3. Families of Functions

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9780521740531c03.xml CUAU156-EVANS January 1, 1904 0:23

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 numberand 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) = xnIn 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 thefamily of functions of the form f (x) = xn where n is a positive integer. It is appropriate todelay the introduction of this family until Chapter 4.

f(x) = xn where n is a negative integerWhen n = 1, f (x) = x1

= 1x

The maximal domain of this function is R\{0}.The graph of the function is as shown.

x0

y

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Michael Evans et al. 2011 Cambridge University Press

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82 Essential Mathematical Methods 3 & 4 CAS

Asymptotes:

There is a horizontal asymptote with equation y = 0As x , 1

x 0+, i.e. from the positive side.

As x , 1x

0, i.e. from the negative direction.There is a vertical asymptote with equation x = 0As x 0+, i.e. from the positive direction, 1

x

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

f (x) = 1

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

When n = 2, f (x) = x2 = 1x2

. The maximal

domain of this function is R\{0}. The graph ofthe function is as shown on the right. x

0

y

Asymptotes:

There is a horizontal asymptote with equation y = 0As x2 , 1

x2 0+, i.e. from the positive side.

There is a vertical asymptote with equation x = 0As x 0+, i.e. from the positive direction, 1

x2

As x 0, i.e. from the negative side, 1x2

f (x) = 1

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

In the diagram on the right, the graphs of f (x) = 1x2

and

f (x) = 1x4

are shown on the one set of axes.

The graphs intersect at the points with

coordinates (1, 1) and (1, 1).Note that

1

x2>

1

x4for x > 1 and x < 1,

and1

x2

1

xmfor x > 1 and x < 1, and 1

xn

1

x3for x > 1 and 1 < x < 0,

and1

x

1

xmfor x > 1 and 1 < x < 0, and 1

xn1

xfor x > 1, and

1x

1

xfor x > 1 and 1 < x < 0, and

13x x2} b Find {x : x

32 < x2}

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

a f (x) = 1x

b f (x) = 1x2

c f (x) = 3x

d f (x) = 13x 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 reection, 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, reections 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 reection 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,

reections 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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Chapter 3 Families of functions 87

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 theordered 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 = 1x2

is 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 ofthe image of (x, y), so x = x, y = 4y

Rearranging gives x = x , y = y

4

Therefore, y = 1x2

becomesy

4= 1

(x )2

So the rule of the transformed function is y = 4x2

x0

(1, 4)

(1, 1)

y

b (x, y) (4x, y)Let (x , y ) be the coordinates of theimage of (x, y), so x = 4x, y = y

Rearranging gives x = x

4, y = y

Therefore, y = 1x2

becomes y = 1(x 4

)2So the rule of the transformed function

is y = 16x2

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 nd the equation of the image of y = f (x), under thedilation (x, y) (x, ay), replace y with y

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

y = f (x) under the dilation (x, y) (x, ay) is the graph with equation ya

= 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 nd the equation of the image of y = f (x), under thedilation (x, y) (bx, y), replace x with x

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 graphof 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 factorof

1

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 = 3ySo the mapping is given by (x, y) (x, 3y)

and the graph of y = x is dilated by a factorof

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 = 4x

b y = 12x

c y = 3x d y = 2x2

2 For y = f (x) = 1x2

, 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

( x4

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

a y = 1x

b y = 3x

c y = 32x

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

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

a i f (x) = 1x2

ii f1(x) = 5x2

b i f (x) = x ii f1(x) = 4xc 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 = 3x c y = 1x3

d y = 1x4

e y = 13x f y = x23 g y = 1

x34

3.3 ReflectionsA transformation which, for example, reects 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 writtenas (x, y) (x, y). This is read as the ordered pair (x, y) is mapped onto the ordered pair(x, y). The reection is a one-to-one mapping from R2 to R2 and uniquely links any orderedpair (a, b) to the ordered pair (a, b).

Similarly, a transformation which reects 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 writtenas (x, y) (x, y). This is read as the ordered pair (x, y) is mapped onto the ordered pair(x, y). The reection is a one-to-one mapping from R2 to R2 and uniquely links any orderedpair (a, b) to the ordered pair (a, b).

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

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

is reected in the line y = x produces the graph of theinverse relation and is discussed separately in Section 3.10.)

Consider reecting 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 reection 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 reection 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 reection:a in the x-axis b in the y-axis

Solution

a Note that a reection 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 = yRearranging 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 reection in the y-axis changes

the x-values, so (x, y) (x, y)Let (x , y ) be the coordinates of the

image of (x, y), so x = x, y = yRearranging 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 reection in the x-axis is the transformation that maps (x, y) (x, y).To nd the equation of the image of y = f (x), under the reection (x, y) (x, y),replace y with y; i.e. the image of the graph with equation y = f (x) under thereection (x, y) (x, y) is the graph with equation y = f (x), which is morecommonly written as y = f (x)

In general, a reection in the y-axis is the transformation that maps (x, y) (x, y).To nd the equation of the image of y = f (x), under the reection (x, y) (x, y),replace x with x; i.e. the image of the graph with equation y = f (x) under thereection (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 reection in the x-axis

ii a reection in the y-axis

a y = |x | b y = 3x c y = 1x3

d y = 1x4

e y = 13x 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) ismapped onto the ordered pair (x, y + 2). The translation is a one-to-one mapping from R2 toR2 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) ismapped onto the ordered pair (x + 3, y). The translation is a one-to-one mapping from R2 toR2 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 unitdown

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 nd the equation of the image of y = f (x), under thetranslation (x, y) (x + h, y + k), replace x with x h and y with y k, i.e. theimage 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 asy = f (x h) + k

Example 5

Find the equation of the image when the graph of y = |x| is transformed by the followingsequence 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)Let (x , y ) be the coordinates of theimage of (x, y), so x = x + 4, y = y 3

Rearranging gives x = x 4, y = y + 3So 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 = 1x

+ 3 b y = 1x2

3 c y = 1(x + 2)2

d y = x 2 e y = 1x 1 f y =

1

x 4

g y = 1x + 2 h y =

1

x 3 i f (x) =1

(x 3)2

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

1

x 1 + 1 l f (x) =1

x 2 + 2

m y = 1x2

4

2 For y = f (x) = 1x, 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 thefollowing:

a i f (x) = x2 ii f1(x) = (x + 5)2b i f (x) = 1

xii f1(x) = 1

x+ 2

c i f (x) = 1x2

ii f1(x) = 1x2

+ 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 = 3x c y = 1x3

d y = 1x4

e y = 13x 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, reections and translations.

Example 6

Find the rule of the image when the graph of the function with rule y = x is translated6 units in the negative direction of the x-axis, reected 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 reection 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). Insummary, (x, y) ((x 6), 2y)

Let (x , y ) be the coordinates of the image of (x, y), so x = (x 6) and y = 2yRearranging gives x = x + 6 and y = y

2

Therefore, y = x becomes y

2= x + 6

The rule of the transformed function is y = 26 x

Example 7

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

transformations:

a reection 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 reection 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 reection 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




Finally, apply the reection 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 = 14

(x 2)2 2

Using the TI-NspireDene f (x) = x2The 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 ClassPadDene f (x) = x2.

The rule for the new function is

f(

1

2(x 2)

) 2.


Michael Evans, Kay Lipson, Peter Jones, Sue Avery 2012 Cambridge University Press

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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 reection 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

reection in the x-axis

iii a reection 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 reection 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 reection in the x-axis, followed by a dilation of

factor 2 from the x-axis

a y = |x | b y = 3x c y = 1x3

d y = 1x4

e y = 13x 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 reection 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 reection 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)

42

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 reection in the x-axis.


5 For the graph of y = x 13 :a Sketch the graph of the image under the sequence of transformations:

a reection 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 = 1x

onto the

graph of the function y = 4x + 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 + 34

. Rearranging gives x = x 5 andy = 4y 3.

So the mapping is given by (x, y) (x 5, 4y 3) which identies the sequenceof 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 = 45

3= 21

5

When y = 0, 4x + 5 3 = 0

4 = 3x + 15 3x = 11

and x = 113

Transformations are shown below:

Original y = 1x

1 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

, 011

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 + 5Solution

The graph is obtained from the graph of y = xthrough 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 5Solution

The graph is obtained from the graph of y = x by:a translation of 4 units in the positive direction

of the x-axis, and

a reection 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 = 1x2

by:

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

a translation of 2 units in the positive direction

of the x-axis.

x

0, 534

y = 5

x = 2

0

y



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Example 13

Sketch the graph of y = 3x + 4

Solution

This is obtained from the graph of y = x by:a reection 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 reection in) the x-axis, followed by atranslation of b units in the positive direction of the y-axis, and

a dilation of factor

1n from (and if n < 0 a reection 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 rst equation into the graph of the second equation:

a y = 1x

, y = 2x 1 + 3 b y =

1

x2, y = 3

(x + 4)2 7

c y = 1x3

, y = 4(1 x)3 5 d y =

3x , y = 2 3x + 1

e y = 1x

, y = 2x + 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) = 3x 1 b g(x) =

2

x + 1 1 c h(x) =3

(x 2)2

d f (x) = 2(x 1)2 1 e h(x) =

1x 3 f f (x) =

1x + 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 3 e y = 5

x + 2 f y = 5x + 2 2

g y = 3x 2 h y =

2(x + 2)2 4 i y =

3

2x 5

j y = 52x

+ 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 = 3x + 2 + 7 b y = 43x 1 + 2 c y = (x + 1)34 6

5 a Show that3x + 2x + 1 = 3

1

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

f : R\{1} R, f (x) = 3x + 2x + 1

b Show that4x 52x + 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 72(x + 12

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

each.

a y = 2x 3 + 4 b y =

4

3 x + 4 c y =2

(x 1)2 + 1

d y = 2x 1 + 2 e y = 3x 4 + 1 f y = 52x + 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 therst column of the table below.

Mapping Rule

Reection in the x-axis x = x = x + 0yy = y = 0x + y

Reection in the y-axis x = x = x + 0yy = y = 0x + y

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

Dilation by factor k from the x-axis x = x = x + 0yy = ky = 0x + ky

Reection in the line y = x x = y = 0x + yy = x = x + 0y

Translation dened by a vector

[a

b

]x = x + ay = y + b

The rst ve mappings are special cases of a general kind of mapping dened by

x = ax + byy = cx + dy

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

These equations can be rewritten as

x = a11x + a12 yy = a21x + a22 y

which yields the equivalent matrix equation[x

y

]=[

a11 a12a21 a22

][x

y

]

A transformation of the form

(x, y) (a11x + a12 y, a21x + a22 y)is called a linear transformation.

The notation T: R2 R2 is often used to indicate that a transformation is a mapping from theCartesian plane into the Cartesian plane. The rule can then be dened through the use of

matrices. Some questions formed in this way are given in Chapter 20.



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The rst ve transformations above can be dened by a 2 2 matrix. This is shown in thetable below.

Mapping Matrix

Reection in the x-axis

[1 0

0 1

]

Reection 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

]

Reection in the line y = x[

0 1

1 0

]

Example 14

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

a a reection 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 theimage of (3, 5).

Solution[a11 a12a21 a22

][1

0

]=[

3

1

]and

[a11 a12a21 a22

][0

1

]=[

2

4

]

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



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i.e. the transformation can be dened by the 2 2 matrix[

3 21 4

]

Let (3, 5) (x , y)

[

x

y

]=[

3 21 4

] [3

5

]=[

3 3 + 2 51 3 + 4 5

]=[

1923

]

(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 dened by (0, 0) (a, b)x = x + ay = y + b

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

suggest

[x

y

]=[

x

y

]+[

a

b

]

where the sum has the denition:

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

[x

y

]+[

a

b

]=[

x + ay + b

]

Composition of mappings

Consider a linear transformation dened by the matrix A =[

a11 a12a21 a22

]composed with a

linear transformation dened by the matrix B =[

b11 b12b21 b22

]

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

transformation of B.

The matrix of the resulting composition is the product

BA =[

b11a11 + b12a21 b11a12 + b12a22b21a11 + b22a21 b21a12 + b22a22

]

Example 16

Find the image of the point (2,3) under a reection in the x-axis followed by a dilation offactor k from the y-axis.



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Solution

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

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

transformation is dened 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 dened by the matrix C =[

a

b

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

matrix equation. Hence nd 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 = XThen AX = X C and hence X = A1(X C)Now A =

[k 0

0 1

]

det(A) = k and therefore A1 = 1k

[1 0

0 k

]= 1k 0

0 1

X = 1k 0

0 1

([ x

y

][

a

b

])= 1k 0

0 1

[ x a

y b

]= 1k (x a)

y b

Hence x = 1k

(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 dened 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 = XNow multiply both sides of the equation by T1.Therefore T1TX = T1X

and X = T1X

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 dened by a 2 2 non-singular matrix.

Example 19

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

0 32 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.

T1T(X + B) = T1XX + B = T1Xand X = T1X B

Therefore

[x

y

]=

0

1

21

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 withequation x

3 2 = 2

(y

2 1

)+ 5

Rearranging gives y = x

3 5



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Exercise 3G

1 Using matrix methods nd the image of the point (1,2) under each of the followingtransformations:

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

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

e reection 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 reection in the line x = 0b reection in the line y = xc reection in the line y = xd 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 dened by the matrix C =[

2

1

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

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

5 A transformation is dened 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 whereT =

[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 31 0

]and B =

[32

]. 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 00 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 sufcient 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 = ax

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

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

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

+ b. Findthe 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 = 3a4

a = 4Substitute in (1) to nd b: 5 = 4 + bTherefore b = 1

and y = 4x

+ 1

Example 21

It is known that the points (2, 1) and (10, 6) lie on a curve with equation y = ax 1 + b.Find the values of a and b.



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Solution

When x = 2, y = 1.Therefore 1 = a1 + b, i.e. 1 = a + b (1)When x = 10, y = 6.Therefore 6 = a9 + 6, i.e. 6 = 3a + b (2)Subtract (1) from (2): 5 = 2a

a = 52

Substitute in (1) to nd b: 1 = 52

+ b

Therefore b = 32

and y = 52

x 1 3

2

Exercise 3H

1 The graph shown has the rule:

y = Ax + 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 = Ax + 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 = Ax2

+ 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 = ax3

+ 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 theform y = a3x + 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 bedened 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 dened by:

f (x) = x + 1and g(x) = 3 2xthen ( f + g)(x) = f (x) + g(x)

= 4 xWe note ( f + g)(2) = f (2) + g(2) = 3 + 1 = 2,i.e. the ordinates are added.

x

1

10112

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 nd 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 dened by theaddition 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) = 1x

+ 1x2

using 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) = 2xb f : (, 2], f (x) = 2 x, g: [2,) R, g(x) = x + 2

3.10 Graphing inverse functionsA transformation which reects each point in the plane in the line y = x can be describedthrough 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 reection 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 reected in the line y = x produces thegraph of the inverse relation.

Consider reecting the graph of the

function shown here in the line

y = x , and observe the effect on ageneral point (x, y) on the graph. 0

x

(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 reection in the line y = x is the transformation that maps(x, y) (y, x). To nd the equation of the image of y = f (x), under the reection(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 reection (x, y) (y, x) is the graph of{(y, x): y = f (x)} and is called the inverse relation of y = f (x).




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 = 1are 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) = 1x, g(x) = 1

x3, h(x) = 1

x5, . . .

f (x) = x 12 , 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) = 1x

13

, h(x) = 1x

14

, r (x) = 1x

32

, s(x) = 1x

34

, v(x) = 1x

35

, w(x) = 1x

53

, . . .

Example 24

Find the inverse of the function with rule f (x) = 3x + 2 + 4 and sketch both functions onone set of axes, clearly showing the exact coordinates of intersection of the two graphs.

Solution

Consider x = 3y + 2 + 4Solve for y:

x 43

=

y + 2

which implies y =(

x 43

)2 2

f 1(x) =(

x 43

)2 2

and as the domain of f 1 = range of ff 1: [4,) R, f 1(x) =

(x 4

3

)2 2

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

y = 3x + 2 + 4, entersolve (x = 3y + 2 + 4, y).




Using the Casio ClassPadTo nd the rule for the inverse of

f (x) = 3x + 2 + 4, enter and highlightx = 3y + 2 + 4.

Tap InteractiveEquation/inequality

solve and set the variable as y.

Note: The graph of f 1 is obtained byreecting the graph of f in the line y = x .The graph of y = f 1(x) is obtainedfrom the graph of y = f (x) by applyingthe transformation (x, y) (y, x).

x

(4, 2)0

(2, 4)

(0, 32 + 4)

y = x

y = 3x + 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 ofy = f (x) and y = f 1(x) are usually found by solving either f (x) = x orf 1(x) = x , rather than the more complicated equation f (x) = f 1(x). (Note thatpoints of intersection between the graphs of y = f (x) and y = f 1(x) that do not lieon 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 43

)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 + 2972

) (17.12, 17.12)



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Example 25

Expressx + 4x + 1 in the form

a

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

f (x) = x + 4x + 1 . Sketch both functions on the one set of axes.

Solution

x + 4x + 1 =

3 + x + 1x + 1 =

3

x + 1 +x + 1x + 1 =

3

x + 1 + 1

Consider x = 3y + 1 + 1

Solve for y : x 1 = 3y + 1

and thus y + 1 = 3x 1

and y = 3x 1 1

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

f 1: R\{1} R, f 1(x) = 3x 1 1

Note: The graph of f 1 is obtained by reecting the graph of f in the line y = x . The twographs 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 = 1are 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) = 1x2

, g(x) = 1x4

, h(x) = 1x6

, . . .

f (x) = x 23 , g(x) = x 25 , h(x) = x 43 , . . .f (x) = x 23 , g(x) = x 45 , 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) = 1x 3

c f : [2,) R, f (x) = x 2 + 4 d f : R\{2} R, f (x) = 3x 2 + 1

e f : R\{1} R, f (x) = 5x 1 1 f f : (, 2] R, f (x) =

2 x + 1

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

a g(x) = 3x

b g(x) = 3x + 2 4 c h(x) = 2 x

d f (x) = 3x

+ 1 e h(x) = 5 2(x 6)3 f 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 31

1

2

2

3

3

2

1

3

y b

0x

1 2 3 41

1

2

2

3

3

2

1

3

y c

0x

1 2 3 41

1

2

2

3

3

2

1

3

y

d

1 2 3 5 641

2

3

2

1

3

0 x

y e

0x

1 2 3 411

2

2

34

34

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 + 1x 1 b 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 + 32x 1 where S = R\

{1

2

}a Show that f f is dened.b Find f f (x) and sketch the graph of f f .c Write down the inverse of f.



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Chapter summary

Functions of the form f (x) = 1xn

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

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

common factor 1, have domain R\{0} and rangeR\{0}. They are one-to-one and odd functionswith asymptotes at x = 0 and y = 0. The points(1, 1) and (1, 1) lie on the graph. Examplesinclude f (x) = 1

x, g(x) = 1

x3, h(x) = 1

x35

and

r (x) = 1x

97

0x

y

Functions of the form f (x) = 1xn

, where n is a positive even integer, and 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\{0} and range R+.They are many-to-one and even functions

with asymptotes at x = 0 and y = 0. Thepoints (1, 1) and (1, 1) lie on the graph.Examples include f (x) = 1

x2, g(x) = 1

x4,

h(x) = 1x

23

and r (x) = 1x

65

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 includef (x) = x3, g(x) = x

13 , 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 aremany-to-one and even functions. The points (1, 1) and (1, 1) lie on the graph. Examplesinclude f (x) = x2, g(x) = x

23 and h(x) = x 43



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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 areone-to-one. The point (1, 1) lies on the graph. Examples include f (x) = x 12 , g(x) = x 14 ,h(x) = x 32 and r (x) = x 34

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

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) = 1x

12

, g(x) = 1x

14

,

h(x) = 1x

32

and r (x) = 1x

54 0

x

y

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

(x, y) (x, ay)To nd the equation of the image of the curve with equation y = f (x) under the dilation(x, y) (x, ay), replace y by y

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 of ya

= 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 nd the equation of the image of the curve with equation y = f (x) under the dilation(x, y) (bx, y), replace x by x

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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To nd the equation of the image of the curve with equation y = f (x) under the reectionin the x-axis, (x, y) (x,y), replace y by y in y = f (x), i.e. the image of the graph ofy = f (x) under the reection (x, y) (x,y) is the graph of y = f (x)To nd the equation of the image of the curve with equation y = f (x) under the reectionin the y-axis, (x, y) (x, y), replace x by x in y = f (x), i.e. the image of the graph ofy = f (x) under the reection (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 nd 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 ofthe graph of y = f (x) under the translation (x, y) (x + h, y + k) is the graph ofy k = f (x h) or y = f (x h) + kTo nd the equation of the inverse relation of y = f (x), under the reection(x, y) (y, x), replace x with y and y with x, i.e. the image of the graph with equationy = f (x) under the reection (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

Reection in the x-axis

[1 0

0 1

]

Reection 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

]

Reection 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) isshown 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 reected in the x-axis and then translated4 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| + 4D y = |x 4| + 3 E y = |x + 4| 3

3 The graph of the function with rule y = ax + 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, 2B 1, 2, 3C 1, 3, 2D 1, 3, 2E 1, 2, 3



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4 The graph of the function f is obtained from the graph of the function with equation y = x 13by a reection in the y-axis followed by a dilation of 5 units from the x-axis. The rule for f

is:

A f (x) = 5x 13 B f (x) = 15

(x) 13 C f (x) = (5x) 13

D f (x) = 15

x13 E f (x) = 5(x) 13

5 A function with rule f (x) = 1x4

can be dened 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 =2D x =1, y =2 E x = 2, y =1



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8 The equations of the vertical and horizontal asymptotes of the graph with equation2

(x + 3)4 5 are:A x = 3, y = 5 B x = 5, y = 3 C x = 3, y = 5D x = 2, y = 5 E x = 3, y = 5

9 The function g: R\{3} R, where g(x) = 1x 3 + 2 has an inverse g

1. The rule and

domain of g1 are:

A g1(x) = 1x 2 + 3, dom g

1 = R\{2}

B g1(x) = 1x 2 + 3, dom g

1 = R\{3}

C g1(x) = 1x + 2 3, dom g

1 = R\{2}

D g1(x) = 1x + 2 3, dom g

1 = R\{3}

E g1(x) = 1x 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) = x3C f : [3, 3] R, f (x) = 9 x D f : R R, f (x) = x 13 + 4E 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) = 1x

+ 2 b f (x) = 3 23x 2 c f (x) = 4(x 2)2 + 3

d h(x) = 4 3x 2 e 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) = 1x

3 b h: (2,) R, f (x) = 1x 2

c f : R\{1} R, f (x) = 2x 1 3 d h: (2,), f (x) =

32 x + 4

e f : R\{1}, h(x) = 1 1x 1

3 Sketch the graphs of each of the following:

a f (x) = 2x 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 = ax

+ 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 2x + 1



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Extended-response questions

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

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

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 g1, the inverse of g.ii Write down the domain of g1.iii Find the values of x for which g(x) = x and hence the values of x for which

g(x) = g1(x).iv Sketch the graphs of y = g(x) and y = g1(x) on the one set of axes.

2 Consider the function f : D R, for which the rule is f (x) = 4 22x + 6, where D isthe maximal domain for f.

a Find D.

b Describe a set of transformations which, when applied to the graph of y = x, producethe graph of y = f (x). Specify the order in which these transformations are to beapplied.

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

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

iii State the rule which takes the reected 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 dened 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.



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c The architect wishes to have exibility 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 currentarch 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-onefunction.

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) = 5020 x , x = 20

b If g(x) = 50x20 x :

i Show that g(x) = 100020 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 g1.

6 When the transformation with rule (x, y) (y, x) (a reection in the line y = x) is appliedto the graph of the one-to-one function, f, the resulting image has rule y = f 1(x), i.e. thegraph of the inverse function is obtained.

a For the graph of y = f (x), nd the rule for the image of f, in terms of f 1(x), for eachof 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 reection in the line y = xii a reection 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 dilation of factor 5 from the y-axis

a reection in the line y = x



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iv a reection in the line y = xa dilation of factor 5 from the y-axis


b Find the image of the graph of y = f (x), in terms of f 1(x), under the transformationwith rule (x, y) (ay + b, cx + d), where a, b, c and d are positive constants, anddescribe this transformation in words.



3. Families of Functions

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