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Functions and Their Graphs
Jackie Nicholas
Janet Hunter
Jacqui Hargreaves

Mathematics Learning Centre
University of Sydney
NSW 2006

c
1999

University of Sydney

Mathematics Learning Centre, University of Sydney

i

Contents
1 Functions
1
...


1

1
...
1

Definition of a function
...
1
...


2

1
...
3

Domain of a function
...
1
...


2

1
...


6

1
...


7

1
...


8

More about functions
2
...
11
2
...
1

Vertical shift
...
1
...
11

2
...
12

2
...
13
2
...
1

Reflection in the x-axis
...
3
...
13

2
...
14

2
...
14

2
...
16

2
...
17

2
...
19

2
...
21

2
...
23
2
...
24
3 Piecewise functions and solving inequalities
3
...
27
3
...
1

Restricting the domain
...
2

Exercises
...
3

Inequalities
...
4

Exercises
...
1

ii
36

Graphs of polynomials and their zeros
...
1
...
36

4
...
2

Polynomial equations and their roots
...
1
...
37

4
...
4

Zeros of cubic polynomials
...
2

Polynomials of higher degree
...
3

Exercises
...
4

Factorising polynomials
...
5

4
...
1

Dividing polynomials
...
4
...
45

4
...
3

The Factor Theorem
...
49

5 Solutions to exercises

50

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Mathematics Learning Centre, University of Sydney

1

Functions

In this Chapter we will cover various aspects of functions
...


1
...
1
...


One way to demonstrate the meaning of this definition is by using arrow diagrams
...
Every element
in X has associated with it exactly one
element of Y
...
The element 1 in set X is assigned two elements,
5 and 6 in set Y
...
This means that there cannot be any repeated x-values
with different y-values
...

F = {(1,5),(3,3),(2,3),(4,2)} is a function
...


The definition we have given is a general one
...
However, in these notes
we will only consider functions where X and Y are subsets of the real numbers
...
This
graphical representation allows us to use a test to decide whether or not we have the
graph of a function: The Vertical Line Test
...
1
...


y

y

x
0
x
0

This is the graph of a function
...


1
...
3

This is not the graph of a function
...


Domain of a function

For a function f : X → Y the domain of f is the set X
...

If only the rule y = f (x) is given, then the domain
√ is taken to be the set of all real x for
which the function is defined
...
This is
sometimes referred to as the natural domain of the function
...
1
...

This corresponds to the set of√y-values when we describe a function as a set of ordered
pairs (x, y)
...


Example
a
...


b
...


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Mathematics Learning Centre, University of Sydney

Solution

√
a
...
We know that square root functions are
only defined for positive numbers so we require that x + 4 ≥ 0, ie x ≥ −4
...

b
...


Example
a
...

y
1
x
–2

0

2

4

6

8

–1
–2
–3

b
...

Solution
a
...

3

b
...
The range is all real y ≥ −3
...

Solution
x2 + y 2 = 16 is not a function as it fails the vertical line test
...


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Mathematics Learning Centre, University of Sydney
y
4

2

x
–2

–4

0

2

4

–2

–4

The graph of x2 + y 2 = 16
...
the domain and range
b
...
f (x2 )
d
...


Solution
y

2

1

x
–1

0

1

2

3

The graph of f (x) = 3x − x2
...
The domain is all real x
...
25
...
f (q) = 3q − q 2

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Mathematics Learning Centre, University of Sydney
2

c
...

f (2 + h) − f (2)
(3(2 + h) − (2 + h)2 ) − (3(2) − (2)2 )
=
h
h
=

6 + 3h − (h2 + 4h + 4) − 2
h

=

−h2 − h
h

= −h − 1

Example
Sketch the graph of the function f (x) = (x − 1)2 + 1 and show that f (p) = f (2 − p)
...

Solution
6

y

4

2

x
–2

0

2

4

The graph of f (x) = (x − 1)2 + 1
...
If p = −1 then
2 − p = 2 − (−1) = 3, and f (−1) = f (3)
...
2

Specifying or restricting the domain of a function

We sometimes give the rule y = f (x) along with the domain of definition
...
For example, if we have the function
y = x2

for

0≤x≤2

then the domain is given as 0 ≤ x ≤ 2
...

Consequently, the range of this function is all real y where 0 ≤ y ≤ 4
...


4

y

2

x
–1

0

1

2

The graph of y = x2 for 0 ≤ x ≤ 2
...
3

The absolute value function

Before we define the absolute value function we will review the definition of the absolute
value of a number
...


or

That is, |4| = 4 since 4 is positive, but | − 2| = 2 since −2 is negative
...

|–2|=2
–2

|4|=4
0

4

More generally, |x − a| can be thought of as the distance of x from a on the numberline
...

The absolute value function is written as y = |x|
...
The graph
of y = |x| is given below
...


2

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Mathematics Learning Centre, University of Sydney

Example
Sketch the graph of y = |x − 2|
...

y
2

1
y = –x + 2 x < 2

y=x–2 x≥2
x

0

1

2

3

4

The graph of y = |x − 2|
...
We will use this fact to sketch graphs of
this type in Chapter 2
...
4

Exercises

1
...
State the domain and range of f (x) =
√
b
...


√

2
...

ψ(x + h) − ψ(x)
h

h = 0
...
Sketch the following functions stating the domain and range of each:

Mathematics Learning Centre, University of Sydney

a
...
y = |2x|
c
...
y = |2x| − 1
...
a
...
If the line x + y + k = 0 cuts the circle x2 + y 2 = 4 in two distinct points, find the
restrictions on k
...
Sketch the following, showing their important features
...
y =

 x
1
2

b
...

6
...
Discuss whether or not y 2 = x3
is a function
...
Sketch the following relations, showing all intercepts and features
...

√
a
...
|x| − |y| = 0
c
...
y =

x
,x
|x|

= 0

e
...

8
...


9
...
f (x) = x2 − 4x
b
...
If φ(x) = log



x
x−1



, find in simplest form:

a
...
φ(3) + φ(4) + φ(5) + · · · + φ(n)
11
...
If y = x2 + 2x and x = (z − 2)2 , find y when z = 3
...
Given L(x) = 2x + 1 and M (x) = x2 − x, find
i
...
M (L(x))

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Mathematics Learning Centre, University of Sydney

12
...
a
...

b
...

14
...


Hence show that S(n) − S(n − 1) =

0

1

...
We will introduce the
concepts of even and odd functions, increasing and decreasing functions and will solve
equations using graphs
...
1
2
...
1

Modifying functions by shifting
Vertical shift

We can draw the graph of y = f (x) + k from the graph of y = f (x) as the addition of
the constant k produces a vertical shift
...
For example, we can sketch the
function y = x2 − 3 from our knowledge of y = x2 by shifting the graph of y = x2 down
by 3 units
...


3

y
y = x2

(1,1)

y = x 2– 3
x

–2

–1

0

1

2

(1,–2)
–3

We can also write y = f (x) − 3 as y + 3 = f (x), so replacing y by y + 3 in y = f (x) also
shifts the graph down by 3 units
...
1
...
If we replace x by x − a inside
the function then the graph will shift to the left by a units if a < 0 and to the right by a
units if a > 0
...

this graph to the right by 2 units
...


2
...
The point (1, 1) has been shifted

Modifying functions by stretching

We can sketch the graph of a function y = bf (x) (b > 0) if we know the graph of y = f (x)
as multiplying by the constant b will have the effect of stretching the graph in the ydirection by a factor of b
...

For example, we can sketch y = 2x2 from our knowledge of y = x2 as follows:
y

y

2

(1,2)

2

(1,1)

1

1

x
–1

0

The graph of y = x2
...
Note, all the yvalues have been multiplied by 2, but the
x-values are unchanged
...


2
...
3
...
Note, all the
y-values have been multiplied by 12 , but
the x-values are unchanged
...
(Think of the
x-axis as a mirror
...

y

y
x

2
–1

0

1

1

x
–1

0

1

The graph of y = |x|
...
3
...
It is the reflection of y = |x| in the x-axis
...


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Mathematics Learning Centre, University of Sydney

For example, we can sketch y = 3−x from our knowledge of y = 3x
...


2
...
It is the reflection
of y = 3x in the y-axis
...
For example,
we can sketch the graph of y = |x2 − 3| from our knowledge of the graph of y = x2 − 3
...


2
...
The negative
values of y = x2 − 3 have been reflected
in the x-axis
...
For example, we
can sketch the graph of y = 2 − (x + 1)2 from the graph of y = x2 provided we can analyse
the combined effects of the modifications
...
The effect of the − sign in front of the brackets turns the
graph up side down
...
We can illustrate
these effects in the following diagrams
...


–1

–1

0

The graph of y = (x + 1)2
...


y
–2

x
–2

y
2

x

0

1

–1

–2

The graph of y = −(x + 1)2
...


x
–2

–1

0

The graph of y = 2−(x+1)2
...


Similarly, we can sketch the graph of (x−h)2 +(y−k)2 = r2 from the graph of x2 +y 2 = r2
...
Replacing y by y − k shifts the
graph up or down k units
...
)
For example, we can use the graph of the circle of radius 3, x2 + y 2 = 9, to sketch the
graph of (x − 2)2 + (y + 4)2 = 9
...

This is a circle centre (0, 0), radius 3
...

This is a circle centre (2, −4), radius 3
...
Replacing y by y + 4 shifts it down 4 units
...
6

Graphing by addition of ordinates

We can sketch the graph of functions such as y = |x| + |x − 2| by drawing the graphs of
both y = |x| and y = |x − 2| on the same axes then adding the corresponding y-values
...


At each point of x the y-values of y = |x| and y = |x − 2| have been added
...

This technique for sketching graphs is very useful for sketching the graph of the sum of
two trigonometric functions
...
7

Using graphs to solve equations

We can solve equations of the form f (x) = k by sketching y = f (x) and the horizontal line
y = k on the same axes
...


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Mathematics Learning Centre, University of Sydney

For example, to solve |x − 3| = 2 we sketch y = |x − 3| and y = 2 on the same axes
...
Therefore |x − 3| = 2 when x = 1
or x = 5
...

y
2

–1

0

1

2

3

4

x

–2

–4

(3,–4)

For what values of k does the equation f (x) = k have
1
...
2 solutions
3
...

Therefore the equation f (x) = k will have

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Mathematics Learning Centre, University of Sydney

1
...
2 solutions if k = 0 or k = −4
3
...


2
...
Sketch the following:
a
...


y = 13 x2

c
...


y = (x + 1)2

2
...


y=

1
x

b
...


y = 3 − (x − 1)3

y = 2|x − 2|

c
...


d
...
Sketch the following:
a
...


4
...


y = |x|

b
...
Sketch the following:
a
...
Sketch the following:
√
a
...
Show that

b
...


y=

x−1
1
=
+ 1
...
Sketch y =



9 − (x − 1)2

c
...


y=

√

9 − x2 − 3

x−1

...

x−1

9
...
y = |x| + x + 1 for −2 ≤ x ≤ 2 [Hint: Sketch by adding ordinates]
b
...
y = 2x + 2−x for −2 ≤ x ≤ 2
d
...

10
...


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Mathematics Learning Centre, University of Sydney

11
...
y = 2f (x)
b
...
y = f (−x)
d
...
y = f (x − 3)
f
...
y = 3 − 2f (x − 3)
h
...
5

–2

12
...
|2x| = 4
b
...
x = x2
d
...
Solve |x − 2| = 3
...
algebraically
b
...

14
...
Find the
coordinates of P
...
Sketch the circle x2 + y 2 − 2x − 14y + 25 = 0
...
] Find
the values of k, so that the line y = k intersects the circle in two distinct points
...
Solve

4
5−x

= 1, using a graph
...
Find all real numbers x for which |x − 2| = |x + 2|
...
Given that Q(p) = p2 − p, find possible values of n if Q(n) = 2
...
Solve |x − 4| = 2x
...
algebraically
b
...


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Mathematics Learning Centre, University of Sydney

2
...

Geometrically, an even function is symmetrical about the y-axis (it has line symmetry)
...
We illustrate this on the following graph
...

Definition:
A function, y = f (x), is odd if f (−x) = −f (x) for all x in the domain of f
...

The function f (x) = x is an odd function as f (−x) = −x = −f (x) for all values of x
...

y
y
1

x

–x
–2

–1

0

1

–1
–y

The graph of y = x
...

1
...
g(x) =

1
2x

3
...

Solution
1
...

2
...

3
...


Example
Sketched below is part of the graph of y = f (x)
...
odd
2
...


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Mathematics Learning Centre, University of Sydney

Solution
y

x
0

y = f (x) is an odd function
...


2
...
In Chapter 5
we will relate these concepts to the derivative of a function
...


The function y = 2x is an example of a function that is increasing over its domain
...


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Mathematics Learning Centre, University of Sydney

y
y

2

4

f(b)
1

2
f(a)
x

x
0

–1

1

The graph of y = 2x
...


a

0

1 b

The graph of y = x2
...


Notice that when a function is increasing it has a positive slope
...

The function y = 2−x is decreasing over its domain
...

y

y

2

2
f(a)

1

1
f(b)
x

–1

0

1

The graph of y = 2−x
...


x
a –1 b

0

The graph of y = x2
...


Notice that if a function is decreasing then it has negative slope
...
11

Exercises

1
...
State the domain and range
...
Where is the graph

1

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Mathematics Learning Centre, University of Sydney

i
...
decreasing?
c
...

ii
...

iv
...


is a constant, find the values of k such that f (x) = k has
no solutions
1 solution
2 solutions
3 solutions
4 solutions
...
Is y = f (x) even, odd or neither?
y

x
–4

0

–2

4

2

–2

2
...


y

x
0
x
0

y = f (x)

y = g(x)

3
...

√
b
...
y = 2x
d
...
y = x4 + 2
e
...


x
x2

y = 2x + 2−x

1
−4

g
...


y=

j
...
Given y = f (x) is even and y = g(x) is odd, prove
a
...
if h(x) = (g(x))2 then h(x) is even

x2

1
+4

h
...
if h(x) =

26

f (x)
, g(x) = 0, then h(x) is odd
g(x)

d
...

5
...
Can you prove that
for every odd function in this set f (0) = 0? If not, give a counter-example
...


3
...
1
...
This is frequently called restricting the domain of the function
...

Sketch the graph of y = 1 − x2 for x ≥ 0
...

Sketch the graph of y = 1 − x for x < 0
...


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Mathematics Learning Centre, University of Sydney

We can now put these pieces together to define a function of the form

f (x) =

⎧
⎪
⎨

1 − x2 for x ≥ 0

⎪
⎩

1−x

for x < 0

We say that this function is defined piecewise
...
This is easy to see if we graph the
function and use the vertical line test
...

2 y

1

x
0

–1

1

–1

The graph shows that f defined in this way is a function
...

The absolute value function
⎧
⎪
⎨

x

⎩

−x for x < 0

f (x) = ⎪

for x ≥ 0

is another example of a piecewise function
...

Notice that we have put an open square (or circle) around the point (0, 2) and a solid
square (or circle) around the point (0, 1)
...
When defining a function piecewise, we must be extremely careful to assign
to each x exactly one value of y
...
2

Exercises

1
...
2f (−1) + f (2)
b
...
For the function given in 1, solve f (x) = 2
...
Below is the graph of y = g(x)
...


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Mathematics Learning Centre, University of Sydney
y
1

x
–3

–2

–1

1

–1

–2

4
...
Sketch the graph of y = f (x) if
⎧
⎪
⎨

√
− 4 − x2 for −2 ≤ x ≤ 0

⎩

x2 − 4

f (x) = ⎪

for x > 0

b
...

c
...
f (x) = 0
ii
...

d
...
0
ii
...
2 solutions
...
Sketch the graph of y = f (x) if

f (x) =

⎧
⎪
⎨

1 − |x − 1| for x ≥ 0

⎪
⎩

|x + 1|

for x < 0

6
...
McMaths burgers are to modernise their logo as shown below
...
e
...

8
...
The following piecewise function is of the form
⎧
⎪
⎨

ax2 + b for 0 < x ≤ 2

⎩

cx + d

f (x) = ⎪

4

for x > 2

y
(4,4)

2

x
–4

–2

0

2

4

–2

–4

Determine the values of a, b, c and d
...
Complete the graph so that f (x) is an odd function defined for all real x, x = 0
...
Write down the equations that now define f (x), x = 0
...
3

Inequalities

We can solve inequalities using both algebraic and graphical methods
...
For the following
examples we will use both, as this allows us to make the connections between the algebra
and the graphs
...
Solve 3 − 2x ≥ 1
...

Remember to reverse the inequality
sign when multiplying or dividing
by a negative number
...

2
...

This is a (2 Unit) quadratic inequality
...


Let y = x2 − 4x + 3
...
We take points in
each interval to determine the sign
of the inequality; eg use x = 0,
x = 2 and x = 4 as test values
...


x
0

1

3

When does the parabola have negative
y-values? OR When is the parabola under the x-axis? From the graph, we see
that this happens when 1 < x < 3
...
Solve x−4
≤ 1
...
There
is a variable in the denominator
...

First multiply by the square of the
denominator

x−4
x−4
0
0

≤
≤
≤
≤

1

2

3

4

5

y

–2

(5,1)

2

6

x

–1

neg positive

positive

1

...
5 and
x = 6
...
It
y = x−4
is a hyperbola with vertical asymptote
at x = 4
...
(5, 1) is the point of intersection
...


Therefore, x < 4 or x ≥ 5
...
Solve x − 3 < 10

...


Sketch y = x − 3 and then y = 10

...

y

x2 − 3x = 10
x2 − 3x − 10 = 0
(x − 5)(x + 2) = 0
Therefore, the critical values are
−2, 0 and 5 which divide the number line into four intervals
...
The points x = −3 and x = 1
satisfy the inequality, so the solution is x < −2 or 0 < x < 5
...
)

6
3

(5,2)

x
–6

–3

0

6
(0,–3)

(–2,–5)

–6

For what values of x does the line lie
under the hyperbola? From the graph,
we see that this happens when x < −2
or 0 < x < 5
...

Hence, where possible,
a
...


|2x − 6| = 2x

ii
...
|2x − 6| = x + 3
iv
...
|2x − 6| = x − 3
b
...

Solution
⎧
⎪
⎨

2x − 6

⎩

−(2x − 6) for x < 3

f (x) = |2x − 6| = ⎪

for x ≥ 3

y
y = 2x

y = |2x – 6|

(9,12)

10
...
0

(1,4)
(1
...
00

x

-2
...
00

4
...
00

8
...
00

y=x–3
-5
...
0

a
...


Mark in the graph of y = 2x
...

It has one point of intersection with y = |2x − 6| = −2x + 6 (x < 3) at x = 1
...


ii
...
5
...
y = x + 3 intersects y = |2x − 6| twice
...
This gives
us the solution x = 9
...

iv
...

v
...

b
...
When k = −3, there is one point of
intersection
...
For k > −3, lines of the form y = x + k will have
two points of intersection
...


3
...
Solve
a
...


4p
p+3

c
...
a
...

b
...

3
...
Find the points of intersection of the graphs y = 5 − x and y = x4
...
On the same set of axes, sketch the graphs of y = 5 − x and y = x4
...
Using part (ii), or otherwise, write down all the values of x for which
5−x>

4
x

4
...
Sketch the graph of y = 2x
...
Solve 2x < 12
...
Suppose 0 < a < b and consider the points A(a, 2a ) and B(b, 2b ) on the graph of
y = 2x
...

Explain why
a+b
2a + 2b
>2 2
2
5
...
Sketch the graphs of y = x and y = |x − 5| on the same diagram
...
Solve |x − 5| > x
...
For what values of m does mx = |x − 5| have exactly
i
...
no solutions
6
...


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Mathematics Learning Centre, University of Sydney

4

Polynomials

Many of the functions we have been using so far have been polynomials
...

Definition
A real polynomial, P (x), of degree n is an expression of the form
P (x) = pn xn + pn−1 xn−1 + pn−2 xn−2 + · · · + p2 x2 + p1 x + p0
where pn = 0, p0 , p1 , · · ·, pn are real and n is an integer ≥ 0
...

We are familiar with the quadratic polynomial, Q(x) = ax2 + bx + c where a = 0
...

√
The function f (x) = x + x is not a polynomial as it has a power which is not an integer
≥ 0 and so does not satisfy the definition
...
1
4
...
1

Graphs of polynomials and their zeros
Behaviour of polynomials when |x| is large

One piece of information that can be a great help when sketching a polynomial is the
way it behaves for values of x when |x| is large
...

The term of the polynomial with the highest power of x is called the leading or dominant
term
...

When |x| is large, the dominant term determines how the graph behaves as it is so much
larger in magnitude than all the other terms
...

There are four possibilities which we summarise in the following diagrams:
y

y

x

1
...


x

2
...


37

Mathematics Learning Centre, University of Sydney

y

y

x

3
...


x

4
...


This gives us a good start to graphing polynomials
...
In Chapter 5 we will use calculus methods to do this
...

4
...
2

Polynomial equations and their roots

If, for a polynomial P (x), P (k) = 0 then we can say
1
...

2
...

3
...

4
...
3

Zeros of the quadratic polynomial

The quadratic polynomial equation Q(x) = ax2 + bx + c = 0 has two roots that may be:
1
...
real (rational or irrational) and equal,
3
...

We will illustrate all of these cases with examples, and will show the relationship between
the nature and number of zeros of Q(x) and the x-intercepts (if any) on the graph
...
Let Q(x) = x2 − 4x + 3
...


y
3

x2 − 4x + 3 = 0
(x − 1)(x − 3) = 0
Therefore x = 1 or 3
...


0
–1 -

|

2

4

38

Mathematics Learning Centre, University of Sydney

2
...

Solving the equation Q(x) = 0 we get,
x2 − 4x − 3 = 0 √
4 ± 16 + 12
x =
√2
Therefore x = 2 ± 7
...


y
0

x
2

4

–2

–4

–6

3
...

Solving the equation Q(x) = 0 we get,
x2 − 4x + 4 = 0
(x − 2)2 = 0
Therefore x = 2
...
Q(x) = 0 has a repeated or double root at x = 2
...


4
...

Solving the equation Q(x) = 0 we get,
x2 − 4x + 5 = 0 √
4 ± 16 − 20
x =
√2
Therefore x = 2 ± −4
...
In this case the
roots are complex
...
That is Q(x) > 0 for all real
x
...


39

Mathematics Learning Centre, University of Sydney

We have given above four examples of quadratic polynomials to illustrate the relationship
between the zeros of the polynomials and their graphs
...


if the quadratic polynomial has two real distinct zeros, then the graph of the polynomial cuts the x-axis at two distinct points;

ii
...
if the quadratic polynomial has no real zeros, then the graph does not intersect the
x-axis at all
...
If we consider polynomials
Q(x) = ax2 + bx + c where a < 0 then we will have a graph which is concave down
...
(See Chapter 2
...


4
...
4

The graph of Q(x) = −(x2 − 4x + 4)
...
It has 3 zeros which may be:
i
...
3 real zeros, all of which are equal (3 equal zeros);
iii
...
1 real zero and 2 complex zeros
...
Let Q(x) = 3x3 − 3x
...

3x3 − 3x = 0
3x(x − 1)(x + 1) = 0
Therefore x = −1 or 0 or 1

1

y

x
–1

The roots are real (in fact rational) and
distinct
...
Let Q(x) = x3
...

We can write this as (x − 0)3 = 0
...


1

y

x
–1

1

0

–1

3
...

Solving the equation Q(x) = 0 we get,
x3 − x2 = 0
x2 (x − 1) = 0
Therefore x = 0 or 1
...


1

0

–1

The graph turns at the double root
...
Let Q(x) = x3 + x
...

There is one real root at x = 0
...


1

y

x
–1

0

1

–1

The graph intersects the x-axis once
only
...
If we consider the polynomial P (x) = −x3 then the graph
of this polynomial is the reflection of the graph of P (x) = x3 in the x-axis
...


The graph of Q(x) = x3
...
2

1

Polynomials of higher degree

We will write down a few rules that we can use when we have a polynomial of degree ≥ 3
...
P (x) = 0 has at most n real roots;
2
...
if P (x) = 0 has a repeated root with an odd power then the graph of P (x) has a
horizontal point of inflection at this repeated root
...
tells us that if we have a quartic polynomial equation f (x) = 0
...

We can illustrate 2
...
Notice how the graph sits
on the x-axis at x = 2
...


42

Mathematics Learning Centre, University of Sydney

We illustrate 3
...
Notice the horizontal point
of inflection at x = 2
...


4
...
Sketch the graphs of the following polynomials if y = P (x) is:
a
...
x(x + 1)(3 − x)
c
...
(x + 1)(x2 − 4x + 5)
2
...
Match the graph
with the polynomial
...
y = x4 b
...
y = x4 + 1 d
...
y = (x − 1)4 f
...


ii
...


y

y
1

1

1

x
–1

x
–1

0

iv
...


y

0

vi
...
Sketch the graphs of the following quartic polynomials if y = C(x) is:
a
...
x(x − 1)(x + 2)(3 − x)
c
...
(x + 1)2 (x − 3)2
e
...
(x + 1)3 (3 − x)
g
...
x2 (x2 − 4x + 5)
...
By sketching the appropriate polynomial, solve:
a
...
(x + 2)(x − 3)(5 − x) > 0
c
...
(x + 2)3 (5 − x) ≥ 0
...
For what values of k will P (x) ≥ 0 for all real x if P (x) = x2 − 4x − 12 + k?
6
...

In each case determine possible values for a, b, c and d
...


b
...


y

y

4
x

0

4

2

4

2

–2

x
–4

2

d
...


y

0

2

x
4

(2,–8)

4

f
...
The graph of the polynomial y = f (x) is given below
...
Use the graph to answer the following questions
...
State the roots of f (x) = 0
...
What is the value of the repeated root
...
For what values of k does the equation f (x) = k have exactly 3 solutions
...
Solve the inequality f (x) < 0
...
What is the least possible degree of f (x)?
f
...

g
...

y (0
...
23)
x
–2

–1

(–1
...
91)

1

2

–10

The graph of the polynomial y = f (x)

4
...
In
this section we will look at methods which will help us factorise polynomials with degree
> 2
...
4
...
Then
P (x)
R(x)
= Q(x) +
,
A(x)
A(x)
where Q(x) is a polynomial called the quotient and R(x) is a polynomial called the
remainder, with the degree of R(x) < degree of A(x)
...

For example: If P (x) = 2x3 + 4x + 3 and A(x) = x − 2, then P (x) can be divided by A(x)
as follows:
2x2 + 4x + 12
x − 2 2x3 + 0x2 + 4x − 3
2x3 − 4x2
4x2 + 4x − 3
4x2 − 8x
12x − 3
12x − 24
21

Mathematics Learning Centre, University of Sydney

45

The quotient is 2x2 + 4x + 12 and the remainder is 21
...

x−2
x−2
This can be written as
2x3 + 4x − 3 = (x − 2)(2x2 + 4x + 12) + 21
...

4
...
2

The Remainder Theorem

If the polynomial f (x) is divided by (x − a) then the remainder is f (a)
...
Since the degree of A(x) is 1, the degree of R(x) is zero
...

f (x) = (x − a)Q(x) + r
f (a) = 0 · Q(a) + r
= r

where r is a constant
...

Example
Find the remainder when P (x) = 3x4 − x3 + 30x − 1 is divided by a
...
2x − 1
...
Using the Remainder Theorem:
Remainder = P (−1)
= 3 − (−1) − 30 − 1
= −27
b
...
What is the
remainder when f (x) is divided by (x − 2)?
Solution
Write f (x) = (x2 − 4) · q(x) + (5x + 6)
...

4
...
3

The Factor Theorem

If x = a is a zero of f (x), that is f (a) = 0, then (x − a) is a factor of f (x) and f (x) may
be written as
f (x) = (x − a)q(x)
for some polynomial q(x)
...

Another useful fact about zeros of polynomials is given below for a polynomial of degree
3
...


ii
...


b
α+β+γ =− ;
a
c
αβ + αγ + βγ = ;
a
d
αβγ = −
...
We will give the partial result
for n = 4
...

a

If a = 1 and the equation P (x) = 0 has a root which is an integer, then that integer must
be a factor of the constant term
...
That is, we look at all the factors of the constant term to see which ones
(if any) are roots of the equation P (x) = 0
...
Factorise f (x)
...
Sketch the graph of y = f (x)
...
Solve f (x) ≥ 0
...
Consider the factors of the constant term, 2
...
Since f (2) = 0, we know that (x − 2) is a
factor of f (x)
...

4x2 − 1
x − 2 4x3 − 8x2 − x + 2
4x3 − 8x2
−x+2
−x+2
So,
f (x) = (x − 2)(4x2 − 1)
= (x − 2)(2x − 1)(2x + 1)

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Mathematics Learning Centre, University of Sydney

b
...

c
...


Example
Show that (x − 2) and (x − 3) are factors of P (x) = x3 − 19x + 30, and hence solve
x3 − 19x + 30 = 0
...
Long division
of P (x) by x2 − 5x + 6 gives a quotient of (x + 5)
...

Solving P (x) = 0 we get (x − 2)(x − 3)(x + 5) = 0
...

Instead of using long division we could have used the facts that
i
...
the product of the zeros must be equal to −30
...

Then 2 · 3 · α = −30, so that α = −5
...


Mathematics Learning Centre, University of Sydney

4
...
When the polynomial P (x) is divided by (x − a)(x − b) the quotient is Q(x) and the
remainder is R(x)
...
Explain why R(x) is of the form mx + c where m and c are constants
...
When a polynomial is divided by (x − 2) and (x − 3), the remainders are 4 and 9
respectively
...

c
...
Also, P (b) = b2
...

2
...
Divide the polynomial f (x) = 2x4 + 13x3 + 18x2 + x − 4 by g(x) = x2 + 5x + 2
...

b
...
(Hint: Assume that α is a
common zero and show by contradiction that α does not exist
...
For the following polynomials,
i
...
solve P (x) = 0
iii
...

a
...
P (x) = x3 − x2 − 16x − 20
c
...
P (x) = x3 − x2 + x − 6
e
...
4

Solutions

√
1
...
The domain of f (x) = 9 − x2 is all real x where −3 ≤ x ≤ 3
...

b
...


2
...
a
...


y

y

4

4

2

2

x

x
0

1

3

5

√
The graph of y = x − 1
...


–2

0

2

The graph of y = |2x|
...


51

Mathematics Learning Centre, University of Sydney

c
...

x−4

4

8 x

6

The domain is all real x = 4 and the range is all real y = 0
...

y
4

2

x
–2

0

2

–1

The graph of y = |2x| − 1
...

4
...
The perpendicular distance d from (0, 0) to x + y + k = 0 is d = | √k2 |
...
For the
line
x+y
+k
=
0
to
cut
the
circle
in
two
distinct
points
d
<
2
...


5
...


b
...

6
...


2

–2

The graph of y 2 = x2
...
a
...


y

y

2

2

x
–2

x

2

0

–2

–2

–2

√
The graph of y = − 4 − x2
...

c
...
This is not
the graph of a function
...


y

y

2

2

x
–2

0

x

2

–2

–2

0

2

–2

The graph of y = x3
...


x
The graph of y = |x|

...
Its domain is all real x = 0, and range
is y = ±1
...

y
2

x
0

2

–2

The graph of |y| = x
...


53

Mathematics Learning Centre, University of Sydney

8
...
a
...

b
...

10
...
φ(3) + φ(4) + φ(5) = log(2
...
φ(3) + φ(4) + φ(5) + · · · + φ(n) = log( n2 )
11
...
y = 3 when z = 3
...
i
...
M (L(x)) = 4x2 + 2x
12
...
a = 2, b = 2 so the equations is y = 2x2 − 2
...
a = 5, b = 1 so the equation is y = x25+1
...
b
...

14
...
8

Solutions

1
...


b
...


0

The graph of y =

1

x2

...


c
...

2
...


b
...

x−2

d
...

c
...


y

–2

–1

−2

...


2

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Mathematics Learning Centre, University of Sydney

3
...


b
...


2

The graph of y = |x3 − 2|
...


y
4

2

x
0

2

4

The graph of y = 3 − (x − 1)3
...


4
...

y

y

4

4

2

2

x
–2

0

x
0

2

The graph of y = |x|
...


c
...


4

4

56

Mathematics Learning Centre, University of Sydney

5
...


b
...


The graph of x2 + (y + 2)2 = 16
...


y

6

4
(1,3)
2

x
–4

0

–2

2

4

The graph of (x − 1)2 + (y − 3)2 = 16
...
a
...


y

3

3

(1,0)

x
–3

0

The graph of y =

3

√

–2

9 − x2
...

–3

y

9 − x2 − 3
...


57

Mathematics Learning Centre, University of Sydney

7
...

x−2

8
...

x−1

b
...
a
...


2

x
–2

0

2

The graph of y = |x| + |x − 1|
for −2 ≤ x ≤ 3
...


d
...


–2

The graph of |x − y| = 1 for
−1 ≤ x ≤ 3
...

y
4

2

x
0

–2

2

–1

The graph of f (x) = |x2 − 1| − 1
...
a
...


y

y
2

x
–3

0

1
...


The graph of y = −f (x)
...
5

59

Mathematics Learning Centre, University of Sydney

c
...


y

y

2
4
x
–1
...


e
...


f
...
5
–2

–2
–4

The graph of y = f (x − 3)
...


The graph of y = f (x + 1) − 2
...


y

y
4

6

2
4

x
–3

2

0

The graph of y = |f (x)|
...


2

60

Mathematics Learning Centre, University of Sydney

12
...


b
...


c
...


y

1
x−2

= −1
...


x = 1 is a solution of x2 = x1
...


a
...
Therefore x = 5 is a solution of the inequality
...
)
For x < 2, |x − 2| = −(x − 2) = −x + 2 = 3
...

(Note that x = −1 is < 2
...

y
4

2

x
–2

0

2

4

The points of intersection are (−1, 3) and (5, 3)
...

14
...

15
...


16
...
Therefore the solution of

4
5−x

= 1 is x = 1
...

y

4

2

x
–4

–2

0

2

4

The point of intersection is (0, 2)
...

18
...

19
...
For x ≥ 4, |x − 4| = x − 4 = 2x when x = −4, but this does not satisfy the
condition of x ≥ 4 so is not a solution
...
x = 43 is < 4 so is a solution
...

b
...
So the solution
of |x − 4| = 2x is x = 43
...
11

Solutions

1
...
The domain is all real x, and the range is all real y ≥ −2
...
i
...
x < −2 or 0 < x < 2
c
...
k < −2
ii
...

iii
...
k = 0
v
...
y = f (x) is even
2
...


b
...


y = f (x) is odd
...


b
...


y = g(x) is odd
...
a
...


even

c
...


odd

e
...


even

g
...


neither

i
...


even

4
...

h(−x) =
=
=
=

f (−x) · g(−x)
f (x) · −g(x)
−f (x) · g(x)
−h(x)

Therefore h is odd
...

h(−x) =
=
=
=
Therefore h is even
...

f (−x)
g(−x)
f (x)
=
−g(x)
f (x)
= −
g(x)
= −h(x)

h(−x) =

Therefore h is odd
...

f (−x) · (g(−x))2
f (x) · (−g(x))2
f (x) · (g(x))2
h(x)

h(−x) =
=
=
=
Therefore h is even
...
If f is defined at x = 0
f (0) =
=
2f (0) =
Therefore f (0) =

3
...


Solutions

1
...
2f (−1) + f (2) = 2(1 − (−1)) + (1 − (2)2 ) = 4 + (−3) = 1
...
f (a2 ) = 1 − (a2 )2 = 1 − a4 since a2 ≥ 0
...
You can see from the graph below that there is one solution to f (x) = 2, and that
this solution is at x = −1
...


⎧
⎪
1
⎪
⎪
⎪
x+1
⎪
⎨ √

1 − x2 for −1 ≤ x ≤ 1

g(x) = ⎪
⎪
⎪
⎪
⎪
⎩

for x < 1

−1

for x > 1

4
...
The domain of f is all real x ≥ −2
...
The range of f is all real y > −4
...
i
...

ii
...

d
...
f (x) = k has no solutions when k ≤ −4
...
f (x) = k has 1 solution when −4 < k < −2 or k > 0
...
f (x) = k has 2 solutions when −2 ≤ k ≤ 0
...
Note that f (0) = 0
...
The domain of g is all real x, x = −2
...

7
...

a
...

b
...

c
...

8
...
Here a = 1, b = −4, c = 2 and d = −4
...
Defining f to be an odd function for all real x, x = 0, we get
y
4

2

x
–4

–2

0

2

4

–2

–4

c
...
4

2x + 4 for x < −2
4 − x2 for −2 ≤ x < 0
x2 − 4 for 0 < x ≤ 2
2x − 4 for x > 2

Solutions

1
...
0 ≤ x ≤ 4
b
...
x < −4 or −3 < x < 3 or x > 4
2
...
The graph of y = 4x(x − 3) is given below
y
5

x
–4

–2

0

–5

2

4

68

Mathematics Learning Centre, University of Sydney

b
...

3
...
The graphs y = 5 − x and y =

4
x

b
...

4
x

y
5

–4

0

–2

2

x

4

–5

c
...

4
...
The graph of y = 2x
...
2x <

1
2

–2

0

2

4

when x < −1
...
The midpoint M of the segment AB has coordinates ( a+b
, 2 +2
)
...
So,
2
a+b
2a + 2b
>2 2
2
y

B
M

5
A

x
–4

–2

0

2

4

69

Mathematics Learning Centre, University of Sydney

5
...

y

5
(2
...
5)
x
–5

0

5

10

b
...
5
...
i
...

ii
...

6
...
3

Solutions

1
...


b
...


The graph of P (x) = x(x + 1)(3 − x)
...


c
...


b
...


2

–5

–10

2
...
iv
...
i
...
iii
...

e
...


f
...


70

Mathematics Learning Centre, University of Sydney

3
...


b
...


y

y
20

5

x
–2

2

The graph of
P (x) = x(x − 1)(x + 2)(3 − x)
...

c
...

e
...

f
...

g
...

h
...


x
–2

0

2

The graph of P (x) = x2 (x2 − 4x + 5)
...
a
...

b
...

c
...

5
...

6
...
P (x) = x(x − 4)
b
...
P (x) = x2 (x − 4)
d
...
P (x) = −x(x − 4)2
f
...
a
...


Mathematics Learning Centre, University of Sydney

72

b
...

c
...
23
...
f (x) < 0 when −2 < x < 0
...
The least possible degree of the polynomial f (x) is 4
...
Since f (0) = 0, the constant in the polynomial is 0
...
f (x) + k ≥ 0 for all real x when k ≥ 9
...


4
...
a
...
So, R(x) is a polynomial of degree ≤ 1
...
Note that if m = 0 the remainder
is a constant
...
Let P (x) = (x2 − 5x + 6)Q(x) + (mx + c) = (x − 2)(x − 3)Q(x) + (mx + c)
...
So, the remainder is
R(x) = 5x − 6
...
Let P (x) = (x − a)(x − b)Q(x) + (mx + c)
...

So, R(x) = (a + b)x − ab
...
a
...
Let α be a common zero of f (x) and g(x)
...

Then since f (x) = g(x)q(x) + r(x) we have
f (α) =
=
=
=

g(α)q(α) + r(α)
(0)q(α) + r(α)
since g(α) = 0
r(α)
0
since f (α) = 0

But, from part b
...

Therefore, f (x) and g(x) do not have a common zero
...

3
...
i
...
x = −1, x = −2 and x = 4 are solutions of P (x) = 0
...


y
10
x
–2

0

2

4

6

–20

The graph of P (x) = x3 − x2 − 10x − 8
...
i
...


ii
...
x = −2 is a double root
...


y
–2

0

x
2

4

6

–20

–40

The graph of P (x) = x3 − x2 − 16x − 20
...
x = −2, x = −1 + 5 and x = −1 − 5 are solutions of P (x) = 0
...
i
...


y
x
0

–2

2

–10

The graph of√P (x) = x3 + 4x2 √
− 8
...

P (x) = x3 − x2 + x − 6 = (x − 2)(x2 + x + 3)
...

ii
...

iii
...
i
...

There is only one real zero at x = 2
...
i
...

ii
...

iii
...


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