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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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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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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?
Sketch the amilies o graphs or these graphs with parameter2. k.
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
What eect does the parameter k have on these graphs? Could you give a
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).
Sketch the amilies o graphs or these graphs with parameter3. k.
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
What eect does the parameter k have on these graphs? Could you give a
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
For x-intercept: y 0=
0 3 5
5 3
1
x
x
x32
=
=
=
-
For y-intercept: x 0=
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?
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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
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227Chapter 5 Functions and Graphs
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
Domain: {all real x}
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.
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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
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229Chapter 5 Functions and Graphs
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?
Use a graphics calculator or a computer with dynamic graphing capability
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?
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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
For x-intercept: y 0=
x
x
x
0 1
1
1
2
2
!
= -
=
=
For y-intercept: x 0=
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-
Domain: xall real" ,
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.
For x-intercept: y 0=
0
1 0
1
x
x
x
1 2= +
+ =
= -
] g
For y-intercept: x 0=
1
y 0 1 2= +
=
] g
-1
-
4
3
2
1
5
-2
-4
-5
-
1 41-2 5-4
y
x
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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
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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,
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233Chapter 5 Functions and Graphs
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
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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
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235Chapter 5 Functions and Graphs
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$= - ] g
-3
4
3
2
1
5
-2
-1
-4
-5
y
3-1-2 421-3-4
x
y=x-1
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236 Maths In Focus Mathematics Extension 1 Preliminary Course
y x 1= - - x 01] g
-3
4
3
2
1
5
-2
-1
-4
-5
y
3-1-2 421-3-4
x
y=-x-1
Draw these on the same number plane and then disregard the dotted
lines to get the graph shown in method 1.
-3
4
3
2
1
5
-2
-1
-4
-5
yy
3-1-2 421-3-4
x
y=-x- 1
y=x-1
Method 3: I you know the shape o the absolute value unctions, fnd the
intercepts.
For x-intercept: 0y=
So f x 0=] g | |
| |
x
x
x
0 1
1
1` !
= -
=
=
Fory-intercept: 0x =
( ) | |f 0 0 11
= -
= -
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237Chapter 5 Functions and Graphs
The graph is V-shaped, passing through these intercepts.
-3
4
3
2
1
5
-2
-1
-4
-5
y
4-2 5321-1-3-4
x
From the graph, notice that x values can be any real number while the
minimum value oyis .1-
Domain: {all real x}
Range: {y:y 1$ - }
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
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238 Maths In Focus Mathematics Extension 1 Preliminary Course
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
= +
= +
- =
Fory-intercept: 0x =
(0) | 0 2 |
2
f = +
=
There is only one
solution or the
equation | x 2 | 0.+ =
Can you see why?
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239Chapter 5 Functions and Graphs
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?
Use a graphics calculator or a computer with dynamic graphing capability
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.
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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.
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241Chapter 5 Functions and Graphs
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
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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?
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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?
Use a graphics calculator or a computer with dynamic graphing capability
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
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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.
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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
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(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
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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
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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
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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
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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.
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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.
Domain: xall real" ,
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:
Domain: xall real" ,
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 # #" ,
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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
1. Find the domain and range o
(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
2. Find the domain and range o
(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.
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259Chapter 5 Functions and Graphs
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= - -
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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
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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
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( )
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!
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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
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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.
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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.
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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!= .
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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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272 Maths In Focus Mathematics Extension 1 Preliminary Course
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