graphs of functions 2

CHAPTER 2
GRAPHS OF FUNCTIONS II
2.1 GRAPHS OF FUNCTIONS
• The graph of a function is a set of points on
the Cartesian Plane that satisfy the function
• Information is presented in the form of graphs
• Graph are widely used in science and technology
• Graphs are very useful to researchers, scientists
and economist

The different type of functions and respective power of x
Type of
function
General form Example Highest
power of variable x
Linear
Quadratic
Cubic
Reciprocal
y = ax + c
y = ax2 + bx + c
y = ax3 + bx2 + cx + d
y = a
x
y = 3x
y = -4x + 5
y = 2x2
y = -3x2 + 2x
y = 2x2 + 5x + 1
y = 2x3
y = -3x3 + 5x
y = 2x3 - 3x + 6
y = 4
x
y = - 2
x
1
3
-1
2
y = ax3
y = ax3 + bx
y = ax3 + bx + c
a ≠ 0

• Using calculator to complete the tables
• Using the scale given to mark the points
on the x-axis and y-axis
• Plotting all the points using the scale given

CALC Memory
Example 1
Calculate the result
for Y = 3X – 5,
when X = 4, and when X = 6
)3
X
- 5
CALC
3X – 5
4 = 7
CALC 6 = 13
ALPHA

CALC Memory
Example 2
for Y = X2 + 3X – 12,
when X = 7, and when X = 8
) 3
X
x2
+
ALPHA
)
X
- 21
CALC
X2 + 3X – 12
7 = 58
CALC 8 = 76
ALPHA

CALC Memory
Example 3
for Y = 2X2 + X – 6,
when X = 3, and when X = -3
)
3
X
x2
+
ALPHA
)
X
- 6
CALC
2X2 + X – 6
3 = 15
CALC (-) = 9
ALPHA
2

CALC Memory
Example 4
for Y = -X3 + 2X + 5,
when X = 2, and when X = -1
)
1
X
x2
+ 2
ALPHA
)
X
+ 5
-X3 + 2X + 5 2 = 1
CALC (-) = 4
ALPHA
(-)
SHIFT x3
CALC

Example 5
for Y = 6 when X = -3,
X
and when X = 0.5
)
XALPHA
3CALC (-) = -2
6
CALC 0 . 125 =
ab/c 6┘x

Example 6
for Y = 6 when X = -3,
X
and when X = 0.5
)
XALPHA
3CALC (-) = -2
6
CALC 0 . 125 =
x-1
6x-1

Y = -2X2 + 40
X 0 0.5 1 1.5 2 3 3.5 4
Y
Y = X3 – 3X + 3
X -3 -2 -1 0 0.5 1 1.5 2
Y
Y = -16
X
X -4 -3 -2 -1 1 2 3 4
Y
40 39.5 38 35.5 32 22 15.5 8
-15 1 5 3 16.25 1.875 51
4 5.33 8 16 -16 -8 -5.33 -4

2(4)2 + 5(4) – 1 = 51
Using Calculator
2 ( )4 x2 + 5 ( 4 )
- 1 =

(-2)3 - 12(-2) + 10 = 26
Using Calculator
2( ) x2
+ 1 0
(-) 1
=
SHIFT - 2
2( )(-)
2( ) 3
+ 1 0
(-) 1
=
V
- 2
2( )(-)
OR

Using Calculator
6
(-3)
= -2
3( )6 ÷ (-) =

x -5 -4 -3 -2 -1 0 1 2 3
y 15 8 0 -1 0 3 15
y = x2 + 2x
y = (
-3
)2
+ 2 (
-3
) = 3
y = (
2
)2
+ 2 (
2
) = 8
3
8

x -4 -2.5 -1 -0.5 0.5 1 2 3.2 4
y 1 1.6 8 -8 -4 -1.25 -1
4
2
y = -4
( )
=
y = -4
( )
=
-1
4 -2-2
y = -4
x
Completing the table of values
-1 2

2.4 3.0 3.9
x-axis scale : 2 cm to 2 units
Marking the points on the x-axis and y-axis
2 4

-4 -2
-3.6 -3.0 -2.1
x-axis scale : 2 cm to 2 units

10
15
12
14.5
13.5
y-axis scale : 2 cm to 5 units
10.75

-20
-15
-18
-15.5
-16.5
y-axis scale : 2 cm to 5 units
-19.25

The x-coordinate and y-coordinate

x
A
B
x
x
x
C
D
(-3,2)
(2,0)
(4,-3)
(0,-4)
E
x
(4,4)
F
x
(-7,-2)

-5
-1 1
x
x
x
x
A
B
C
D
Ex (-0.5,2)
(-1,-3)
(0,3)
(0.3,1.5)
(0.5,-1.5)

-2
-1 1
x
x
x
x
A
B
C
D (-1,-1.2)
(0,1.2)
(0.3,0.6)
(0.2,-1)

-10
-1 1
x
x
x
x
A
B
C
D(-1,-6)
(0,5)
(0.3,3)
(0.5,-3)

2.1 A Drawing the Graphs
 Construct a table for a chosen range of x values, for example
-4 ≤ x ≤ 4
 Draw the x-axis and the y-axis and suitable scale for each axis
starting from the origin
 Plot the x and y values as coordinate pairs on the Cartesian Plane
 Join the points to form a straight line (using ruler) or smooth curve
(using French Curve/flexible ruler) with a sharp pencil
 Label the graphs
To draw the graph of a function, follow these step

2.1 A Drawing the Graphs
Draw the graph of y = 3x + 2
for -2 ≤ x ≤ 2
solution
x
y
-2 0
-4 2 8
0-2-4 2 4
-2
-4
2
4
6
8
x
y
GRAPH OF A LINEAR FUNCTION
8
3 + 2
22

Draw the graph of y = x2 + 2x for -5 ≤ x ≤ 3
solution
x -5 -4 -3 -2 -1 0 1 2 3
y 15 8 0 -1 0 3 15
y = x2 + 2x
3 83
+ 2
2
-3 -3

x
x
x
x
x
x
x
x
x
y
x0 1 2
2
4
6
8
10
-1-3-4-5
16
14
12
-2 3
y = x2 + 2x
GRAPH OF A QUADRATIC FUNCTION

Draw the graph of y = x3 - 12x + 3 for -4 ≤ x ≤ 4
solution
x -4 -3 -2 -1 0 1 2 3 4
y -13 12 19 14 3 -8 -13 -6 19
y = x3 - 12x + 3
-13
- 12
3
-4 -4 + 3

0 1 2 3 4-1-2-3-4
-5
-10
-15
5
10
15
20
25
y
x
x
x
x
x
x
x
x
x
x
GRAPH OF A CUBIC FUNCTION

Draw the graph of y = -4 for -4 ≤ x ≤ 4.
x
solution
y = -4
x
x -4 -2.5 -1 -0.5 0.5 1 2 3.2 4
y 1 1.6 4 8 -8 -4 -2 -1.25 -14
-4
-1

1 2 3 4-1-2-3-4 0
y
x
2
4
6
8
-2
-4
-6
-8
X
X
X
X
X
X
X
X
X
GRAPH OF A RECIPROCAL FUNCTION

x
x
x
x
x
x
x
x
x
y
x0 1 2
2
4
6
8
10
-1-3-4-5
16
14
12
-2 3-3.5
5
y = 11
-4.4 2.5
2.1 B Finding Values of Variable from a Graph
y = x2 + 2x
Find
(a)the value of
y when
x = -3.5
(b) the value of
x when
y = 11
solution
From the graph;
(a)y = 5
(b) X = -4.4, 2.5

x
x
x
x
x
x
x
x
x
y
x0 1 2
2
4
6
8
10
-1-3-4-5
16
14
12
-2 3-3.5
5
y = x2 + 2x

x
x
x
x
x
x
x
x
x
y
x0 1 2
2
4
6
8
10
-1-3-4-5
16
14
12
-2 3
5
11
y = x2 + 2x
-4.4 2.5

x
y
0 1 2 3 4
8
6
4
2
-2
-4
-6
-8
-4 -3 -2 -1
x
x
x
x
x
x
x
x
-2.2
-1.2
1.8
3.4
( a ) y = -2.2
( b ) x = -1.2
Find
(a)the value of
y when
x = 1.8
(b) the value of
x when
y = 3.4
solution
y = -4
x
Values obtained from
the graphs are
approximations
Notes

2.1 C Identifying the shape of a Graph from
a Given Function
LINEAR a
y
x
y = x
0
b
x
y = -x + 2
0
2
y

a Given Function
QUADRATIC a
y
x
y = x2
0
b
x
y = -x2
0
y

a Given Function
CUBIC a
y
x
y = x3
0
b
x
y = -x3 + 2
0
2
y

a Given Function
RECIPROCAL a
y
x
0
b
x
0
y
y = 1
x
y = -1
x

2.1 D Sketching Graphs of Function
• Sketching a graph means drawing a graph without
the actual data
• When we sketch the graph, we do not use
a graph paper, however we must know the important
characteristics of the graph such as its general form
(shape), the y-intercept and x-intercept
• It helps us to visualise the relationship of the variables

EXAMPLE y = 2x + 4
4
-2 0
y
x
 find the x-intercept of
y = 2x + 4.
Substitute y = 0
2x + 4 = 0
2x = -4
x = -2
Thus, x-intercept = -2
 find the y-intercept of
y = 2x + 4.
Substitute x = 0
y = 2(0) + 4
y = 4
Thus, y-intercept = 4
 draw a straight line that
passes x-intercept and y-intercept
y = 2x + 4
A Sketching The Graph of A Linear Function

B Sketching The Graph of A Quadratic Function
EXAMPLE y = -2x2 + 8
 a < 0
 the shape of the graph is
 y-intercept is 8
 find the x-intercept of
y = -2x2 + 8.
Substitute y = 0
-2x2 + 8 = 0
-2x2 = -8
x2 = 4
Thus, x-intercept = -2 and 2
x
0
y
-2 2
8

B Sketching The Graph of A Cubic Function
EXAMPLE y = -3x3 + 5
 a < 0
 the shape of the graph is
 y-intercept is 5
x
0
y
5

2.2 The Solution of An Equation By Graphical
Method
Solve the equation x2 = x + 2
Solution
x2 = x + 2
x2 - x – 2 = 0
(x– 2)(x + 1) = 0
x = 2, x = -1

2
y
1
3
4
0-1 1-2 2 x
y = x2
y = x + 2
A
B
• Let y = x + 2
and y = x2
• Draw both
graphs on
the same
axes
• Look at the
points of
intersection:
A and B.
Read the
values of the
coordinates
of x.
x = -1 and
x = 2
Solve the equation x2 = x + 2 by using the Graphical Method

Method
12. (a) Complete Table 1 for the equation y = x2 + 2x by writing down the values of y
when x = -3 and 2.
x -5 -4 -3 -2 -1 0 1 2 3
y 15 8 0 -1 0 3 15
(b) By using scale of 2 cm to 1 unit on the x-axis and 2 cm to 2 units on the y-axis,
draw the graph of y = x2 + 2x for -5 ≤ x ≤ 3.
(c) From your graph, find
(i) the value of y when x = -3.5,
(ii) the value of x when y = 11.
(d) Draw a suitable straight line on your graph to find a value of x which satisfies
the equation of x 2 + x – 4 = 0 for -5 ≤ x ≤ 3.
TABLE 1

Method
The graph y = x2 - 5x - 3 is drawn. Determine the suitable straight line
to be drawn to solve each of the following equations.
a) x2 - 5x - 3 = 4
b) x2 - 5x - 3 = 2x + 4
c) x2 - 5x - 2 = x + 4
d) x2 - 5x - 10 = 0
e) x2 - 7x - 2 = 0
EXAMPLE 1

solution
a) x2 - 5x - 3 = 4 x2 - 5x - 3 = y4
Therefore, y = 4 is the suitable straight line
b) x2 - 5x - 3 = 2x + 2 x2 - 5x - 3 = y2x + 2 y
Therefore, y = 2x + 2 is the suitable straight line
to be drawn to solve the equation: x2 - 5x - 3 = 4
y

solution
c) x2 - 5x - 2 = x + 4
- 1
-1 on both sides
Therefore, y = x + 3 is the suitable straight line
x2 - 5x - 2 = x + 4 - 1
x2 - 5x - 3 = x + 3x + 3
to be drawn to solve the equation: x2 - 5x – 2 = x + 4

solution
d) x2 - 5x - 10 = 0 Rearrange the equation
Therefore, y = 7 is the suitable straight line
x2 - 5x = 10
x2 - 5x = 10 - 3- 3
x2 - 5x - 3 = 77
-3 on both sides

solution
e) x2 - 7x - 2 = 0 Rearrange the equation
Therefore, y = 2x - 1 is the suitable straight line
x2 = 7x + 2
x2 = 7x + 2- 5x - 3- 5x - 3
x2 - 5x - 3 = 2x -12x -1
-5x - 3 on both sides

Alternative Method
Since a straight line is needed, we used to eliminate the term, x2.
The following method can be used
y = x2 - 5x - 3 1
0 = x2 - 7x - 2 2
1 - 2 y-0 = -5x - (-7x) - 3 - ( -2)
y = 2x - 1
e

Method
The graph y = 8 is drawn. Determine the suitable straight line
x
to be drawn to solve each of the following equations.
a) 4 = x + 1
x
b) -8 = -2x - 2
x
EXAMPLE 2

solution
4 = x + 1
x
Multiply both sides by 2a
We get 8 = 2x + 2
x
2x + 2
x
to be drawn to solve each the equation: 4 = x + 1
x

solution
-8 = -2x - 2
x
Multiply both sides by -1b
We get 8 = 2x + 2
x
2x + 2
x
to be drawn to solve each the equation: - 8 = -2x - 2
x

12. (a)
x -5 -4 -3 -2 -1 0 1 2 3
y 15 8 0 -1 0 3 15
y = x2 + 2x
y = (-3 )2
+ 2 (-3) = 3
y = ( 2 )2
+ 2 ( 2 ) = 8
8
solution
3

12. (a)
x
x
x
x
x
x
x
x
x
y
x0 1 2
2
4
6
8
10
-1-3-4-5
16
14
12
-2 3

12. (c)
x
x
x
x
x
x
x
x
x
y
x0 1 2
2
4
6
8
10
-1-3-4-5
16
14
12
-2 3-3.5
5
y = 11
-4.4 2.5
Answer:
(i) y = 5.0
(ii) x = -4.4
x = 2.5

12. (d)
x
x
x
x
x
x
x
x
x
y
x0 1 2
2
4
6
8
10
-1-3-4-5
16
14
12
-2 3
y = x2 + 2x + 0
0 = x2 + x - 4-
y = x + 4
x 0 -4
y 4 0
x
x 1.5-2.5
Answer:
(d) x = 1.5
x = -2.5

ax2 + bx + c = 0
x2 + x – 4 = 0
a = 1 b = 1 c = -4
MODE EQN
1 1
Unknowns ?
2 3
Degree?
2 3
2 a ? 1 = b ? 1 = c ?
(-) 4 x1 = 1.561552813 = x2 = -2.561552813
Press 3x
=

2.3 REGION REPRESENTING INEQUALITIES IN TWO
VARIABLES
2.3 A Determining Whether a Given Point Satisfies y = ax + b,
y > ax + b or y < ax + b
How can we determine whether a given point satisfies
y = 3x + 1, y < 3x + 1or y > 3x + 1 ?

VARIABLES
Let us consider the point (3,5). The point can only satisfies one of the
following relations:
(a) y = 3x + 1 (b) y < 3x + 1 (c) y > 3x + 1
y 3x + 1
5 3(3) + 1
5 10
=
<
>
<
Since the y-coordinate of the point (3,5) is 5, which is less than 10,
we conclude that y < 3x + 1 . Therefore, the point (3,5) satisfies the relation
y < 3x + 1
<

VARIABLES
Determine whether the following points satisfy y = 3x - 1, y < 3x - 1 or
y > 3x - 1.
(a) (1,-1) (b) (3,10) (c) (2,9)
EXAMPLE

VARIABLES
For point (1,-1)
When x = 1, y = 3(1) - 1 = 2
Since the y-coordinate of the point (1,-1) is -1, which is less than 2,
we conclude that y < 3x - 1 . Therefore, the point (1,-1) satisfies the relation
y < 3x - 1
solution a

2.3
REGION REPRESENTING INEQUALITIES IN TWO
VARIABLES
For point (3,10)
When x = 3, y = 3(3) - 1 = 8
Since the y-coordinate of the point (3,10) is 10, which is greater than 8,
we conclude that y > 3x - 1 . Therefore, the point (3,10) satisfies the relation
y > 3x - 1
solution b

VARIABLES
2.3 A
For point (-1,-4)
When x = -1, y = 3(-1) - 1 = -4
Since the y-coordinate of the point (-1,-4) is -4, which is equal to -4,
we conclude that y = 3x - 1 . Therefore, the point (-1,-4) satisfies the relation
y = 3x - 1
solution c
Determining Whether a Given Point Satisfies y = ax + b,

VARIABLES
2.3 B
Determining The Position of A Given Point Relative to
y = ax + b
 All the points satisfying y < ax + b are below the graph
 All the points satisfying y = ax + b are on the graph
 All the points satisfying y > ax + b are above the graph

VARIABLES
2.3 B
y = ax + b
-2-4
2
4
6
8
x
y
20
-2
-4
4-6
-8
6
P(4,8)
Q(4,2)
y < xy > x
The point P(4,8) lies
above the line y = x.
This region is represented
by y > x
The point Q(4,2) lies
below the line y = x.
by y < x
Q(4,4)
on the line y = x.
by y = x

VARIABLES
2.3 B
y = ax + b
-2-4
2
4
6
8
x
y
20
-2
-4
4-8
-8
8
P(-8,6)
Q(4,4)
y < 3x + 2y > 3x + 2
The point P(-8,6) lies
above the line y = 3x + 2.
by y > 3x + 2
below the line y = 3x + 2.
by y < 3x + 2
Q(2,8)

VARIABLES
2.3 C Identifying The Region Satisfying y > ax + b or y < ax + b
-2-4
2
4
6
8
x
y
20
-2
-4
4-8
-8
8
Determine whether the
shaded region in the graph
satisfies y < 3x + 2 or
y > 3x + 2
EXAMPLE
solution
The shaded region is
below the graph, y = 3x + 2.
Hence, this shaded region
satisfies y< 3x + 2

VARIABLES
2.3 D Shading The Regions Representing Given Inequalities
Symbol Type of
Line
< or > Dashed
Line
≤ or ≥ Solid line
The type of line to be drawn depends
on inequality symbol
The table above shows the
symbols of inequality and the
corresponding type of line
to be drawn
HoT TiPs
The dashed line indicates that all points
are not included in the region. The solid
line indicates that all points on the line
are included

VARIABLES
0
x
y
b
0
x
y
b
0
x
y
b
0
x
y
b
y > ax + b
a > 0
y < ax + b
a > 0
y ≥ ax + b
a > 0
y ≤ ax + b
a > 0

VARIABLES
0
x
y
by >ax + b
a < 0
0
x
y
b y ≥ ax + b
a < 0
x
y
x
y
y ≤ ax + b
a < 0
b
0
y < ax + b
a < 0
0
b

VARIABLES
0
y
a
x >a
a > 0
x
y
a
x > a
a < 0
x
y
x
y
x ≤ a
a < 0
x ≤ a
a > 0
0a
x
0
0 a

VARIABLES
2.3 E
Determine The Region which Satisfies Two or More
Simultaneous Linear Inequalities
y
x
2
0 2 3-3
1
EXAMPLE
Shade the region that satisfies
3y < 2x + 6, 2y ≥ -x + 2 and x ≤ 3.
X = 3

VARIABLES
y
x
2
0 2 3-3
1
X = 3
3y < 2x + 6, 2y ≥ -x + 2 and x ≤ 3.

VARIABLES
y
x
2
0 2 3-3
1
X = 3A
Region A satisfies 2y ≥ -x + 2, 3y < 2x + 6, and x ≤ 3

VARIABLES
2.3 E Determine The Region which Satisfies Two or More
y
x
2
0 2 3-3
1
X = 3A
Region A satisfies
2y ≥ -x + 2,
3y < 2x + 6, and x ≤ 3

VARIABLES
y
x
2
0 2 3-3
1
X = 3A
Region A satisfies
2y > -x + 2,
3y ≤ 2x + 6, and x < 3

VARIABLES
y
x
3
0
x = 3
Region B satisfies
y ≥ -x + 3,
y < x , and x ≤ 3
B

VARIABLES
y
x
3
0
x = 3
Region B satisfies
y > -x + 3,
y ≤ x , and x < 3
B

VARIABLES
y
x
3
0
x = -3
Region C satisfies
y > -x + 3,
y ≤ -2x , and x >-3
-3
C

x
x
x
x
x
x
x
y
x0 1 2
2
4
6
8
10
-1-3-4-5
16
14
12
-2 3
x 1.5-2.5
y = x2 + 2x
y = x + 4
x
x
x
Region A satisfies
y ≥ x2 + 2x,
y ≤ x + 4,
and x ≥ 0
A

x
x
x
x
x
x
x
y
x0 1 2
2
4
6
8
10
-1-3-4-5
16
14
12
-2 3
x 1.5-2.5
y = x2 + 2x
y = x + 4
x
x
x
Region A satisfies
y ≥ x2 + 2x,
y ≤ x < 4,
and x ≥ 0
A

y ≤ 2x + 8, y ≥ x, and y < 8
y
x0
y = 8
y = x
y = 2x + 8
SPM Clone
y ≤ 2x + 8
y ≥ x
y < 8

y ≤ 2x + 8, y ≥ x, and y < 8
y
x0
y = 8
y = x


K1
y = 2x + 8 3
SPM Clone
P2

3.
y
x0
y = 8
y = x

K2
y = 2x + 8
2

3.
y
x0
y = x

K1
y = 2x + 8
1

y
x
O
y = 2x6
On the graphs provided, shade the region which satisfies
the three inequalities x < 3, y ≤ 2x – 6 and y ≥ -6
[3 marks]
y = 6

y
x
O
y = 2x6
y = 6
Solution:
x = 3
K3
x-intercept = -(-6 ÷2) = 3
x < 3, y ≤ 2x – 6 and y ≥ -6

On the graphs provided, shade the region which satisfies
the three inequalities y ≤ x - 4, y ≤ -3x + 12 and y > -4
[3 marks]
y
x
y = 3x+12
O
y = x4

y
x
y = 3x+12
O
y = x4
y = 4
Solution:
K3
y-intercept =-4
y ≤ x - 4, y ≤ -3x + 12 and y > -4

SPM 2003 PAPER 2
REFER TO QUESTION
NO. 12

12. ( a )
x -4 -2.5 -1 -0.5 0.5 1 2 3.2 4
y 1 1.6 8 -8 -4 -1.25 -1
4
2
  K1K1
y = -4
( )
=
y = -4
( )
=
-1
4 -2-2

1 2 3 4-1-2-3-4 0
y
x
2
4
6
8
-2
-4
-6
-8
X
X
X
X
X
X
X
X
X

K1
 


K1N1


 
K1N1

12(b)

12. ( a )
x -4 -2.5 -1 -0.5 0.5 1 2 3.2 4
y 1 1.6 8 -8 -4 -1.25 -14 2
 K1K0

1 2 3 4-1-2-3-4 0
y
x
2
4
6
8
-2
-4
-6
-8
X
X
X
X
X
X
X
X
X

K1
 


K1N1



K1 N0

12(b)

x
y
0 1 2 3 4
8
6
4
2
-2
-4
-6
-8
-4 -3 -2 -1
x
x
x
x
x
x
x
12. ( c )
x
-2.2
-1.2
1.8
3.4
( i ) y = -2.2
( ii ) x = -1.2
 P1
 P1

12. (d)
x
y
0 1 2 3 4
8
6
4
2
-2
-4
-6
-8
-4 -3 -2 -1
x
x
x
x
x
x
x
x
y = -2x - 3
-2.4
0.8
 K1K1
x = - 2.4
x = 0.8

 N1
4 = 2x + 3
x
- 4 = -2x - 3
x

graphs of functions 2

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