Ch5 Functions and Graphs

7/28/2019 Ch5 Functions and Graphs

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TERMINOLOGY

5Functions andGraphs

Arc of a curve: Part or a section o a curve between twopoints

Asymptote: A line towards which a curve approaches butnever touches

Cartesian coordinates: Named ater Descartes. A system olocating points (x, y) on a number plane. Point (x, y) hasCartesian coordinates xand y

Curve: Another word or arc. When a unction consistso all values o xon an interval, the graph o y f x=

] gis

called a curve y f x= ] gDependent variable: A variable is a symbol that canrepresent any value in a set o values. A dependentvariable is a variable whose value depends on the valuechosen or the independent variable

Direct relationship: Occurs when one variable variesdirectly with another i.e. as one variable increases, sodoes the other or as one variable decreases so doesthe other

Discrete: Separate values o a variable rather than acontinuum. The values are distinct and unrelated

Domain: The set o possible values o xin a given domainor which a unction is defned

Even function: An even unction has line symmetry(reection) about the y-axis, and f x f x =- -] ]g gFunction: For each value o the independent variablex,there is exactly one value o y, the dependent variable.A vertical line test can be used to determine i arelationship is a unction

Independent variable: A variable is independent i it maybe chosen reely within the domain o the unction

Odd function: An odd unction has rotational symmetry

about the origin (0, 0) and where f x f x =- -] ]g gOrdered pair: A pair o variables, one independent andone dependent, that together make up a single point inthe number plane, usually written in the orm (x, y)

Ordinates: The vertical or ycoordinates o a point arecalled ordinates

Range: The set o real numbers that the dependentvariableycan take over the domain (sometimes calledthe image o the unction)

Vertical line test: A vertical line will only cut the graph oa unction in at most one point. I the vertical line cuts

the graph in more than one point, it is not a unction


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205Chapter 5 Functions and Graphs

INTRODUCTION

FUNCTIONS AND THEIR GRAPHS are used in many areas, such as mathematics,

science and economics. In this chapter you will study functions, function

notation and how to sketch graphs. Some of these graphs will be studied inmore detail in later chapters.

DID YOU KNOW?

The number plane is called the Cartesian plane after Rene

Descartes (15961650). He was known as one of the first

modern mathematicians along with Pierre de Fermat

(16011665). Descartes used the number plane to develop

analytical geometry. He discovered that any equation

with two unknown variables can be represented by a line.The points in the number plane can be called Cartesian

coordinates.

Descartes used letters at the beginning of the

alphabet to stand for numbers that are known, and letters

near the end of the alphabet for unknown numbers. This is

why we still use xand yso often!

Do a search on Descartes to find out more details of

his life and work.

Descartes

Functions

Definition of a function

Many examples of functions exist both in mathematics and in real life. These

occur when we compare two different quantities. These quantities are called

variables since they vary or take on different values according to some pattern.

We put these two variables into a grouping called an ordered pair.


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206 Maths In Focus Mathematics Extension 1 Preliminary Course

EXAMPLES

1. Eye colour

Name Anne Jacquie Donna Hien Marco Russell Trang

Colour Blue Brown Grey Brown Green Brown Brown

Ordered pairs are (Anne, Blue), (Jacquie, Brown), (Donna, Grey), (Hien,

Brown), (Marco, Green), (Russell, Brown) and (Trang, Brown).

2. y x 1= +

x 1 2 3 4

y 2 3 4 5

The ordered pairs are (1, 2), (2, 3), (3, 4) and (4, 5).

3.

A

B

C

D

E

1

2

3

4

The ordered pairs are (A, 1), (B, 1), (C, 4), (D, 3) and (E, 2).

Notice that in all the examples, there was only one ordered pair or each

variable. For example, it would not make sense or Anne to have both blueand brown eyes! (Although in rare cases some people have one eye thats a

dierent colour rom the other.)

A relation is a set o ordered points (x, y) where the variables x and yare

related according to some rule.

A function is a special type o relation. It is like a machine where or

every INPUT there is only one OUTPUT.

INPUT PROCESS OUTPUT

The frst variable (INPUT) is called the independent variable and the

second (OUTPUT) the dependent variable. The process is a rule or pattern.


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For example, in ,y x 1= + we can use any number or x (the independent

variable), say x 3= .

When x

y

3

3 1

4

=

= +

=

As this value oydepends on the number we choose or x, yis called thedependent variable.

A unction is a relationship between two variables where or

every independent variable, there is only one dependent variable.

This means that or every x value, there is only one yvalue.

While we oten call the

independent variable

x and the dependent

variable y, there are other

pronumerals we could

use. You will meet some

o these in this course.

Investigation

When we graph unctions in mathematics, the independent variable

(usually the x-value) is on the horizontal axis while the dependent

variable (usually the y-value) is on the vertical axis.

In other areas, the dependent variable goes on the horizontal axis. Find

out in which subjects this happens at school by surveying teachers or

students in dierent subjects. Research dierent types o graphs on the

Internet to fnd some examples.

Here is an example o a relationship that is NOT a unction. Can you see the

dierence between this example and the previous ones?

A

B

C

D

E

1

2

3

4

In this example the ordered pairs are (A, 1), (A, 2), (B, 1), (C, 4), (D, 3)

and (E, 2).

Notice that A has two dependent variables, 1 and 2. This means that it is

NOT a unction.


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Here are two examples o graphs on a number plane.

1.

x

y

2.

x

y

There is a very simple test to see i these graphs are unctions. Notice that

in the frst example, there are two values oywhen x 0= . The y-axis passes

through both these points.

x

y


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I a vertical line cuts a graph only once anywhere along the graph, the

graph is a unction.

y

x

I a vertical line cuts a graph in more than one place anywhere along the

graph, the graph is not a unction.

x

y

There are also other x values that give two yvalues around the curve. I

we drew a vertical line anywhere along the curve, it would cross the curve in

two places everywhere except one point. Can you see where this is?

In the second graph, a vertical line would only ever cross the curve in one

place.

So when a vertical line cuts a graph in more than one place, it shows thatit is not a unction.


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EXAMPLES

1. Is this graph a unction?

Solution

A vertical line only cuts the graph once. So the graph is a unction.

2. Is this circle a unction?

Solution

A vertical line can cut the curve in more than one place. So the circle is

not a unction.

You will learn how to sketch these

graphs later in this chapter.


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3. Does this set o ordered pairs represent a unction?

, , , , , , , , ,2 3 1 4 0 5 1 3 2 4- -^ ^ ^ ^ ^h h h h hSolution

For each x value there is only one yvalue, so this set o ordered pairs is a

unction.

4. Is this a unction?

y

x

3

Solution

y

x

3

Although it looks like this is not a unction, the open circle at x 3= on

the top line means that x 3= is not included, while the closed circle on

the bottom line means that x 3= is included on this line.

So a vertical line only touches the graph once at x 3= . The graph is

a unction.


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1.

2.

3.

4.

5.

6.

7.

8.

9. , , , ,, , ,1 3 2 1 3 3 4 0-^ ^ ^ ^h h h h10. , , , , , ,,1 3 2 1 2 7 4 0-^ ^ ^ ^h h h h11.

1

2

3

4

5

1

2

3

4

5

12. 1

2

3

4

5

1

2

3

4

5

13.1

2

3

4

5

1

2

3

4

5

5.1 Exercises

Which o these curves are unctions?


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14. Name Ben Paul Pierre Hamish Jacob Lee Pierre Lien

Sport Tennis Football Tennis Football Football Badminton Football Badminton

15.A 3

B 4

C 7

D 3

E 5

F 7

G 4

Function notation

Iydepends on what value we give x in a unction, then we can say that yis aunction ox. We can write this as y f x= ] g.

Notice that these two examples are asking or the same value and f(3) is

the value o the unction when x 3= .

EXAMPLES

1. Find the value oywhen x 3= in the equation y x 1= + .

Solution

When :x

y x

3

1

3 1

4

=

= +

= +

=

2. If x x 1= +] g , evaluate f(3).Solution

f x x

f

1

3 3 1

4

= +

= +

=

]]

gg

Iy f x= ] g then f(a) is the value oyat the point on the unction where x a=


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EXAMPLES

1. I ,f x x x3 12= + +] g fnd .f 2-] g Solution

( ) ( )f 2 2 3 2 1

4 6 1

1

2- = - + - +

= - +

= -

] g

2. I ,f x x x3 2= -] g fnd the value o .f 1-] g Solution

( )

( )

f x x x

f 1 1 1

1 12

3 2

3

= -

- = - - -

= - -

= -

2] ]g g

3. Find the values ox or which ,f x 0=] g given that .f x x x3 102= + -] g Solution

( )

i.e.

( ) ( )

,

f x

x x

x x

x x

x x

0

3 10 0

5 2 0

5 0 2 0

5 2

2

=

+ - =

+ - =

+ = - =

= - =

4. Find , ,f f f3 2 0] ] ]g g g and if f x4-] ]g g is defned aswhen

when .f x

x x

x x

3 4 2

2 21

$=

+

-

] g )

Solution

since 4 21-

( ) ( ) since

( ) ( ) since

( ) ( ) since

( ) ( )

f

f

f

f

3 3 3 4 3 2

13

2 3 2 4 2 2

10

0 2 0 0 2

0

4 2 4

8

1

$

$

= +

=

= +

=

= -

=

- = - -

=

5. Find the value og g g1 2 3+ - -] ] ]g g g iwhen

when

when

x

x

x

2

1 2

1

2

1

# #-

-

g x

x

x2 1

5

2

= -] g

*

This is the same as fnding y

when 2.x -=

Putting (x) 0= is dierent

rom fnding (0) . Follow

this example careully.

Use (x) 3x 4= + when

x is 2 or more, and use

(x) 2x = - when x is less

than 2.


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Solution

( ) ( )

( )

( )

g

g

g

1 2 1 1 1 1 2

1

2 5 2 1

3 3 3 2

9

since

since

since21

2

# #= - -

=

- = - -

=

=

( ) ( ) ( )g g g1 2 3 1 5 9

3

So + - - = + -

= -

DID YOU KNOW?

Leonhard Euler (170783), from Switzerland, studied functions and invented the termf(x) for function notation. He studied theology, astronomy, medicine, physics and oriental

languages as well as mathematics, and wrote more than 500 books and articles on

mathematics. He found time between books to marry and have 13 children, and even when

he went blind he kept on having books published.

1. Given ,f x x 3= +] g fnd f 1] g and.f 3-] g

2. I ,h x x 22= -] g fnd ,h h0 2] ]g gand .h 4-] g

3. I ,f x x2= -] g fnd , ,f f f5 1 3-] ] ]g g gand .f 2-] g

4. Find the value of f0 2+ -] ]g g i.f x x x 14 2= - +] g

5. Find f 3-] g i .f x x x2 5 43= - +] g 6. I ,f x x2 5= -] g fnd x when

.f x 13=] g 7. Given ,f x x 32= +] g fnd any

values ox or which .f x 28=] g 8. I ,f x 3x=] g fnd x when

.f x271

=] g 9. Find values oz or which

f z 5=] g given .f z z2 3= +] g

10. I ,f x x2 9= -] g fnd f p^ h and.f x h+] g

11. Find g x 1-] g when.g x x x2 32= + +] g

12. I ,f x x 13= -] g fnd f k] g as aproduct o actors.

13. Given ,f t t t 2 12= + +] g fndtwhen .f t 0=] g Also fnd anyvalues otor which .f t 9=] g

14. Given ,f t t t 54 2

= + -] g fnd thevalue o .f b f b- -] ]g g 15. f x

x x

x x

1

1

or

or

32

#=] g )

Find ,f f5 1] ]g g and .1-] g

16. f x

x x

x x

x x

2 4 1

3 1 1

1

i

i

i21 1

$

#

=

-

+ -

-

] gZ

[

\

]]

]]

Find the values o

.f f f2 2 1- - + -] ] ]g g g

5.2 Exercises

We can use pronumerals

other than or unctions.


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17. Find g g g3 0 2+ + -] ] ]g g g ig x

x x

x x

1 0

2 1 0

when

when 1

$=

+

- +

] g ) 18. Find the value o

f f f3 2 2 3- + -

] ] ]g g g whenf x

x x

x x

x

2

2 2

4 2

or

or

or

2

2

1

# #= -

-

] g * 19. Find the value of f1 3- -] ]g g

i ( )1 2

2 3 1 2f x

x x

x x x

or

or

3

21

$=

-

+ -

*

20. If xx

x x

3

2 32=

-

- -] g evaluate(a) f(2)explain why the unction(b)

does not exist or x 3=

by taking several(c) x values

close to 3, fnd the value oythat

the unction is moving towards

as x moves towards 3.

21. I f x x x5 42= +] g , fndf x h f x+ -] ]g g in its simplestorm.

22. Simpliyh

f x h f x+ -] ]g gwhere

f x x x22

= +] g 23. If x x5 4= -] g , fnd f x f c -] ]g g

in its simplest orm.

24. Find the value of k2^ h if x

x x

x x

3 5 0

0

or

or2 1

$=

+] g *

25. I

f x

x x

x x x

3

2 0

when

when

3

2

$

#

=

- +

x5 0 3when 1 1] gZ

[

\

]]

]]

evaluate

(a) f(0)

(b) f f2 1-] ]g g(c) f n2-^ h

Graphing Techniques

You may have previously learned how to draw graphs by completing a table

o values and then plotting points. In this course, you will learn some other

techniques that will allow you to sketch graphs by showing their important

eatures.

Intercepts

One o the most useul techniques is to fnd the x- and y-intercepts.

For x-intercept, y 0=

For y-intercept, x 0=

Everywhere on the x-axis,0=y and everywhere on

the y-axis 0=x .


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EXAMPLE

Find the x- and y-intercepts o the unction .f x x x7 82= + -] g

Solution

For x-intercept: y 0=

,

,

x x

x x

x x

x x

0 7 8

8 1

8 0 1 0

8 1

2= + -

= + -

+ = - =

= - =

] ]g g

For y-intercept: x 0=

y 0 7 0 8

8

2= + -

= -

] ]g g

This is the same as.y x x7 82= + -

You will use the intercepts

to draw graphs in the next

section in this chapter.

Domain and range

You have already seen that the x-coordinate is called the independent variable

and the y-coordinate is the dependent variable.

The set o all real numbers x or which a unction is defned is called the

domain.

The set o real values or yor f(x) as x varies is called the range (or

image) of.

EXAMPLE

Find the domain and range o .f x x2=] g

Solution

You can see the domain and range rom the graph, which is the parabola .y x2=

x

y

CONTINUED


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Notice that the parabola curves outwards gradually, and will take on any

real value or x. However, it is always on or above the x-axis.

Domain: {all real x}

Range: {y: y 0$ }

You can also fnd the domain and range rom the equation y x2= . Notice

that you can substitute any value or x and you will fnd a value oy.

However, all the y-values are positive or zero since squaring any number

will give a positive answer (except zero).

Odd and even functions

When you draw a graph, it can help to know some o its properties, or

example, whether it is increasing or decreasing on an interval or arc o thecurve (part o the curve lying between two points).

I a curve is increasing, as x increases, so does y, and the curve is moving

upwards, looking rom let to right.

I a curve is decreasing, then as x increases,ydecreases and the curve

moves downwards rom let to right.


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EXAMPLES

1. State the domain over which each curve is increasing and decreasing.

xx3x2x1

y

Solution

The let-hand side o the parabola is decreasing and the right side is

increasing.

So the curve is increasing or x2x2

and the curve is decreasing when

x1x2.

2.

xx3

x2x1

y

Solution

The let-hand side o the curve is increasing until it reaches they-axis

(where x 0= ). It then turns around and decreases until x3

and then

increases again.

So the curve is increasing or ,x x x03

1 2 and the curve is

decreasing or .x x03

1 1

The curve isnt increasing or

decreasing at x2. We say that it is

stationary at that point. You will

study stationary points and urther

curve sketching in the HSC Course.

Notice that the curve is

stationary at x 0= and .x x3

=


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Functions are odd i they have point symmetry about the origin. A graph

rotated 180 about the origin gives the original graph.

This is an odd unction:

x

y

For even unctions, f x f x= -] ]g g or all values ox.

For odd unctions, f x f x- = -] ]g g or all values ox in the domain.

As well as looking at where the curve is increasing and decreasing, we can

see i the curve is symmetrical in some way. You have already seen that the

parabola is symmetrical in earlier stages o mathematics and you have learned

how to fnd the axis o symmetry. Other types o graphs can also be symmetrical.

Functions are even i they are symmetrical about the y-axis. They have

line symmetry (reection) about the y-axis.This is an even unction:

x

y


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EXAMPLES

1. Show that f x x 32= +] g is an even unction.

Solution

f x x

x

f x

f x x

3

3

3 is an even unction

2

2

2`

- = - +

= +

=

= +

] ]

]

]

g g

g

g

2. Show that f x x x3= -] g is an odd unction.

Solution

f x x x

x x

x x

f x

f x x x is an odd unction

3

3

3

3`

- = - - -

= - +

= - -

= -

= -

] ] ]

^

]

]

g g g

h

g

g

Investigation

Explore the amily o graphs of x xn=] g .

For what values on is the unction even?

For what values on is the unction odd?

Which amilies o unctions are still even or odd given k? Let k take on

dierent values, both positive and negative.

1. f x kxn=] g

2. f x x kn= +] g

3. f x x k n= +] ]g g

k is called a parameter.

Some graphics calculators

and computer programs use

parameters to show how

changing values o k change the

shape o graphs.

1. Find the x- and y-intercept o

each unction.

(a) y x3 2= -

(b) x y2 5 20 0- + =

(c) x y3 12 0+ - =

(d) f x x x32= +] g

(e) f x x 42= -] g

() p x x x5 62= + +] g

(g) y x x8 152= - +

(h) p x x 5

3= +

] g

5.3 Exercises


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(i) y xx

x3

0!=+ ] g

(j) g x x9 2= -] g

2. Show that f x f x= -] ]g g where

f x x 22= -] g . What type o

unction is it?

3. If x x 13= +] g , fnd

(a) f x2^ h

(b) ( )f x 26 @

(c) f x-] g

Is it an even or odd unction?(d)

4. Show that g x x x x3 28 4 2= + -] g is

an even unction.

5. Show that f(x) is odd, where

.f x x=] g

6. Show that f x x 12= -] g is an even

unction.

7. Show that f x x x4 3= -] g is an

odd unction.

8. Prove that f x x x4 2= +] g is an

even unction and hence fnd

.f x f x- -] ]g g

9. Are these unctions even, odd or

neither?

(a) yx x

x4 2

3

=

-

(b) yx 1

13

=

-

(c) f xx 4

32

=

-

] g

(d) yx

x

33

=+

-

(e) f x x x

x5 2

3

=-] g

10. In is a positive integer, or

what values on is the unction

f x xn=] g

even?(a)

odd?(b)

11. Can the unction f x x xn= +] g

ever be

even?(a)

odd?(b)

12. For the unctions below, state

(i) the domain over which the

graph is increasing

(ii) the domain over which

the graph is decreasing

(iii) whether the graph is odd,even or neither.

x

y(a)

x

4

y(b)

2-2

x

y(c)


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Investigation

Use a graphics calculator or a computer with graphing sotware to sketchgraphs and explore what eect dierent constants have on each type o

graph.

I your calculator or computer does not have the ability to use parameters

(this may be called dynamic graphing), simply draw dierent graphs by

choosing several values or k. Make sure you include positive and negative

numbers and ractions or k.

Alternatively, you may sketch these by hand.

Sketch the amilies o graphs or these graphs with parameter1. k.

y kx

y kx

y kx

y kx

y xk

(a)

(b)

(c)

(d)

(e)

2

3

4

=

=

=

=

=

What eect does the parameter k have on these graphs? Could you give a

general comment about y k f x= ] g?


y x k

y x k

y x k

y x k

y x k1

(a)

(b)

(c)

(d)

(e)

2

2

3

4

= +

= +

= +

= +

= +

] g


general comment about y f x k= +] g ?

-2

1 2

-4

-1-2

2

4

y

x

(d) y

x

(e)

CONTINUED


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Gradient orm:

y mx b= + has gradient m and y-intercept b

General orm:ax by c 0+ + =

Investigation

Are straight line graphs always unctions? Can you fnd an example o a

straight line that is not a unction?

Are there any odd or even straight lines? What are their equations?

For the amily o unctions y k f x= ] g, as k varies, the unction changes

its slope or steepness.

For the amily o unctions ,y f x k= +] g as k varies, the graph moves up

or down (vertical translation).For the amily o unctions y f x k= +] g, as k varies, the graph moves let

or right (horizontal translation).


y x k

y x k

y x k

y x ky

x k

1

(a)

(b)

(c)

(d)(e)

2

3

4

= +

= +

= +

= +

=+

]

]

]

g

g

g


general comment about y f x k= +] g?

When 0 ,k2 the graphmoves to the let and when

0 ,k1 the graph moves to

the right.

Notice that the shape o most graphs is generally the same regardless o the

parameter k. For example, the parabola still has the same shape even though it

may be narrower or wider or upside down.

This means that i you know the shape o a graph by looking at its

equation, you can sketch it easily by using some o the graphing techniques in

this chapter rather than a time-consuming table o values. It also helps you to

understand graphs more and makes it easier to fnd the domain and range.You have already sketched some o these graphs in previous years.

Linear Function

A linear unction is a unction whose graph is a straight line.


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EXAMPLE

Sketch the unction f x x3 5= -] g and state its domain and range.

Solution

This is a linear unction. It could be written as .y x3 5= -

Find the intercepts


0 3 5

5 3

1

x

x

x32

=

=

=

-


3 5

5

y 0=

= -

-] g

-1

-2

y

5

4

3

2

1 1 23

6

-3

-4

-5

1 4-1-2 32-3-4

x

Notice that the line extends over the whole o the number plane, so that

it covers all real numbers or both the domain and range.Domain: {all real x}

Range: {all real y}

Notice too, that you can

substitute any real number

into the equation o the

unction or x, and any real

number is possible or y.

The linear unction ax by c 0+ + = has domain {all real x}and range {all real y} where a and b are non-zero

Special lines

Horizontal and vertical lines have special equations.

Use a graphics calculator or a computer with dynamic graphing capability

to explore the eect o a parameter on a linear unction, or choose

dierent values ok (both positive and negative).

Sketch the amilies o graphs or these graphs with parameter k

1. y kx=

2. y x k= +

3. y mx b= + where m and b are both parameters

What eect do the parameters m and b have on these graphs?


23/86


EXAMPLES

1. Sketch y 2= on a number plane. What is its domain and range?

Solution

x can be any value and yis always 2.

Some o the points on the line will be (0, 2), (1, 2) and (2, 2).

This gives a horizontal line with y-intercept 2.

-1

-3

y

4

3

2

1

5

-2

-4

-5

1 4-1-2

x

32-3-4

Domain: xall real" ,

Range: : 2y y =" ,

2. Sketch x 1= -

on a number plane and state its domain and range.

Solution

ycan be any value and x is always .1-

Some o the points on the line will be , , ,1 0 1 1- -^ ^h h and , .1 2-^ h

This gives a vertical line with x-intercept .1-

Domain: : 1x x = -" ,

Range: yall real" ,

-

-

3

-

-4

-

---

y

x


24/86


x a= is a vertical line with x-intercept a

Domain: :x x a=! +

Range: {all real y}

y b= is a horizontal line with y-intercept b


Range: :y y b=" ,

5.4 Exercises

1. Find the x- and y-intercepts o

each unction.

(a) y x 2= -

(b) f x x2 3= +] g (c) x y2 1 0+ =-

(d) x y 3 0+ =-

(e) x y3 6 2 0=- -

2. Draw the graph o each straight

line.

(a) x 4=

(b) x 3 0=-

(c) y 5=

(d) y 1 0+ =

(e) f x x2 1= -] g () y x 4= +

(g) f x x3 2= +] g

(h) x y 3+ =

(i) x y 1 0=- -

(j) x y2 3 0+ =-

3. Find the domain and range o

(a) x y3 2 7 0+ =-

(b) y 2=

(c) x 4= - (d) x 2 0=-

(e) y3 0=-

4. Which o these linear unctions

are even or odd?

(a) y x2=

(b) y 3=

(c) x 4=

(d) y x= -

(e) y x=

5. By sketching x y 4 0=- - and

x y2 3 3 0+ =- on the same set

o axes, fnd the point where they

meet.


25/86


Applications

The parabola shape is used in many different applications as it has specialproperties that are very useful. For example if a light is placed inside the parabola

at a special place (called the focus), then all light rays coming from this light and

bouncing off the parabola shape will radiate out parallel to each other, giving a

strong light. This is how car headlights work. Satellite dishes also use this property

of the parabola, as sound coming in to the dish will bounce back to the focus.

The pronumeral

a is called the

coefcient o .x2

Quadratic Function

The quadratic unction gives the graph o a parabola.

f x ax bx c 2= + +] g is the general equation o a parabola.

Ia 02 the parabola is concave upwards

Ia 01 the parabola is concave downwards


26/86


The lens in a camera and glasses are also parabola shaped. Some bridges look

like they are shaped like a parabola, but they are often based on the catenary.

Research the parabola and catenary on the Internet for further information.

Investigation

Is the parabola always a unction? Can you fnd an example o a parabola

that is not a unction?


to explore the eect o a parameter on a quadratic unction, or choose

dierent values ok (both positive and negative).

Sketch the amilies o graphs or these graphs with parameter k.1. y kx2=

2. y x k2= +

3. y x k 2= +] g

4. y x kx2= +

What eect does the parameter k have on these graphs?

Which o these amilies are even unctions? Are there any odd quadratic

unctions?


27/86


EXAMPLES

1. (a) Sketch the graph of ,y x 12= - showing intercepts.

(b) State the domain and range.

Solution

This is the graph of a parabola. Since(a) ,a 02 it is concave upward


x

x

x

0 1

1

1

2

2

!

= -

=

=


0 1

1

y 2= -

= -

From the graph, the curve is moving outwards and will extend(b)

to all real x values. The minimum yvalue is .1-


Range: :y y 1$ -" ,

2. Sketch .f x x 1 2= +] ]g g

Solution

This is a quadratic function. We find the intercepts to see where the

parabola will lie.

Alternatively, you may know from your work on parameters that

f x x a 2= +] ]g g will move the function f x x2=] g horizontally a units to the

left.

So f x x 1 2= +] ]g g moves the parabola f x x2=] g 1 unit to the left.


0

1 0

1

x

x

x

1 2= +

+ =

= -

] g


1

y 0 1 2= +

=

] g

-1

-

4

3

2

1

5

-2

-4

-5

-

1 41-2 5-4

y

x


28/86


3. For the quadratic unction f x x x 62= + -] g Find the(a) x- andy-intercepts

Find the minimum value o the unction(b)

State the domain and range(c)

For what values o(d) x is the curve decreasing?

Solution

For(a) x-intercept:y 0=

This means f x 0=] g

,

,

x x

x xx x

x x

0 6

3 23 0 2 0

3 2

2= + -

= + -

+ = - =

= - =

] ]g g

Fory-intercept: x 0=

f 0 0 0 66

2= + -

= -

] ] ]g g g

Since(b) ,a 02 the quadratic unction has a minimum value.

Since the parabola is symmetrical, this will lie halway between the

x-intercepts.

Halway between 3x = - and 2:x =

23 2

21- + = -

Minimum value is f21

-c m

f21

21

21

6

41

21

6

641

2

- = - + - -

= - -

= -

c c cm m m

So the minimum value is .641

-

CONTINUED

You will learn more

about this in Chapter 10.

-1

-3

4

3

2

1

5

-2

-4

-5

1 4-1-2 32-3-4

y

x


29/86


Sketching the quadratic unction gives a concave upward parabola.(c)

From the graph, notice that the parabola is gradually going outwards and

will include all real x values.

Since the minimum value is 641

- , allyvalues are greater than this.

Domain: xall real" ,Range: : 6yy

41

$ -' 1The curve decreases down to the minimum point and then(d)

increases. So the curve is decreasing or all .x

2

11-

4. (a) Find the x- andy-intercepts and the maximum value o the

quadratic unction .f x x x4 52= - + +] g (b) Sketch the unction and state the domain and range.

(c) For what values ox is the curve increasing?

Solution

For(a) x-intercept: 0y=

So f x 0=] g 0 4 5

4 5 0

0

x xx x

x x5 1

2

2

= - + +

=

+ =

- -

-] ]g g

,

,

x x

x x

5 0 1 0

5 1

- = + =

= = -

Fory-intercept: 0x =

f 0 0 4 0 5

5

2= - + +

=

] ] ]g g g

-

-3

4

-2

-

-

-

y

--

x

-1

2

1

4,


30/86


Since ,a 01 the quadratic unction is concave downwards and has a

maximum value halway between the x-intercepts 1x = - and .x 5=

21 5

2- +

=

f 2 2 4 2 59

= - + +

=

2] ] ]g g g So the maximum value is 9.

Sketching the quadratic unction gives a concave downward parabola.(b)

From the graph, the unction can take on all real numbers or x, but the

maximum value oryis 9.

Domain: xall real" ,Range: : 9y y#" ,

From the graph, the unction is increasing on the let o the(c)

maximum point and decreasing on the right.

So the unction is increasing when .x 21

1. Find the x- andy-intercepts o

each unction.

(a) 2y x x2= +

(b) 3y x x2= - +

(c) f x x 12= -] g (d) y x x 22= - -

(e) y x x9 82= +-

2. Sketch

(a) 2y x2= +

(b) y x 12= - +

(c) f x x 42= -] g (d) 2y x x

2= +

(e) y x x2= - -

() f x x 3= - 2] ]g g

5.5 Exercises

-1

9

8

7

5

4

3

2

6

1

-2

-3

-4

-5

y

2 51 643-1-2-3-4

x


31/86


EXAMPLES

1. Sketch f x x 1= -] g and state its domain and range.Solution

Method 1: Table o values

When sketching any new graph or the frst time, you can use a table o

values. A good selection o values is x3 3# #- but i these dont give

enough inormation, you can fnd other values.

Absolute Value Function

You may not have seen the graphs o absolute unctions beore. I you are not

sure about what they look like, you can use a table o values or look at the

defnition o absolute value.

(g) f x x 1 2= +] ]g g (h)y x x3 42= + -

(i) y x x2 5 32= - +

(j) f x x x3 22= - + -] g

3. For each parabola, fndthe(i) x- andy-intercepts

the domain and range(ii)

(a) y x x7 122= +

(b) f x x x42= +] g (c) y x x2 82= - -

(d) y x x6 92= +-

(e) f t t4 2= -] g 4. Find the domain and range o

(a) y x 52= -

(b) f x x x6

2= -

] g (c) f x x x 22= - -] g (d) y x2= -

(e) f x x 7 2= -] ]g g 5. Find the range o each unction

over the given domain.

(a) y x2= or x0 3# #

(b) y x 42= - + or x1 2# #-

(c) f x x 12= -] g or x2 5# #- (d) y x x2 32= + - or x2 4# #-

(e)y x x

22= - +- orx

0 4# #

6. Find the domain over which each

unction is

increasing(i)

decreasing(ii)

(a) y x2=

(b) y x2

= - (c) f x x 92= -] g (d) y x x42= - +

(e) f x x 5 2= +] ]g g 7. Show that f x x2= -] g is an even

unction.

8. State whether these unctions are

even or odd or neither.

(a) y x 12= +

(b) f x x 32= -

] g

(c) y x2 2= -

(d) f x x x32= -] g (e) f x x x2= +] g () y x 42= -

(g) y x x2 32= - -

(h)y x x5 42= +-

(i) p x x 1 2= +] ]g g (j) y x 2= - 2] g


32/86


CONTINUED

e.g. When :x 3= -

| |y 3 13 12

= - -

= -

=

x -3 -2 -1 0 1 2 3

y 2 1 0 -1 0 1 2

This gives a v-shaped graph.

y

-2

4

3

2

1

5

-1

-3

-4

-5

1 4-1-2 32-3-4

x

Method 2: Use the defnition o absolute value

| |y x

x x

x x1

1 0

1 0

when

when 1$

= - =-

- -& This gives 2 straight line graphs:

y x x1 0$= - ] g

-3

4

3

2

1

5

-2

-1

-4

-5

y

3-1-2 421-3-4

x

y=x-1


33/86


y x 1= - - x 01] g

-3

4

3

2

1

5

-2

-1

-4

-5

y

3-1-2 421-3-4

x

y=-x-1

Draw these on the same number plane and then disregard the dotted

lines to get the graph shown in method 1.

-3

4

3

2

1

5

-2

-1

-4

-5

yy

3-1-2 421-3-4

x

y=-x- 1

y=x-1

Method 3: I you know the shape o the absolute value unctions, fnd the

intercepts.

For x-intercept: 0y=

So f x 0=] g | |

| |

x

x

x

0 1

1

1` !

= -

=

=


( ) | |f 0 0 11

= -

= -


34/86


The graph is V-shaped, passing through these intercepts.

-3

4

3

2

1

5

-2

-1

-4

-5

y

4-2 5321-1-3-4

x

From the graph, notice that x values can be any real number while the

minimum value oyis .1-


Range: {y:y 1$ - }

2. Sketch .| |y x 2= +

Solution

Method 1: Use the defnition o absolute value.

| | ( )y xx x

x x22 2 0

2 2 0whenwhen 1

$= + = + +- + +'

This gives 2 straight lines:

2y x= + when x 2 0$+

x 2$ -

-3

4

3

2

1

5

-2

-1

-4

-5

y

3-1-2 421-3-4

x

y=x+2

I you already know how

to sketch the graph o

y | x |= , translate the

graph oy | x | 1= -

down 1 unit, giving it a

y-intercept o .1-

CONTINUED


35/86


2y x= - +] g when x 2 01+ i.e.y x 2= - - when x 21 -

-3

4

3

2

1

5

-2

-1

-4

-5

y

3-1-2 421-3-4

x

y=-x- 2

Draw these on the same number plane and then disregard the dotted lines.

-3

4

3

2

1

5

-2

-1

-4

-5

y

3-1-2 421-3-4

x

y= -x-2

y= x+ 2

Method 2: Find intercepts

For x-intercept: 0y=

So 0f x =] g 0 | 2 |

0 2

2

x

x

x

= +

= +

- =


(0) | 0 2 |

2

f = +

=

There is only one

solution or the

equation | x 2 | 0.+ =

Can you see why?


36/86


The graph is V-shaped, passing through these intercepts.

-3

4

3

2

1

5

-2

-1

-4

-5

y

3-1-2 421-3-4

x

I you know how to

sketch the graph o

y | x |= , translate it 2

places to the let or the

graph o .y | x 2 |= +

Investigation

Are graphs that involve absolute value always unctions? Can you fnd an

example o one that is not a unction?

Can you fnd any odd or even unctions involving absolute values? What

are their equations?


to explore the eect o a parameter on an absolute value unction, or

choose dierent values ok (both positive and negative).

Sketch the amilies o graphs or these graphs with parameter k

1. | |f x k x=] g 2. | |f x x k= +] g 3. | |f x x k= +] g What eect does the parameter k have on these graphs?

The equations and inequations involving absolute values that you studied in

Chapter 3 can be solved graphically.


37/86


EXAMPLES

Solve

1. |2 1 | 3x - =

Solution

Sketch | 2 1 |y x= - and 3y= on the same number plane.

The solution o|2 1 | 3x - = occurs at the intersection o the graphs, that

is, , .x 1 2= -

2. |2 1 | 3 2x x= -+

Solution

Sketch | 2 1 |y x= + and 3 2y x= - on the same number plane.

The solution is 3.x =

3. | 1 | 2x 1+

Solution

Sketch | 1 |y x= + and 2y= on the same number plane.

The graph shows that

there is only one solution.

Algebraically, you need to

fnd the 2 possible solutions

and then check them.


38/86


The solution o| 1 | 2x 1+ is where the graph | 1 |y x= + is below the

graph 2,y= that is, .x3 11 1-

1. Find the x- andy-intercepts o

each unction.

(a) | |y x=

(b) | |f x x 7= +] g (c) | |f x x 2= -] g (d) 5 | |y x=

(e) | |f x x 3= - +] g () | 6 |y x= +

(g) | |f x x3 2= -] g (h) | 5 4 |y x

= +

(i) | 7 1 |y x= -

(j) | |f x x2 9= +] g 2. Sketch each graph on a number

plane.

(a) | |y x=

(b) | |f x x 1= +] g (c) | |f x x 3= -] g (d) 2 | |y x=

(e) | |f x x= -] g () | 1 |y x

= +

(g) | |f x x 1= - -] g (h) | 2 3 |y x= -

(i) | 4 2 |y x= +

(j) | |f x x3 1= +] g 3. Find the domain and range o

each unction.

(a) | 1 |y x= -

(b) | |f x x 8= -] g

(c) | |f x x2 5= +] g (d) 2 | | 3y x= -

(e) | |f x x 3= - -] g 4. Find the domain over which each

unction is

increasing(i)

decreasing(ii)

(a) | 2 |y x= -

(b) | |f x x 2= +

] g

(c) | |f x x2 3= -] g (d) 4 | | 1y x= -

(e) | |f x x= -] g 5. For each domain, fnd the range

o each unction.

(a) | |y x= or x2 2# #-

(b) | |f x x 4= - -] g orx4 3# #-

(c) | |f x x 4= +] g or x7 2# #- (d) | 2 5 |y x= - or x3 3# #-

(e) | |f x x= -] g or x1 1# #- 6. For what values ox is each

unction increasing?

(a) | 3 |y x= +

(b) | |f x x 4= - +] g (c) | |f x x 9= -] g (d) | |y x 2 1= - -

(e) | |f x x 2= - +] g

5.6 Exercises


39/86


7. Solve graphically

(a) | | 3x =

(b) | |x 12

(c) | |x 2#

(d) | 2 | 1x + =

(e) | 3 | 0x- =

() |2 3 | 1x - =

(g) | |x 1 41-

(h) | |x 1 3#+

(i) | |x 2 22-

(j) | |x 3 1$-

(k) | |x2 3 5#+

(l) | |x2 1 1$-

(m) |3 1 | 3x x- = +

(n) |3 2 | 4x x- = -

(o) | 1 | 1x x- = +

(p) | 3 | 2 2x x+ = + (q) |2 1 | 1x x+ = -

(r) |2 5 | 3x x- = -

(s) | 1 | 2x x- =

(t) |2 3 | 3x x- = +

The Hyperbola

A hyperbola is a unction with its equation in the orm .xy a y xa

or= =

EXAMPLE

Sketch1

.y x=

Solution

1y x= is a discontinuous curve since the unction is undefned at .x 0=

Drawing up a table o values gives:

x -3 -2 -121-

41- 0

41

21 1 2 3

y3

1-

2

1- -1 -2 -4 4 2 1

2

1

3

1

Class Discussion

What happens to the graph as x becomes closer to 0? What happens as x

becomes very large in both positive and negative directions? The value o

yis never 0. Why?


40/86


To sketch the graph of a more general hyperbola, we can use the domain and

range to help find the asymptotes (lines towards which the curve approaches

but never touches).

The hyperbola is an example of a discontinuous graph, since it has a gap

in it and is in two separate parts.

Investigation

Is the hyperbola always a function? Can you find an example of a

hyperbola that is not a function?

Are there any families of odd or even hyperbolas? What are their

equations?


to explore the effect of a parameter on a hyperbola, or choose differentvalues ofk (both positive and negative).

Sketch the families of graphs for these graphs with parameter k

1. y xk

=

2.1

y x k= +

3.1

yx k

=+

What effect does the parameter k have on these graphs?

EXAMPLES

1. (a) Find the domain and range of .f xx 3

3=

-

] g

Hence sketch the graph of the function.(b)

Solution

This is the equation of a hyperbola.

To find the domain, we notice that .x 3 0!-

So x 3!

Also ycannot be zero (see example on page 242).

Domain: {all real x: x 3! }

Range: {all real y: y 0! }

The lines 3x = and 0y = (the x-axis) are called asymptotes.

The denominator cannot

be zero.

CONTINUED


41/86


To make the graph more accurate we can fnd another point or two. The

easiest one to fnd is the y-intercept.

For y-intercept, 0x =

1

y0 3

3=

-

= -

-3

4

3

2

1

5

-2

-1

-4

-5

y

-1-2 4 521-3-4

x

x=3

y=0

Asymptotes

3

2. Sketch .yx2 4

1= -

+

Solution

This is the equation o a hyperbola. The negative sign turns the hyperbolaaround so that it will be in the opposite quadrants. I you are not sure

where it will be, you can fnd two or three points on the curve.

To fnd the domain, we notice that .x2 4 0!+

x

x

2 4

2

!

!

-

-

For the range, ycan never be zero.

Domain: {all real x: x 2! - }

Range: {all real y: y 0! }

So there are asymptotes at x 2= - and y 0= (the x-axis).

To make the graph more accurate we can fnd the y-intercept.For y-intercept, x 0=

( )y

2 0 41

41

= -+

= -

Notice that this graph is

a translation of3

yx

=

three units to the right.


42/86


y

-2

x

-1

4

The unction f xbx c

a=

+

] g is a hyperbola with

domain :x xb

call real ! -& 0 and

range {all real :y y 0! }

1. For each graph

State the domain and range.(i)

Find the(ii) y-intercept i it

exists.

Sketch the graph.(iii)

(a)2

y x=

(b)1

y x= -

(c) f xx 1

1=

+

] g

(d) f xx 2

3=

-

] g

(e)3 6

1y

x=

+

() f xx 3

2= -

-

] g

(g) f xx 1

4=

-

] g

(h)1

2y

x= -

+

(i) f xx6 3

2=

-

] g

(j)2

6y

x= -

+

2. Show that f x x2

=] g is an oddunction.

3. Find the range o each unction

over the given domain.

(a) f xx2 5

1=

+

] g or x2 2# #-

(b)3

1y

x=

+or x2 0# #-

(c) f xx2 4

5=

-

] g or x3 1# #-

5.7 Exercises


43/86


(d) f xx 4

3= -

-

] g or x3 3# #-

(e)3 1

2y

x= -

+or x0 5# #

4. Find the domain o each unction

over the given range.

(a)3

y x= or y1 3# #

(b)2

y x= - or y2 21

# #- -

(c) f xx 1

1=

-

] g or y171

# #- -

(d) f xx2 1

3= -

+

] g or

y131

# #- -

(e)3 2

6y

x=

-or y1

21

6# #

Circles and Semi-circles

The circle is used in many applications, including building and design.

Circle gate

A graph whose equation is in the orm 0x ax y by c 2 2

+ + + + = has theshape o a circle.

There is a special case o this ormula:

The graph ox y r2 2 2+ = is a circle, centre 0, 0^ h and radius r

Proof

ry

x

(x, y)

y

x


44/86


Given the circle with centre (0, 0) and radius r:

Let (x, y) be a general point on the circle, with distances rom the origin x

on the x-axis and yon the y-axis as shown.

By Pythagoras theorem:

c a b

r x y

2 2 2

2 2 2

`

= +

= +

EXAMPLE

Sketch the graph o(a) 4.x y2 2+ = Is it a unction?

State its domain and range.(b)

Solution

This is a circle with radius 2 and centre (0, 0).(a)

y

x

-2

-2 2

2

The circle is not a unction since a vertical line will cut it in more than

one place.

y

x

2

2

-2

-2

The radius is .4

CONTINUED


45/86


Notice that the(b) x-values or this graph lie between 2- and 2 and

the y-values also lie between 2- and 2.

Domain: : 2 2{ }x x# #-

Range: : 2 2{ }y y# #-

The circle x y r2 2 2+ = has domain: :x r x r # #-! + and

range: :y r y r# #-" ,

The equation o a circle, centre (a, b) and radius ris x a y b r 2 2 2+ =] ^g h

We can use Pythagoras theorem to fnd the equation o a more general circle.

Proof

Take a general point on the circle, (x, y) and draw a right-angled triangle as

shown.

y

x

(a, b)

x

y

r

(x,y)

a

bx- a

y - b

Notice that the small sides o the triangle are x a and y b and the

hypotenuse is r, the radius.

By Pythagoras theorem:

c a b

r x a y b

2 2 2

2 2 2

= +

= +] ^g h


46/86


EXAMPLES

1. (a) Sketch the graph o .x y 812 2+ =

(b) State its domain and range.

Solution

The equation is in the orm(a) .x y r2 2 2+ =

This is a circle, centre (0, 0) and radius 9.

y

x9

9

-9

-9

From the graph, we can see all the values that are possible or(b) x

and yor the circle.

Domain: : 9 9{ }x x# #-

Range: : 9 9{ }y y# #-

2. (a) Sketch the circle .x y1 2 42 2+ + =] ^g h (b) State its domain and range.

Solution

The equation is in the orm(a) .x a y b r 2 2 2+ =] ^g h

x y

x y

1 2 4

1 2 2

2

2 2

+ + =

+ - =

2

2

] ^

] ]_

g h

g gi

So 1, 2a b= = - and 2r =

CONTINUED


47/86


This is a circle with centre ,1 2-^ h and radius 2.To draw the circle, plot the centre point ,1 2-^ h and count 2 units up,down, let and right to fnd points on the circle.

y

x

1

1

-2 2 3 4-1

-3

-4

-5

2

3

4

5

-1

-2

-3-4

(1, -2)

From the graph, we can see all the values that are possible or(b) x

and yor the circle.

Domain: : 1 3{ }x x# #-

Range: : 4 0{ }y y# #-

3. Find the equation o a circle with radius 3 and centre ,2 1-^ h in

expanded orm.

Solution

This is a general circle with equation x a y b r 2 2 2+ =] ^g h where,a b2 1= - = and .r 3=

Substituting:

x a y b r

x y

x y

2 1 3

2 1 9

2 2 2

2 2 2

2 2

+ =

- - + =

+ + =

] ^]] ^

] ^

g hg g hg h

Remove the grouping symbols.

a b a ab b

x x x

x x

a b a ab b

y y y

y y

2

2 2 2 2

4 4

2

1 2 1 1

2 1

So

So

2 2 2

2 2 2

2

2 2 2

2 2 2

2

+ = + +

+ = + +

= + +

= - +

= - +

= - +

]] ] ]

]^ ^ ]

gg g g

gh h g

The equation o the circle is:

x x y y

x x y y

x x y y

x x y y

4 4 2 1 9

4 2 5 9

4 2 5 9

4 2 4 0

9 9

2

2

2

2

+ + + - + =

+ + - + =

+ + + =

+ + - - =

- -

You may need to revise this

in Chapter 2.


48/86


Investigation

The circle is not a unction. Could you break the circle up into

two unctions?

Change the subject o this equation to y.

What do you notice when you change the subject to y? Do you get two

unctions? What are their domains and ranges?

I you have a graphics calculator, how could you draw the graph o a

circle?

The equation o the semi-circle above the x-axis with centre (0, 0)

and radius ris y r x2 2= -

The equation o the semi-circle below the x-axis with centre (0, 0)

and radius ris y r x2 2= - -

y r x2 2= - is the semi-circle above the x-axis since its range is y$ 0

or all values.

y

xr

r

-r

The domain is { :x r x r # #- } and the range is { :y y r0 # # }

Proof

x y r

y r xy r x

2 2 2

2 2 2

2 2!

+ =

=

= -

This gives two unctions:

By rearranging the equation o a circle, we can also fnd the equations o

semi-circles.


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y r x2 2= - - is the semi-circle above the x-axis since its range is

y 0# or all values.

y

xr

-

r

-r

The domain is { :x r x r # #- } and the range is { : }y r y 0# #-

EXAMPLES

Sketch each unction and state the domain and range.

1. f x x92

= -] g Solution

This is in the orm f x r x2 2= -] g where .r 3= It is a semi-circle above the x-axis with centre (0, 0) and radius 3.

y

x3

3

-3

Domain: : 3 3{ }x x# #-

Range: : 0 3{ }y y# #


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2. y x4 2= - -

Solution

This is in the orm y r x2 2= - - where .r 2=

It is a semi-circle below the x-axis with centre (0, 0) and radius 2.

y

x2

-2

-2

Domain: : 2 2{ }x x# #-

Range: : 2 0{ }y y# #-

1. For each o the ollowing

sketch each graph(i)

state the domain and(ii)

range.

(a) 9x y2 2+ =

(b) x y 16 02 2+ =-

(c) x y2 1 42 2+ =] ^g h (d) 1 9x y2 2+ + =

] g

(e) x y2 1 12 2+ + =] ^g h

2. For each semi-circle

state whether it is above or(i)

below the x-axis

sketch the unction(ii)

state the domain and(iii)

range.

(a) 25y x2= - -

(b) 1y x2= -

(c) 36y x2= -

(d) 64y x2= - -

(e) 7y x2= - -

3. Find the length o the radius and

the coordinates o the centre o

each circle.

(a) 100x y2 2+ =

(b) 5x y2 2+ =

(c) x y4 5 162 2+ =] ^g h (d) x y5 6 492 2+ + =] ^g h (e) x y 3 812 2+ =^ h

5.8 Exercises


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4. Find the equation o each circle

in expanded orm (without

grouping symbols).

Centre (0, 0) and radius 4(a)

Centre (3, 2) and radius 5(b)

Centre(c) ,1 5-

^ h and radius 3Centre (2, 3) and radius 6(d)

Centre(e) ,4 2-^ h and radius 5Centre() ,0 2-^ h and radius 1Centre (4, 2) and radius 7(g)

Centre(h) ,3 4- -^ h and radius 9Centre(i) ,2 0-^ h and radius 5Centre(j) ,4 7- -^ h and radius 3

Other Graphs

There are many other dierent types o graphs. We will look at some o these

graphs and explore their domain and range.

Exponential and logarithmic functions

EXAMPLES

1. Sketch the graph of x 3x=] g and state its domain and range.

Solution

I you do not know what this graph looks like, draw up a table o values.

You may need to revise the indices that you studied in Chapter 1.

e.g. When 0:x =

y 3

1

c=

=

:x

y

1

3

3

1

31

When1

1

= -

=

=

=

-

x 3- 2- 1- 0 1 2 3

y271

91

31

1 3 9 27

I you already know what the shape o the graph is, you can draw it

just using 2 or 3 points to make it more accurate.

You will meet these

graphs again in the

HSC Course.


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This is an exponential unction with y-intercept 1. We can fnd one

other point.

When

x

y

1

3

3

1

=

=

=

y

x

1

2

1

3

From the graph, x can be any real value (the equation shows this as well

since any x value substituted into the equation will give a value or y).

From the graph, yis always positive, which can be confrmed by

substituting dierent values ox into the equation.


Range: :y y 02" ,

2. Sketch logf x x=] g and state the domain and range.

Solution

Use the LOG key on your calculator to complete the table o values.

Notice that you cant fnd the log o 0 or a negative number.

x 2 1 0 0.5 1 2 3 4

y # # # 0.3 0 0.3 0.5 0.6

y

x

1

2

1 2 3 4

-1

From the graph and by trying dierent values on the calculator, ycan be

any real number while x is always positive.

Domain: :x x 02! +

Range: yall real" ,

You learned about

exponential graphs in earlier

stages of maths.


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The exponential unction y ax= has domain {all real x} and

range { :y y 02 }

The logarithmic unction logy xa

= has domain :x x 02! + and

range {all real y}

Cubic function

A cubic unction has an equation where the highest power ox is .x3

EXAMPLE

1. Sketch the unction f x x 23= +

] gand state its domain and range.

Solution

Draw up a table o values.

x 3 2 1 0 1 2 3

y 25 6 1 2 3 10 29

y

x

1

1

-2 2 3 4

-1

-3

-4

-5

2

3

4

5

-1

-2

-3-4

The unction can have any real x or yvalue:


Range: yall real" ,

If you already

know the shape of

, ( )y x f x x 23= = +3 hasthe same shape as ( )f x x= 3

but it is translated 2 units up

(this gives a y-intercept of 2).


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Domain and range

Sometimes there is a restricted domain that aects the range o a unction.

EXAMPLE

1. Find the range of x x 23= +] g over the given domain o .x1 4# #-

Solution

The graph o f x x 23= +] g is the cubic unction in the previous example.From the graph, the range is {all real y}. However, with a restricted

domain o x1 4# #- we need to see where the endpoints o this

unction are.

f

f

1 1 2

1 21

4 4 2

64 2

66

3

3

- = - +

= - +

=

= +

= +

=

] ]

] ]

g g

g g

Sketching the graph, we can see that the values oyall lie between

these points.

y

x

(-1, 1)

(4, 66)

Range: 1 66: yy # #" ,


55/86


You may not know what a unction looks like on a graph, but you can still

fnd its domain and range by looking at its equation.

When fnding the domain, we look or values ox that are impossible.

For example, with the hyperbola you have already seen that the denominator

o a raction cannot be zero.

For the range, we look or the results when dierent values ox aresubstituted into the equation. For example, x2 will always give zero or a

positive number.

EXAMPLE

Find the domain and range o .f x x 4= -] g

Solution

We can only fnd the square root o a positive number or zero. 4 0x

x 4So $

$

When you take the square root o a number, the answer is always positive

(or zero). So y 0$

Domain: :x x 4$! +

Range: :y y 0$" ,

5.9 Exercises


(a) 4 3y x= +

(b) f x 4= -] g (c) 3x =

(d) f x x4 12=] g (e) p x x 23=] g () f x xx 12 2= - -] g (g) 64x y2 2+ =

(h) f tt 4

3=

-

] g

(i) ( )g2

5zz

= +

(j) | |f x x=] g


(a) y x=

(b) 2y x= -

(c) | |f x x2 3= -] g (d) | | 2y x= -

(e) f x x2 5= - +] g () | |y x5= -

(g) 2y x=

(h) y 5x= -

(i) f x xx 1

=+] g

(j)2

4 3yx

x= -

3. Find the x-intercepts o

(a) y x x 5 2= -] g (b) f x x x x1 2 3= +] ] ] ]g g g g(c) y x x x6 83 2= +-

(d) g x x x164 2= -] g (e) 49x y2 2+ =

You may like to

simplify the function

by dividing byx.


56/86


4. (a) Solve .x1 02 $-

(b) Find the domain o

.f x x1 2= -] g

5. Find the domain o

(a) 2y x x2= - -

(b) g t t t62= +] g

6. Each o the graphs has a

restricted domain. Find the range

in each case.

(a) y x2 3= - in the domain

x3 3# #-

(b) y x2= in the domain

x2 3# #-

(c) f x x3=] g in the domainx2 1# #-

(d)1

y x= in the domain

x1 5# #

(e) | |y x= in the domain

0 4x# #

() y x x22= - in the domain

x3 3# #-

(g) y x2= - in the domain

x1 1# #-

(h) y x 12= - in the domain

x2 3# #-

(i) y x x2 32= - - in the domain

x4 4# #-

(j) y x x7 62= - + - in the

domain 0 7x# #

7. (a) Find the domain or the

unction .yx 1

3=

+

Explain why there is no(b)

x- intercept or the unction.

State the range o the(c)unction.

8. Given the unction f x x

x=] g

fnd the domain o the(a)

unction

fnd its range.(b)

9. Draw each graph on a number

plane

(a) f x x4=] g (b) y x3= -

(c) y x 34= -

(d) 2p x x3=] g (e) 1g x x3= +] g () 100x y2 2+ =

(g) 2 1y x= +

10. (a) Find the domain and range o

.y x 1= -

(b) Sketch the graph o .y x 1= -

11. Sketch the graph o .y 5x=

12. For each unction, state

its domain and range(i)

the domain over which the(ii)

unction is increasing

the domain over which the(iii)

unction is decreasing.(a) y x2 9= -

(b) f x x 22= -] g (c)

1y x=

(d) f x x3=] g (e) f x 3x=] g

13. (a) Solve .x4 02 $-

(b) Find the domain and range o

(i) 4y x2= -

(ii) .y x4

2= - -


57/86


DID YOU KNOW?

A lampshade can produce a hyperbola

where the light meets the fat wall.

Can you nd any other shapes made by

a light?

Lamp casting its light

Limits and Continuity

Limits

The exponential function and the hyperbola are examples of functions that

approach a limit. The curve y ax= approaches the x-axis when x approaches

very large negative numbers, but never touches it.

That is, when , .x a 0x" "3-

Putting a 3- into index form gives

aa1

1

03

Z

=

=

3

3

-

We say that the limit ofax as x approaches 3- is 0. In symbols, we write

.lim a 0x =x " 3-

A line that a graph approaches

but never touches is called an

asymptote.

EXAMPLES

1. Find .lim x

x x5x 0

2+

"

Solution

Substituting 0x = into the function gives0

0, which is undefined.

Factorising and cancelling help us find the limit.

( )

lim lim

lim

x

x x

x

x x

x

5 5

5

5

x x

x

0

2

0 1

1

0

+=

+

= +

=

" "

"

] g


58/86


2. Find .limx

x

4

22

-

-

x 2"

Solution

Substituting 2x = into the unction gives 00, which is undefned.

lim lim

lim

x

x

x x

x

x

4

2

2 2

2

21

41

2 1

1

-

-=

+ -

-

=+

=

x x2 2" "

x 2"

^ _h i

3. Find .limh

h x hx h2 72 2+ -h 0"

Solution

lim lim

lim

h

h x hx h

h

h hx x

hx x

x

2 7 2 7

2 7

7

2 2 2

2

2

+ -=

+ -

= + -

= -

h 0"

h h0 0" "

^ h

Continuity

Many unctions are continuous. That is, they have a smooth, unbroken curve(or line). However, there are some discontinuous unctions that have gaps in

their graphs. The hyperbola is an example.

I a curve is discontinuous at a certain point, we can use limits to fnd the

value that the curve approaches at that point.

EXAMPLES

1. Find lim

x

x

1

12

-

-

x 1"and hence describe the domain and range o the curve

1

1.y

x

x2=

-

-Sketch the curve.

Solution

Substituting 1x = into1

1xx2

-

-gives

0

0

CONTINUED


59/86


( )

lim lim

limx

x

x

x x

x11

1

1 1

1

2

x x

x

1

2

1

1

-

-=

-

+ -

= +

=

"

"

-

] ]g g

1

1y

xx2

=-

-is discontinuous at 1x = sinceyis undefned at that point.

This leaves a gap in the curve. The limit tells us thaty 2" as 1,x " so

the gap is at , .1 2^ h Domain: : , 1x x xall real !" ,Range: : , 2y y yall real !" ,

yxx

x

x x

x

1

1

1

1 1

1

2

=-

-

=

+

= +

-

-^ ^h h

` the graph isy x 1= + where x 1!

2. Find limx

x x

2

2x 2

2

+

+ -

" -

and hence sketch the curve .yx

x x

2

22=

+ -

+

Solution

Substituting x 2= - intox

x x2

22

+

+ -gives

00

lim lim

lim

xx x

x

x x

x

22

2

1 2

1

3

x x

x

2

2

2

2

+

+ -=

+

- +

=

= -

-

" "

"

- -

-

^

^ ^

^h

h h

h

2yx

x xx

yx

x

x

x

2

2

2

2

1

1

is discontinuous at2

=+

+ -= -

=+

= -

+ -^ ^h h

So the unction isy x 1= - where .x 2! - It is discontinuous at , .2 3- -^ h

Remember that .x 1!


60/86


1. Find(a) lim x 52 +

x 4"

(b) lim t 7-t 3" -

(c) lim x x2 43 + -x 2"

(d) lim xx x32 +

x 0"

(e) limh

h h

2

22

-

- -

h 2"

() limy

y

5

1253

-

-

y 5"

(g) limx

x x

12 12

+

+ +

x 1"-

(h) limx

x x4

2 82

+

+ -

x 4" -

(i) limc

c

4

22

-

-

c 2"

(j) limx x

x 12

-

-

x 1"

(k) lim h

h h h2 73 2+ -h 0"

(l) limh

hx hx h32 2- +h 0"

(m) limh

hx h x hx h2 3 53 2 2- + -h 0"

(n) lim x cx c3 3

-

-

x c"

2. Determine which o theseunctions are discontinuous and

fnd x values or which they are

discontinuous.

(a) 3y x2= -

(b)1

1y

x=

+

(c) f x x 1= -] g

(d)4

1y

x2=

+

(e)4

1

y x2=

-

3. Sketch these unctions, showing

any points o discontinuity.

(a)3

y xx x2

=+

(b)3

3y

xx x2

=+

+

(c)1

5 4y

xx x2

=+

+ +

5.10 Exercises


61/86


A special limit is lim x1

0=x " 3

EXAMPLES

1. Find .limx x

x

2 3

32

2

- +x " 3

Solution

lim lim

lim

x x

x

x

x

x

x

x

x

x

x

x x

2 3

3

2 3

3

12 3

3

1 0 0

3

3

(dividing by the highest power o )2

2

2

2

2 2

2

2

2

- +

=

- +

=

- +

=- +

=

x x

x

" "

"

3 3

3

2. Find

(a) limx x

x

4 42 + +x " 3

(b) limx x

x

4 42 + +x " 3-

Solution

(a) lim limx x

x

x

x

x

x

x

x

x

4 4 4 42

2

2

2 2

2

+ +

=

+ +

x x" "3 3

lim

x x

x

14 4

1

1 0 0

0

0

2

=

+ +

=+ +

=

x " 3

Further Graphs

Graphs o unctions with asymptotes can be difcult to sketch. It is important

to fnd the limits as the unction approaches the asymptotes.


62/86


Since1

0x " from the positive side when ,x " 3+ we can write

0limx x

x

4 42 + +=

+

x " 3

(b) lim limx x

x

x x

x

4 4 14 4

1

0

2

2

+ +

=

+ +

=

x x" "3 3- -

Since1

0x " from the negative side when ,x " 3- we can write

0limx x

x

4 42 + +=

-

x " 3-

3. Find .lim

x

x

1

3 2

-x " 3

Solution

Dividing by x2 will give0

3.

Divide by x.

lim lim

lim

x

x

xx

x

xx

x

x

x

x

1

3

1

3

11

3

1 03

3

2

2

-=

-

=

-

=-

=

x x

x

" "

"

3 3

3

1x

4

x

42

+ + is positive

whether x is or .-+ Can

you see why?

General graphs

It is not always appropriate to sketch graphs, for example, a hyperbola or

circle, from a table of values. By restricting the table of values, important

features of a graph may be overlooked.

Other ways of exploring the shape of a graph include:

intercepts

The x-intercept occurs when 0.y =

The y-intercept occurs when 0.x =

even and odd functions

Even functions ( )f x f x- =^ h6 @ are symmetrical about the y-axis.

Odd functions ( )f x f x- = -^ h6 @ are symmetrical about the origin.


63/86


asymptotes

Vertical asymptotes occur when f x 0!] g and ,h x 0=] g given

.f xh x

g x=]

]

]g

g

g

Horizontal and other asymptotes are ound (i they exist) when

fnding .lim f xx "!3

] g domain and range

The domain is the set o all possible x values or a unction.

The range is the set o all possibleyvalues or a unction.

EXAMPLES

1. Sketch .yx

x

92

2

=

-

Solution

Intercepts:

For x-intercept, 0y=

0

0

x

x

x

x

092

2

2

=

-

=

=

So the x-intercept is 0.

Fory-intercept, 0x=

y

0 2

0

0

2

=-

=

So they-intercept is 0

Type of function:

( )

( )

f xx

x

x

x

f x

9

9

2

2

2

2

- =

- -

-

=

-

=

]]

gg

The unction is even so it is symmetrical about they-axis.

Vertical asymptotes:

0

3 0, 3 0

3, 3

x

x x

x x

x x

9 0

3 3

2

!

! !

! !

!-

+ -

+ -

-

] ]g g

So there are asymptotes at x 3!= .


64/86


As x" 3 rom LHS:

(3 )f3 9

32

2

=

-

=-

+

= -

-

-

-

^^h

h

Soy" 3-

As x " 3 rom RHS:

(3 )f3 9

32

2

=

-

=+

+

= +

+

+

+

^^h

h

Soy" 3

As x" 3- rom LHS:

( )f 33 9

32

2

- =

- -

-

=+

+

= +

-

-

-

^^

hh

Soy" 3

As 3x " - rom RHS:

( )f 33 9

32

2

- =

- -

-

=-

+

= -

+

+

+

^^

hh

Soy" 3-

Horizontal asymptotes:

lim lim

lim

x

x

x

x

x

x

x

x

9 9

19

1

1 0

1

1

2

2

2

2

2

2

2

2

-

=

-

=

-

=

-

=

x x

x

" "

"

3 3

3

As x" 3

( )f9

1

2

2

3

3

3

2

=

-

So as x" 3,y" 1 rom above

You could substitute values

close to 3 on either side into

the equation, say 2.9 on LHS

and 3.1 on RHS.

You could substitute values

close to 3- on either side

into the equation, say .3 1-

on LHSand .2 9- on RHS.

CONTINUED


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As x" 3-

( )f9

1

3

3

3

2

- =

- -

-

2

2

]]

gg

So as x" 3- ,y" 1 rom above

Domain: {x: all real x 3!! }

Range:

When 3, 1x y2 2

When ,x y3 3 01 1 #-

When ,x y3 11 2-

So the range is {y:y2 1,y# 0}.

All this inormation put together gives the graph below.

2. Sketch .( )f xx

x

2

2

=-

Solution

Intercepts:

For x-intercept, 0y=

0

0

0

xx

x

x

2

2

2

=-

=

=

So the x-intercept is 0

Fory-intercept, 0x =

0

y 0 20

2

=-

=

So they-intercept is 0.

Type of function:

( )

( )

f xx

x

xx

x

x

f x

2

2

2

2

2

2

!

- =

-

-

=- -

= -+

-

-]]

gg

The unction is neither even nor odd.

You could substitute

values such as 1000 and

1000- to see what

ydoes as x

approaches .!3


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Vertical asymptotes:

xx2 0

2!

!

-

So there is an asymptote at 2x = .

As x" 2 rom LHS:

(2 )f2 2

22

=-

=-

+

= -

-

-

-^ h

Soy" 3-

As x" 2 rom RHS:

(2 )f2 2

22

=

-

=+

+

= +

+

+

+^ h

Soy" 3

You could substitute values close to 2 on either side into the equation, say

1.9 on LHS and 2.1 on RHS.

e.g. When 2.1x =

(2.1).

.

44.1

f2 1 2

2 1 2=

-

=

] g

Horizontal asymptotes:

lim lim

lim

x

x

xx

x

xx

x

x

x

x

2 2

12

1 0

2

2

-=

-

=

-

=-

=

x x

x

" "

"

3 3

3

This means that as x approaches !3, the unction approachesy x= .

As x" 3

( )fx

2

2

33

3

2

= -

So as x" 3,y"x rom above.

As x" 3-

( )f

x2

33

3

1

- =- -

-2] g

So as x" 3- ,y"x rom above.

Note: If we divide everything

by ,x2 we get .0

1Divide by x.

CONTINUED


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This is not easy to see, so substitute values such as 1000 and 1000- to see

whatydoes as x approaches 3.

e.g. When 1000x = -

( )f 1000 1000 2

1000

998

2

- = - -

-

= -

] g

The point ,1000 998- -^ h is just above the liney x= .Domain: {x: all real x 2! }

Range:

When x 22 we fnd that an approximate range isy 352 (substituting

dierent values ox)

When ,x y2 01 #

So the range is { : ,y y y35 02 # }

Putting all this inormation together gives the graph below.

y

x

2

There is a method that combines all these eatures to make sketching easier.

EXAMPLES

1. Sketch .yx

x

92

2

=

-

Solution

First fnd the critical points (x-intercepts and vertical asymptotes).

3 3y

x x

x x#=

+ -] ]g g


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- 0

0

0

x

x

x

x

x

y

9

0

intercepts:

2

2

2

=

=

-

=

=

( 3)( 3) 0x x

x 3

asymptotes:

!

+ - =

=

These critical points, , ,x 0 3!= divide the number plane into our regions.

Then sketch ,y x y x 3= = + and 3y x= - on your graph.

Look at the sign o the curve in each region.

gion :Re y x

y x

y x

yx x

x x

1

3

3

3 3`

#

#

#

= +

= + +

= - +

=

+ -

=+ +

+ +

= +

] ]g g

gion :Re y x

y x

y x

yx x

x x

2

3

3

3 3`

#

#

#

= +

= + +

= - -

=

+ -

=+ -

+ +

=-

+

= -

] ]g g

The curve is above the x-axis

in this region.

The curve is below the x-axis

in this region.

A graph is positive if it is

above the x-axis.

These are straight lines at the

critical points.

CONTINUED


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gion :Re y x

y x

y x

yx x

x x

3

3

3

3 3`

#

#

#

= -

= + +

= - -

=

+ -

=+ -

- -

=-

+

= -

] ]g g

gion :Re y x

y x

y x

yx x

x x

4

3

3

3 3`

#

#

#

= -

= + -

= - -

=

+ -

=- -

- -

=+

+

= +

] ]g g

Find any horizontal asymptotes.

rom above

1 rom above

lim lim

lim

x

x

x

x

x

9 19

1

1

9

2

2

2

2

2

-

=

-

=

-

=

x x

x

" "

"

3 3

3-

All this inormation put together gives the ollowing graph.

2. Sketch .yx x

x

2 1

1=

+ -

+

] ]g g

Solution

Find the critical points.

1 ( )

vertical asymptotes

x x

xx

21

intercept= - -

= -

=^ h0

The curve is below the

x-axis in this region.

The curve is above the

x-axis in this region.

Check these!


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Use these to divide the number plane into 4 regions and sketch

1, 2 1.y x y x y xand= + = + = -

gion :Re y

x x

x1

2 1

1

#

=

+ -

+

=+ +

+

= +

] ]g g

gion :Re yx x

x3

2 1

1

#

=

+ -

+

=+ -

-

= +

] ]g g

gion :Re yx x

x2

2 1

1

#

=

+ -

+

=+ -

+

= -

] ]g g

gion :Re yx x

x4

2 1

1

#

=

+ -

+

=- -

-

= -

] ]g g

For horizontal asymptotes

lim lim

lim

lim

x x

x

x x

x

x x

x x

x x

x

2 1

1

2

1

11 2

1 1

0

2 1

10

2

2

2

+ -

+=

+ -

+

=

+ -

+

=

+ -

+=

+

-

x " 3-

x x

x

" "

"

3 3

3

] ]

] ]

g g

g g

All this inormation put together gives the ollowing graph.

The y-intercept is .2

1-


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Class Investigation

You can explore graphs o this type on a graphical calculator or by using

computer sotware designed to draw graphs.

5.11 Exercises

1. Find

limx

x2x " 3

(a)

(b) lim

x 4

2+x " 3

(c) limx

x

1

52

+x " 3

(d) limx x

x23

3

-x " 3

(e) limx x

x

7 12

2

+ +x " 3

() limx x

x

2 7

65

5

- -x " 3

(g) limx

x x

3 1

2 3 6

3

3

+

- -

x " 3

(h) limx x

x

4 27 93

2

+ -x " 3

(i) limx

x

25 2

+x " 3

(j) limx

x

1

3

-x " 3

2. (a) Show that

3

1

1 3

x

Ch5 Functions and Graphs

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Transcript of Ch5 Functions and Graphs