −1 = 2,

i.e. the ordinates are added.

x

1

10–1–1–2

–2 2(2, –1)

y = f (x)

y = ( f + g)(x)

y = g(x)

(2, 3)(2, 2)

3 4

2

3

4

y

Now check that the same principle applies for other points on the graphs. A table of

values can be a useful aid to find points that lie on the graph of y = ( f + g)(x).

x f (x) g(x) ( f + g)(x)

−1 0 5 5

0 1 3 4

3

2

5

20

5

2

2 3 −1 2

The table shows that the points (−1, 5), (0, 4),

(3

2,

5

2

)and (2, 2) lie on the graph of

y = ( f + g)(x)

Example 23

Sketch the graph of y = ( f + g)(x) where f (x) = √x and g(x) = x


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Solution

It can be seen that the function with rule

( f + g)(x) = √x + x is defined by the

addition of the two functions f and g.

10

1

2

x

y = ( f + g)(x)

y = g(x)

y = f (x)

y

Exercise 3I

1 Sketch the graph of f : R+ ∪ {0} → R, f (x) = √x + x using addition of ordinates.

2 Sketch the graph of f : [−2, ∞) → R, f (x) = √x + 2 + x using addition of ordinates.

3 Sketch the graph of f : R+ ∪ {0} → R, f (x) = −√x + x using addition of ordinates.

4 Sketch the graph of f : R\(0) → R, f (x) = 1

x+ 1

x2using addition of ordinates.

5 For each of the following sketch the graph of f + g:

a f : [−2, ∞) → R, f (x) = √2 + x, g: R → R, g(x) = −2x

b f : (−∞, 2], f (x) = √2 − x, g: [−2, ∞) → R, g(x) = √

x + 2

3.10 Graphing inverse functionsA transformation which reflects each point in the plane in the line y = x can be described

through ‘interchanging the x- and y-coordinates of each point in the plane’ and can be written

as (x, y) → (y, x). This is read as ‘the ordered pair (x, y) is mapped onto the ordered pair

(y, x)’. The reflection is a one-to-one mapping from R2 to R2 and uniquely links any ordered

pair (a, b) to the ordered pair (b, a).

This special case where the graph of a function is reflected in the line y = x produces the

graph of the inverse relation.

Consider reflecting the graph of the

function shown here in the line

y = x , and observe the effect on a

general point (x, y) on the graph. 0x

(x, y)

y

x

(x, y)

(y, x)

y = x

0

y

The line y = x acts as a ‘mirror’ line. The point (x, y) is mapped onto (y, x), i.e.

(x, y) → (y, x)

In general, a reflection in the line y = x is the transformation that maps

(x, y) → (y, x). To find the equation of the image of y = f (x), under the reflection

(x, y) → (y, x), replace x with y and y with x; i.e. the image of the graph of

{(x, y): y = f (x)} under the reflection (x, y) → (y, x) is the graph of

{(y, x): y = f (x)} and is called the inverse relation of y = f (x).


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Families of functions, with A, B, b ∈ R, A �= 0One-to-one functions

f (x) = A

(x + b)n+ B, where n is a positive odd integer, and

g(x) = A(x + b)pq + B where p and q are integers, p odd

and where the highest common factor of p, q = 1

are one-to-one functions, and therefore have inverses that are functions. See Section 1.7.

Simple examples of such functions (other than polynomials which will be explored in

Chapter 4) are:

f (x) = 1

x, g(x) = 1

x3, h(x) = 1

x5, . . .

f (x) = x12 , g(x) = x

13 , h(x) = x

14 , r (x) = x

32 , s(x) = x

34 , v(x) = x

35 , w(x) = x

53 , . . .

f (x) = 1

x12

, g(x) = 1

x13

, h(x) = 1

x14

, r (x) = 1

x32

, s(x) = 1

x34

, v(x) = 1

x35

, w(x) = 1

x53

, . . .

Example 24

Find the inverse of the function with rule f (x) = 3√

x + 2 + 4 and sketch both functions on

one set of axes, clearly showing the exact coordinates of intersection of the two graphs.

Solution

Consider x = 3√

y + 2 + 4

Solve for y:x − 4

3=

√y + 2

which implies y =(

x − 4

3

)2

− 2

∴ f −1(x) =(

x − 4

3

)2

− 2

and as the domain of f −1 = range of f

f −1: [4, ∞) → R, f −1(x) =(

x − 4

3

)2

− 2

Using the TI-NspireTo find the rule for the inverse of

y = 3√

x + 2 + 4, enter solve(x = 3

√y + 2 + 4, y

).


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Using the Casio ClassPadTo find the rule for the inverse of

f (x) = 3√

x + 2 + 4, enter and highlight

x = 3√

y + 2 + 4.

Tap Interactive > Equation/inequality

> solve and set the variable as y.

Note: The graph of f −1 is obtained by

reflecting the graph of f in the line y = x .

The graph of y = f −1(x) is obtained

from the graph of y = f (x) by applying

the transformation (x, y) → (y, x).x

(4, –2)0

(–2, 4)

(0, 3√2 + 4)

y = x

y = 3√x + 2 + 4

y =x – 4

3

y

2 – 2

The graphs meet where f (x) = x = f −1(x). Points of intersection of the graphs of

y = f (x) and y = f −1(x) are usually found by solving either f (x) = x or

f −1(x) = x , rather than the more complicated equation f (x) = f −1(x). (Note that

points of intersection between the graphs of y = f (x) and y = f −1(x) that do not lie

on the line y = x also sometimes exist.)

In this particular example, it is simpler to solve f −1(x) = x

That is,

(x − 4

3

)2

− 2 = x, x > 4(x − 4

3

)2

= x + 2

x2 − 17x − 2 = 0

∴ x = 17 ± √172 + (4 × 2)

2

As x > 4, only the positive solution is valid.

The two graphs meet at the point

(17 + √

297

2,

17 + √297

2

)≈ (17.12, 17.12)


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Example 25

Expressx + 4

x + 1in the form

a

x + b+ c and hence find the inverse of the function

f (x) = x + 4

x + 1. Sketch both functions on the one set of axes.

Solution

x + 4

x + 1= 3 + x + 1

x + 1= 3

x + 1+ x + 1

x + 1= 3

x + 1+ 1

Consider x = 3

y + 1+ 1

Solve for y : x − 1 = 3

y + 1

and thus y + 1 = 3

x − 1

and y = 3

x − 1− 1

The range of f is R\{1}, and thus:

f −1: R\{1} → R, f −1(x) = 3

x − 1− 1

Note: The graph of f −1 is obtained by reflecting the graph of f in the line y = x . The two

graphs meet where

3

x + 1+ 1 = x, x �= −1,

i.e. where x = ±2

x

y = –1

x = –1

(–2, –2)

(–4, 0)

y = 1

x = 1

y = x

(0, –4)

0 (4, 0)

(2, 2)

(0, 4)

y

The two graphs meet at the points with

coordinates (2, 2) and (−2, −2).

Many-to-one functionsf (x) = A

(x + b)n+ B, where n is a positive even integer, and

g(x) = A(x + b)pq + B where p and q are integers, p even, q odd

and where the highest common factor of p, q = 1

are many-to-one functions, and therefore have inverse relations that are not functions.


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Simple examples of such functions (other than polynomials which will be explored in

Chapter 4) are:

f (x) = 1

x2, g(x) = 1

x4, h(x) = 1

x6, . . .

f (x) = x23 , g(x) = x

25 , h(x) = x

43 , . . .

f (x) = x− 23 , g(x) = x− 4

5 , h(x) = x− 65 , . . .

Inverse relations will be discussed in Chapter 7.

Exercise 3J

1 Find the inverse function of each of the following, and sketch the graph of the inverse

function:

a f : R+ ∪ {0} → R, f (x) = √x + 2 b f : R\{3} → R, f (x) = 1

x − 3

c f : [2, ∞) → R, f (x) = √x − 2 + 4 d f : R\{2} → R, f (x) = 3

x − 2+ 1

e f : R\{1} → R, f (x) = 5

x − 1− 1 f f : (−∞, 2] → R, f (x) = √

2 − x + 1

2 For each of the following functions, find the inverse function and state its domain:

a g(x) = 3

xb g(x) = 3√x + 2 − 4 c h(x) = 2 − √

x

d f (x) = 3

x+ 1 e h(x) = 5 − 2

(x − 6)3f g(x) = 1

(x − 1)34

+ 2

3 For each of the following, copy the graph onto a grid and sketch the graph of the inverse on

the same set of axes. In each case state whether the inverse is or is not a function.

a

0x

1 2 3–1

–1

–2

–2

–3

–3

2

1

3

y b

0x

1 2 3 4–1

–1

–2

–2

–3

–3

2

1

3

y c

0x

1 2 3 4–1

–1

–2

–2

–3

–3

2

1

3

y

d

1 2 3 5 64–1

–2

–3

2

1

3

0 x

y e

0x

1 2 3 4–1–1

–2

–2

–3–4

–3–4

2

1

34

y


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4 Find the rule for the inverse of the following functions:

a f : R\{1} → R, f (x) = x + 1

x − 1b f : [2, ∞) → R, f (x) = √

x − 2

c f : R\{

2

3

}→ R, f (x) = 2x + 3

3x − 2

5 Let f : S → R be given by f (x) = x + 3

2x − 1where S = R\

{1

2

}a Show that f ◦ f is defined.

b Find f ◦ f (x) and sketch the graph of f ◦ f .

c Write down the inverse of f.


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Review


Chapter summary

Functions of the form f (x) = 1

xn, where n is a positive odd integer, and f (x) = x

− pq ,

where p and q are both positive, odd integers with highest

common factor 1, have domain R\{0} and range

R\{0}. They are one-to-one and odd functions

with asymptotes at x = 0 and y = 0. The points

(−1, −1) and (1, 1) lie on the graph. Examples

include f (x) = 1

x, g(x) = 1

x3, h(x) = 1

x35

and

r (x) = 1

x97

0x

y

Functions of the form f (x) = 1

xn, where n is a positive even integer, and f (x) = x

− pq ,

where p is a positive even integer and q is a

positive odd integer with highest common

factor 1, have domain R\{0} and range R+.

They are many-to-one and even functions

with asymptotes at x = 0 and y = 0. The

points (−1, 1) and (1, 1) lie on the graph.

Examples include f (x) = 1

x2, g(x) = 1

x4,

h(x) = 1

x23

and r (x) = 1

x65

x0

y

Functions of the form f (x) = xpq , where p and q are both positive odd integers with

highest common factor 1, have domain R and range R. They are one-to-one and odd

functions. The points (−1, −1), (0, 0) and (1, 1) lie on the graph. Examples include

f (x) = x3, g(x) = x13 , h(x) = x

37 and r (x) = x

53

x0

y

x0

y

Functions of the form f (x) = xpq , where p is a positive even integer and q is a positive odd

integer with highest common factor 1, have domain R and range [0, ∞). They are

many-to-one and even functions. The points (−1, 1) and (1, 1) lie on the graph. Examples

include f (x) = x2, g(x) = x23 and h(x) = x

43


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Rev

iew


x0

y

0x

y

Functions of the form f (x) = xpq , where p is a positive odd integer and q is a positive even

integer with highest common factor 1, have domain [0, ∞) and range [0, ∞) and are

one-to-one. The point (1, 1) lies on the graph. Examples include f (x) = x12 , g(x) = x

14 ,

h(x) = x32 and r (x) = x

34

x0

y

0x

y

Functions of the form f (x) = x− p

q , where p is a positive odd integer and q is a positive

even integer with highest common factor 1, have domain

R+ and range R+ and asymptotes at x = 0 and y = 0.

They are one-to-one functions. The point (1, 1) lies on

the graph. Examples include f (x) = 1

x12

, g(x) = 1

x14

,

h(x) = 1

x32

and r (x) = 1

x54 0

x

y

The dilation of factor a (a > 0) from the x-axis is the transformation which maps

(x, y) → (x, ay)

To find the equation of the image of the curve with equation y = f (x) under the dilation

(x, y) → (x, ay), replace y byy

ain y = f (x), i.e. the image of the graph of y = f (x) under

the dilation (x, y) → (x, ay) is the graph ofy

a= f (x) or y = a f (x)

The dilation of factor b (b > 0) from the y-axis is the transformation which maps

(x, y) → (bx, y)

To find the equation of the image of the curve with equation y = f (x) under the dilation

(x, y) → (bx, y), replace x byx

bin y = f (x), i.e. the image of the graph of y = f (x)

under the dilation (x, y) → (bx, y) is the graph of y = f( x

b

)


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Review


To find the equation of the image of the curve with equation y = f (x) under the reflection

in the x-axis, (x, y) → (x, −y), replace y by −y in y = f (x), i.e. the image of the graph of

y = f (x) under the reflection (x, y) → (x, −y) is the graph of y = − f (x)

To find the equation of the image of the curve with equation y = f (x) under the reflection

in the y-axis, (x, y) → (−x, y), replace x by −x in y = f (x), i.e. the image of the graph of

y = f (x) under the reflection (x, y) → (−x, y) is the graph of y = f (−x)

The translation of h units (h > 0) in the positive direction of the x-axis and k units (k > 0)

in the positive direction of the y-axis is the transformation that maps

(x, y) → (x + h, y + k)

To find the equation of the curve with the equation y = f (x) under the translation

(x, y) → (x + h, y + k), replace x by x − h and y by y − k in y = f (x), i.e. the image of

the graph of y = f (x) under the translation (x, y) → (x + h, y + k) is the graph of

y − k = f (x − h) or y = f (x − h) + k

To find the equation of the inverse relation of y = f (x), under the reflection

(x, y) → (y, x), replace x with y and y with x, i.e. the image of the graph with equation

y = f (x) under the reflection (x, y) → (y, x) is the graph with equation x = f (y)

A function which is one-to-one has an inverse function. A function which is many-to-one

has an inverse relation that is not a function.

Transformation Matrix

Reflection in the x-axis

[1 0

0 −1

]

Reflection in the y-axis

[−1 0

0 1

]

Dilation by factor k from the y-axis

[k 0

0 1

]

Dilation of factor k from the x-axis

[1 0

0 k

]

Reflection in the line y = x

[0 1

1 0

]


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Multiple-choice questions

1 The graph of the function with rule y = f (x) is

shown below.

x

2

20

–2

–2

y

Which one of the following is most likely to be the graph of the inverse function?

A

x

2

20

–2

–2

y B

x

2

20

–2

–2

y C

x

2

20

–2

–2

y

D

x

2

20

–2

–2

y E

x

2

20

–2

–2

y

2 The graph of the function with rule y = |x | is reflected in the x-axis and then translated

4 units in the negative direction of the x-axis and 3 units in the negative direction of the

y-axis. The rule for the new function is:

A y = |−x + 4| − 3 B y = −|x − 4| + 3 C y = −|x − 3| + 4

D y = |−x − 4| + 3 E y = −|x + 4| − 3

3 The graph of the function with rule y = a

x + b+ c

is shown on the right.

x0

–2

–3

y

A possible set of values for a, b and c respectively is:

A −1, 3, 2

B 1, 2, −3

C −1, −3, −2

D −1, 3, −2

E 1, 2, −3


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Review


4 The graph of the function f is obtained from the graph of the function with equation y = x13

by a reflection in the y-axis followed by a dilation of 5 units from the x-axis. The rule for f

is:

A f (x) = −5x13 B f (x) = 1

5(−x)

13 C f (x) = (−5x)

13

D f (x) = −1

5x

13 E f (x) = −5(−x)

13

5 A function with rule f (x) = 1

x4can be defined on different domains. Which one of the

following does not give the correct range for the given domain?

A dom f = [−1, −0.5], ran f = [1, 16]

B dom f = [−0.5, 0) ∪ (0, 0.5], ran f = [16, ∞)

C dom f = (−0.5, 0.5)\{0}, ran f = (16, ∞)

D dom f = [−0.5, 1]\{0}, ran f = [1, 16]

E dom f = [0.5, 1), ran f = (1, 16]

6 The function with rule y = f (x) is shown on the right.

x

2

–2

–2 20

y

Which one of the following could be the graph of the function with rule y = f (−x)?

A

x

2

–2

–2 20

y B

x

2

–2

–2 20

y C

x

2

–2

–2 20

y

D

x

2

–2

–2 20

y E

x

2

–2

–2 20

y

7 Let g(x) = 3

(x + 1)3− 2. The equations of the asymptotes of the inverse function g−1 are:

A x = −2, y = 1 B x = −2, y = −1 C x = 1, y = −2

D x = −1, y = −2 E x = 2, y = −1


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iew


8 The equations of the vertical and horizontal asymptotes of the graph with equation−2

(x + 3)4− 5 are:

A x = 3, y = −5 B x = −5, y = −3 C x = −3, y = −5

D x = −2, y = −5 E x = −3, y = 5

9 The function g: R\{3} → R, where g(x) = 1

x − 3+ 2 has an inverse g−1. The rule and

domain of g−1 are:

A g−1(x) = 1

x − 2+ 3, dom g−1 = R\{2}

B g−1(x) = 1

x − 2+ 3, dom g−1 = R\{3}

C g−1(x) = 1

x + 2− 3, dom g−1 = R\{2}

D g−1(x) = −1

x + 2− 3, dom g−1 = R\{3}

E g−1(x) = 1

x − 2+ 3, dom g−1 = R\{−3}

10 Which one of the following functions does not have an inverse function?

A f : [0, ∞) → R, f (x) = |x − 2| B f : R → R, f (x) = x3

C f : [−3, 3] → R, f (x) = √9 − x D f : R → R, f (x) = x

13 + 4

E f : R → R, f (x) = 3x + 7

Short-answer questions (technology-free)

1 State the maximal domain and range of each of the following:

a f (x) = 1

x+ 2 b f (x) = 3 − 2

√3x − 2 c f (x) = 4

(x − 2)2+ 3

d h(x) = 4 − 3

x − 2e f (x) = √

x − 2 − 5

2 Sketch the graphs of each of the following. Label any asymptotes and axes intercepts. State

the range of each:

a f : R\{0} → R, f (x) = 1

x− 3 b h: (2, ∞) → R, f (x) = 1

x − 2

c f : R\{1} → R, f (x) = 2

x − 1− 3 d h: (2, ∞), f (x) = −3

2 − x+ 4

e f : R\{1}, h(x) = 1 − 1

x − 13 Sketch the graphs of each of the following:

a f (x) = 2√

x − 3 + 1 b g(x) = 3

(x − 2)2− 1 c h(x) = −3

(x − 2)2− 1

4 The points with coordinates (1, 3) and (3, 7) lie on the curve with equation of the form

y = a

x+ b. Find the values of a and b.

5 Find the inverse of the function with the rule f (x) = √x − 2 + 4 and sketch both functions

on the one set of axes.

6 Find the inverse of the function with the rule f (x) = x − 2

x + 1


SAMPLE

P1: FXS/ABE P2: FXS


Review


Extended-response questions

1 Consider the function f : D → R, for which the rule is f (x) = 24

x + 2− 6, where D is the

maximal domain for f.

a Find D.

b Describe a set of transformations which, when applied to the graph of y = 1

x, produce

the graph of y = f (x). Specify the order in which these transformations are to be

applied.

c Find the coordinates of the points where the graph of f cuts the axes.

d Let g: (−2, ∞) → R, g(x) = f (x).

i Find the rule for g−1, the inverse of g.

ii Write down the domain of g−1.

iii Find the values of x for which g(x) = x and hence the values of x for which

g(x) = g−1(x).

iv Sketch the graphs of y = g(x) and y = g−1(x) on the one set of axes.

2 Consider the function f : D → R, for which the rule is f (x) = 4 − 2√

2x + 6, where D is

the maximal domain for f.

a Find D.

b Describe a set of transformations which, when applied to the graph of y = √x, produce

the graph of y = f (x). Specify the order in which these transformations are to be

applied.

c Find the coordinates of the points where the graph of f cuts the axes.

d Find the rule for f −1, the inverse of f. e Find the domain for f −1.

f Find the value(s) of x for which f (x) = x and hence the values of x for which

f (x) = f −1(x).

g Sketch the graphs of y = f (x) and y = f −1(x) on the one set of axes.

3 a i Find the dilation from the x-axis which takes y = x2 to the parabola with its vertex

at the origin and which passes through the point (25, 15).

ii State the rule which reflects this dilated parabola in the x-axis.

iii State the rule which takes the reflected parabola of ii to a parabola with

x-intercepts (0, 0) and (50, 0) and vertex (25, 15).

iv State the rule which takes the curve of y = x2 to the parabola defined in iii.

b The plans for the entrance of a new building

involve twin parabolic arches as

shown in the diagram.

x

15 m

50 m 50 m

Arch 1 Arch 2

y

0

i From the results of a, give the equation

for the curve of arch 1.

ii Find the translation which maps the curve

of arch 1 to the curve of arch 2.

iii Find the equation of the curve of arch 2.


SAMPLE

P1: FXS/ABE P2: FXS


Rev

iew


c The architect wishes to have flexibility in her planning and so wants to develop an

algorithm for determining the equations of the curves given arch width m metres and

height n metres.

i Find the rule for the transformation which takes the graph of y = x2 to the current

arch 1 with these dimensions.

ii Find the equation for the curve of arch 1.

iii Find the equation for the curve of arch 2.

4 Consider the function g: D → R, for which the rule is g(x) = 3

(3x − 4)2+ 6, where D is

the maximal domain of g.

a Find D.

b Find the smallest value of a such that f: (a, ∞) → R, f (x) = g(x) is a one-to-one

function.

c Find the inverse function for f.

d Find the value of x for which f (x) = f −1(x)

e On the one set of axes sketch the graph of y = f (x) and y = f −1(x)

5 a Sketch the curve with equation f (x) = 50

20 − x, x �= 20

b If g(x) = 50x

20 − x:

i Show that g(x) = 1000

20 − x− 50

ii Sketch the graph of y = g(x)

iii Show that g(x) = 20 f (x) − 50

c Find the rule for the function g−1.

6 When the transformation with rule (x, y) → (y, x) (a reflection in the line y = x) is applied

to the graph of the one-to-one function, f, the resulting image has rule y = f −1(x), i.e. the

graph of the inverse function is obtained.

a For the graph of y = f (x), find the rule for the image of f, in terms of f −1(x), for each

of the following sequences of transformations:

i a translation of 3 units in the positive direction of x-axis

a translation of 5 units in the positive direction of the y-axis

a reflection in the line y = x

ii a reflection in the line y = x

a translation of 3 units in the positive direction of x-axis

a translation of 5 units in the positive direction of the y-axis

iii a dilation of factor 3 from the x-axis


a reflection in the line y = x


SAMPLE

P1: FXS/ABE P2: FXS


Review


iv a reflection in the line y = x



b Find the image of the graph of y = f (x), in terms of f −1(x), under the transformation

with rule (x, y) → (ay + b, cx + d), where a, b, c and d are positive constants, and

describe this transformation in words.


SAMPLE

Documents

P1: FXS/ABE P2: FXS CHAPTER 3...P1: FXS/ABE P2: FXS 0521842344c03.xml CUAT030-EVANS August 27, 2008 1:39 Chapter3—Families of functions 83 In the diagram below, the graphs of f (x)