2.0 graphs of functions 2
TRANSCRIPT
-
8/3/2019 2.0 Graphs of Functions 2
1/123
-
8/3/2019 2.0 Graphs of Functions 2
2/123
-
8/3/2019 2.0 Graphs of Functions 2
3/123
-
8/3/2019 2.0 Graphs of Functions 2
4/123
CHAPTER 2GRAPHS 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
-
8/3/2019 2.0 Graphs of Functions 2
5/123
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 + 2xy = 2x2 + 5x + 1
y = 2x3
y = -3x3 + 5x
y = 2x
3
- 3x + 6
y = 4
x
y = - 2
x
1
3
-1
2
y = ax3
y = ax3 + bxy = ax3 + bx + c
a 0
-
8/3/2019 2.0 Graphs of Functions 2
6/123
LINEAR FUNCTION
-
8/3/2019 2.0 Graphs of Functions 2
7/123
LINEAR FUNCTION
-
8/3/2019 2.0 Graphs of Functions 2
8/123
QUADRATIC FUNCTION
-
8/3/2019 2.0 Graphs of Functions 2
9/123
QUADRATIC FUNCTION
-
8/3/2019 2.0 Graphs of Functions 2
10/123
QUBIC FUNCTION
-
8/3/2019 2.0 Graphs of Functions 2
11/123
QUBIC FUNCTION
-
8/3/2019 2.0 Graphs of Functions 2
12/123
RECIPROCAL FUNCTION
-
8/3/2019 2.0 Graphs of Functions 2
13/123
RECIPROCAL FUNCTION
-
8/3/2019 2.0 Graphs of Functions 2
14/123
Using calculator to complete the tables
Using the scale given to mark the pointson the x-axis and y-axis
Plotting all the points using the scale given
-
8/3/2019 2.0 Graphs of Functions 2
15/123
COMP
-
8/3/2019 2.0 Graphs of Functions 2
16/123
CALC MemoryCALC MemoryExample 1Example 1Calculate the resultforY = 3X 5,
when X = 4, and when X = 6
)3X
-5
CALC
3X 5
4 = 7CALC 6 = 13
ALPHA
-
8/3/2019 2.0 Graphs of Functions 2
17/123
CALC MemoryCALC MemoryExample 2Example 2Calculate the resultforY = X2 + 3X 12,when X = 7, and when X = 8
) 3X
x2 +ALPHA
)X
- 21CALC X2 + 3X 12
7 = 58CALC
8 = 76
ALPHA
-
8/3/2019 2.0 Graphs of Functions 2
18/123
CALC MemoryCALC MemoryExample 3Example 3Calculate the resultforY = 2X2 + X 6,when X = 3, and when X = -3
)
3
X
x2 +ALPHA
)X
- 6CALC 2X2
+ X 63 = 15
CALC (-) =9
ALPHA
2
-
8/3/2019 2.0 Graphs of Functions 2
19/123
CALC MemoryCALC MemoryExample 4Example 4Calculate the resultforY = -X3 + 2X + 5,when X = 2, and when X = -1
)
1
X
x2
+ 2
ALPHA
)
X
+ 5-X3 + 2X + 5 2 = 1CALC
(-) = 4
ALPHA
(-)SHIFT x3
CALC
-
8/3/2019 2.0 Graphs of Functions 2
20/123
Example 5Example 5Calculate the resultforY = 6 when X = -3,
X
and when X = 0.5
)XALPHA
3CALC (-) = -2
6
CALC 0 . 125 =
ab/c 6x
-
8/3/2019 2.0 Graphs of Functions 2
21/123
Example 6Example 6Calculate the resultforY = 6 when X = -3,
X
and when X = 0.5
)XALPHA
3CALC (-) = -2
6
CALC 0 . 125 =
x-1 6x-1
-
8/3/2019 2.0 Graphs of Functions 2
22/123
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 2Y
Y = -16XX -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
-
8/3/2019 2.0 Graphs of Functions 2
23/123
2(4)2 + 5(4) 1 = 51Using Calculator
2 ( )4 x2 + 5 ( 4 )
- 1 =
-
8/3/2019 2.0 Graphs of Functions 2
24/123
(-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
-
8/3/2019 2.0 Graphs of Functions 2
25/123
Using Calculator6
(-3)= -2
3( )6 (-) =
-
8/3/2019 2.0 Graphs of Functions 2
26/123
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
-
8/3/2019 2.0 Graphs of Functions 2
27/123
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
-
8/3/2019 2.0 Graphs of Functions 2
28/123
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
-
8/3/2019 2.0 Graphs of Functions 2
29/123
-4 -2
-3.6 -3.0 -2.1
x-axis scale : 2 cm to 2 units
-
8/3/2019 2.0 Graphs of Functions 2
30/123
10
15
12
14.5
13.5
y-axis scale : 2 cm to 5 units
10.75
-
8/3/2019 2.0 Graphs of Functions 2
31/123
-20
-15
-18
-15.5
-16.5
y-axis scale : 2 cm to 5 units
-19.25
-
8/3/2019 2.0 Graphs of Functions 2
32/123
The x-coordinate and y-coordinate
-
8/3/2019 2.0 Graphs of Functions 2
33/123
xA
B
x
x
x
C
D
(-3,2)
(2,0)
(4,-3)
(0,-4)
Ex
(4,4)
F
x
(-7,-2)
-
8/3/2019 2.0 Graphs of Functions 2
34/123
-5
-1 1
x
x
xx
A
B
C
D
Ex (-0.5,2)
(-1,-3)
(0,3)
(0.3,1.5)
(0.5,-1.5)
-
8/3/2019 2.0 Graphs of Functions 2
35/123
-2
-1 1
x x
xx
A
B
CD (-1,-1.2)
(0,1.2)
(0.3,0.6)
(0.2,-1)
-
8/3/2019 2.0 Graphs of Functions 2
36/123
-10
-1 1
x
x
xx
A
B
C
D(-1,-6)
(0,5)
(0.3,3)
(0.5,-3)
-
8/3/2019 2.0 Graphs of Functions 2
37/123
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 steps;
-
8/3/2019 2.0 Graphs of Functions 2
38/123
2.1 A Drawing the Graphs
Draw the graph of y = 3x + 2for -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
-
8/3/2019 2.0 Graphs of Functions 2
39/123
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
+ 22
-3 -3
y
-
8/3/2019 2.0 Graphs of Functions 2
40/123
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
-
8/3/2019 2.0 Graphs of Functions 2
41/123
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
- 123
-4 -4 + 3
-
8/3/2019 2.0 Graphs of Functions 2
42/123
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
xx
GRAPH OF A CUBIC FUNCTION
-
8/3/2019 2.0 Graphs of Functions 2
43/123
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
-
8/3/2019 2.0 Graphs of Functions 2
44/123
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
-
8/3/2019 2.0 Graphs of Functions 2
45/123
xx
USING FRENCH CURVE
-
8/3/2019 2.0 Graphs of Functions 2
46/123
x
x
x
x
USING FRENCH CURVE
-
8/3/2019 2.0 Graphs of Functions 2
47/123
x
x
x
USING FRENCH CURVE
-
8/3/2019 2.0 Graphs of Functions 2
48/123
-10
-15
-20
-
8/3/2019 2.0 Graphs of Functions 2
49/123
-
8/3/2019 2.0 Graphs of Functions 2
50/123
-
8/3/2019 2.0 Graphs of Functions 2
51/123
-
8/3/2019 2.0 Graphs of Functions 2
52/123
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 whenx = -3.5
(b) the value of
x when
y = 11
solution
From the graph;
(a)y = 5
(b) X = -4.4, 2.5
-
8/3/2019 2.0 Graphs of Functions 2
53/123
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
2.1 B Finding Values of Variable from a Graph
y = x2 + 2x
-
8/3/2019 2.0 Graphs of Functions 2
54/123
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
2.1 B Finding Values of Variable from a Graph
y = x2 + 2x-4.4 2.5
-
8/3/2019 2.0 Graphs of Functions 2
55/123
x =1.5
y
-
8/3/2019 2.0 Graphs of Functions 2
56/123
x
y
0 1 2 3 4
8
6
4
2
-2
-4
-6
-8
-4 -3 -2 -1
xx
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 fromthe graphs are
approximations
Notes
Id tif i th h f G h f
-
8/3/2019 2.0 Graphs of Functions 2
57/123
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
Id tif i g th h f G h f
-
8/3/2019 2.0 Graphs of Functions 2
58/123
2.1 C Identifying the shape of a Graph from
a Given Function
QUADRATIC a
y
x
y = x2
0
bx
y = -x2
0
y
Id tif i g th h f G h f
-
8/3/2019 2.0 Graphs of Functions 2
59/123
2.1 C Identifying the shape of a Graph from
a Given Function
CUBIC a
y
x
y = x3
0
b
x
y = -x3 + 2
0
2
y
Id tif i g th h f G h f
-
8/3/2019 2.0 Graphs of Functions 2
60/123
2.1 C Identifying the shape of a Graph from
a Given Function
RECIPROCAL a
y
x0
b
x0
y
y = 1x
y = -1
x
-
8/3/2019 2.0 Graphs of Functions 2
61/123
2.1 D Sketching Graphs ofFunction
Sketching a graph means drawing a graph withoutthe 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
-
8/3/2019 2.0 Graphs of Functions 2
62/123
EXAMPLE y = 2x + 4
4
-2 0
y
x
find the x-intercept ofy = 2x + 4.
Substitute y = 0
2x + 4 = 0
2x = -4
x = -2Thus, x-intercept = -2
find the y-intercept of
y = 2x + 4.
Substitute x = 0
y = 2(0) + 4y = 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
-
8/3/2019 2.0 Graphs of Functions 2
63/123
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 = 4Thus, x-intercept = -2 and 2
x0
y
-2 2
8
-
8/3/2019 2.0 Graphs of Functions 2
64/123
B Sketching The Graph of A Cubic Function
EXAMPLE y = -3x3 + 5
a < 0
the shape of the graph is
y-intercept is 5
x0
y
5
-
8/3/2019 2.0 Graphs of Functions 2
65/123
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
-
8/3/2019 2.0 Graphs of Functions 2
66/123
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 thepoints 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
The Solution of An Equation By Graphical
-
8/3/2019 2.0 Graphs of Functions 2
67/123
2.2 The Solution of An Equation By Graphical
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
2 2 The Solution of An Equation By Graphical
-
8/3/2019 2.0 Graphs of Functions 2
68/123
2.2 The Solution of An Equation By Graphical
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
-
8/3/2019 2.0 Graphs of Functions 2
69/123
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
The graph y = x2 - 5x - 3 is drawn. Determine the suitable straight line
to be drawn to solve the equation: x2 - 5x - 3 = 4
y
-
8/3/2019 2.0 Graphs of Functions 2
70/123
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
The graph y = x2 - 5x - 3 is drawn. Determine the suitable straight line
to be drawn to solve the equation: x2 - 5x 2 = x + 4
-
8/3/2019 2.0 Graphs of Functions 2
71/123
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
The graph y = x2 - 5x - 3 is drawn. Determine the suitable straight line
to be drawn to solve the equation: x2 - 5x - 10 = 0
-
8/3/2019 2.0 Graphs of Functions 2
72/123
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
The graph y = x2 - 5x - 3 is drawn. Determine the suitable straight line
to be drawn to solve the equation: x2 - 7x - 2 = 0
-
8/3/2019 2.0 Graphs of Functions 2
73/123
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
The graph y = x2 - 5x - 3 is drawn. Determine the suitable straight line
to be drawn to solve the equation: x2 - 7x - 2 = 0
2 2 The Solution of An Equation By Graphical
-
8/3/2019 2.0 Graphs of Functions 2
74/123
2.2 The Solution of An Equation By Graphical
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 - 2x
EXAMPLE 2
-
8/3/2019 2.0 Graphs of Functions 2
75/123
solution
4 = x + 1xMultiply both sides by 2a
We get 8 = 2x + 2
x
Therefore, y = 2x + 2 is the suitable straight line
2x + 2
The graph y = 8 is drawn. Determine the suitable straight line
x
to be drawn to solve each the equation: 4 = x + 1
x
-
8/3/2019 2.0 Graphs of Functions 2
76/123
solution
-8 = -2x - 2
x
Multiply both sides by -1
b
We get 8 = 2x + 2
x
Therefore, y = 2x + 2 is the suitable straight line
2x + 2
The graph y = 8 is drawn. Determine the suitable straight line
x
to be drawn to solve each the equation: - 8 = -2x - 2
x
2 2 The Solution of An Equation By Graphical
-
8/3/2019 2.0 Graphs of Functions 2
77/123
2.2 The Solution of An Equation By Graphical
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
-
8/3/2019 2.0 Graphs of Functions 2
78/123
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) y
-
8/3/2019 2.0 Graphs of Functions 2
79/123
12. (a)
x
x
x
xx
x
x
x
x
x0 1 2
2
4
6
8
10
-1-3-4-5
16
14
12
-2 3
12. (c)
-
8/3/2019 2.0 Graphs of Functions 2
80/123
( )
x
x
x
xx
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)
-
8/3/2019 2.0 Graphs of Functions 2
81/123
( )
x
x
x
xx
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 + 00 = x2+ x - 4-
y = x +4
x 0 -4
y 4 0
x
x1.5-2.5
Answer:
(d) x = 1.5
x = -2.5
-
8/3/2019 2.0 Graphs of Functions 2
82/123
a 2 + b + c 0
-
8/3/2019 2.0 Graphs of Functions 2
83/123
ax2 + bx + c = 0
x2+ x 4 = 0
a = 1 b = 1 c = -4
MODE EQN
1 1Unknowns ?
2 3Degree?
2 3
2 a ? 1 = b ? 1 = c ?
(-) 4 x1 = 1.561552813 = x2 = -2.561552813
Press 3x
=
-
8/3/2019 2.0 Graphs of Functions 2
84/123
2.3 REGIONREPRESENTING INEQUALITIESIN 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 satisfiesy = 3x + 1, y < 3x + 1or y > 3x + 1 ?
2 3 REGION REPRESENTING INEQUALITIES IN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
85/123
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
2.3 A Determining Whether a Given Point Satisfies y = ax + b,
y > ax + b or y < ax + b
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
=
3x - 1.
(a) (1,-1) (b) (3,10) (c) (2,9)
EXAMPLE
2 3 REGION REPRESENTING INEQUALITIES IN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
87/123
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
2.3 A Determining Whether a Given Point Satisfies y = ax + b,
y > ax + b or y < ax + b
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 relationy < 3x - 1
solution a
REGION REPRESENTING INEQUALITIES IN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
88/123
2.3REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
2.3 A Determining Whether a Given Point Satisfies y = ax + b,
y > ax + b or y < ax + b
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 relationy > 3x - 1
solution b
-
8/3/2019 2.0 Graphs of Functions 2
89/123
2.3 REGIONREPRESENTING INEQUALITIESIN TWO 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 relationy = 3x - 1
solution c
Determining Whether a Given Point Satisfies y = ax + b,
y > ax + b or y < ax + b
-
8/3/2019 2.0 Graphs of Functions 2
90/123
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
2.3 BDetermining 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
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
91/123
Q
2.3 BDetermining The Position of A Given Point Relative to
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) liesabove the line y = x.
This region is represented
by y > x
The point Q(4,2) liesbelow the line y = x.
This region is represented
by y < x
Q(4,4)
The point Q(4,4) lies
on the line y = x.This region is represented
by y = x
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
92/123
2.3 BDetermining The Position of A Given Point Relative to
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.
This region is representedby y > 3x + 2
The point Q(4,4) lies
below the line y = 3x + 2.
This region is represented
by y < 3x + 2
Q(2,8)
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
93/123
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 ory > 3x + 2
EXAMPLE
solution
The shaded region is
below the graph, y = 3x + 2.
Hence, this shaded region
satisfies y< 3x + 2
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
94/123
2.3 D Shading The Regions Representing Given Inequalities
Symbol Type ofLine
< or > Dashed
Line
or Solid line
The type of line to be drawn depends
on inequality symbol
The table above shows thesymbols of inequality and thecorresponding type of line
to be drawn
HoT TiPsThe dashed line indicates that all points
are not included in the region. The solid
line indicates that all points on the line
are included
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
95/123
2.3 D Shading The Regions Representing Given Inequalities
0
x
y
b
0x
y
b
0x
y
b
0x
y
b
y > ax + b
a > 0y < ax + b
a > 0
y ax + b
a > 0y ax + b
a > 0
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
96/123
2.3 D Shading The Regions Representing Given Inequalities
0x
y
by >ax + b
a < 0
0x
y
b y ax + b
a < 0
x
y
x
y
y ax + b
a < 0b
0
y < ax + b
a < 0
0
b
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
97/123
2.3 D Shading The Regions Representing Given Inequalities
0
y
a
x >a
a > 0
x
y
a
x > a
a < 0
x
y
x
y
x a
a < 0x a
a > 0
0a
x0
0 a
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
98/123
2.3 EDetermine The Region which Satisfies Two or More
Simultaneous Linear Inequalities
y
x
2
0 2 3-3
1
EXAMPLEShade the region that satisfies
3y < 2x + 6, 2y -x + 2 and x 3.
X = 3
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
99/123
y
x
2
0 2 3-3
1
X = 3
Shade the region that satisfies
3y < 2x + 6, 2y -x + 2 and x 3.
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
100/123
y
x
2
0 2 3-3
1
X = 3A
Region A satisfies 2y -x + 2, 3y < 2x + 6, and x 3
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
101/123
2.3 E Determine The Region which Satisfies Two or More
Simultaneous Linear Inequalities
y
x
2
0 2 3-3
1
X = 3A
Region A satisfies2y -x + 2,
3y < 2x + 6, and x 3
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
102/123
2.3 E Determine The Region which Satisfies Two or More
Simultaneous Linear Inequalities
y
x
2
0 2 3-3
1
X = 3A
Region A satisfies
2y > -x + 2,3y 2x + 6, and x < 3
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
103/123
2.3 E Determine The Region which Satisfies Two or More
Simultaneous Linear Inequalities
y
x
3
0
x = 3
Region B satisfies
y -x + 3,
y < x , and x 3
B
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
104/123
2.3 E Determine The Region which Satisfies Two or More
Simultaneous Linear Inequalities
y
x
3
0
x = 3
Region B satisfies
y > -x + 3,
y x , and x < 3
B
2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES
-
8/3/2019 2.0 Graphs of Functions 2
105/123
2.3 E Determine The Region which Satisfies Two or More
Simultaneous Linear Inequalities
y
x
3
0
x = -3Region C satisfies
y > -x + 3,
y -2x , and x >-3
-3
C
y
Region A satisfies
-
8/3/2019 2.0 Graphs of Functions 2
106/123
x
x
x
xx
x
x
x0 1 2
2
4
6
8
10
-1-3-4-5
16
14
12
-2 3x
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
Region A satisfies
-
8/3/2019 2.0 Graphs of Functions 2
107/123
x
x
x
xx
x
x
x0 1 2
2
4
6
8
10
-1-3-4-5
16
14
12
-2 3x
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
ySPM Clone
-
8/3/2019 2.0 Graphs of Functions 2
108/123
Shade the region that satisfies
y 2x +8, y x, and y < 8
x0
y = 8
y = x
y = 2x +8
y 2x + 8
y x
y < 8
ySPM Clone
-
8/3/2019 2.0 Graphs of Functions 2
109/123
Shade the region that satisfies
y 2x +8, y x, and y < 8
x0
y = 8
y = x
K1
y = 2x +8 3
SPM Clone
P2
3y
-
8/3/2019 2.0 Graphs of Functions 2
110/123
3.
x0
y = 8
y = x
K2
y = 2x +82
-
8/3/2019 2.0 Graphs of Functions 2
111/123
-
8/3/2019 2.0 Graphs of Functions 2
112/123
-
8/3/2019 2.0 Graphs of Functions 2
113/123
y
xO
y = 2x6
y = 6
Solution:
x = 3
K3
x-intercept = -(-6 2) = 3
x < 3, y 2x 6 and y -6
-
8/3/2019 2.0 Graphs of Functions 2
114/123
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 = x4
-
8/3/2019 2.0 Graphs of Functions 2
115/123
y
x
y = 3x+12
O
y = x4
y = 4
Solution:
K3
y-intercept =-4
y x - 4, y -3x + 12 and y > -4
SPM 2003 PAPER 2
-
8/3/2019 2.0 Graphs of Functions 2
116/123
SPM 2003 PAPER 2
REFER TO QUESTIONNO. 12
12 ( a )
-
8/3/2019 2.0 Graphs of Functions 2
117/123
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
-
8/3/2019 2.0 Graphs of Functions 2
118/123
12. ( a )
-
8/3/2019 2.0 Graphs of Functions 2
119/123
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
y
8X 12(b)
-
8/3/2019 2.0 Graphs of Functions 2
120/123
1 2 3 4-1-2-3-4 0 x
2
4
6
8
-2
-4
-6
-8
X
X
XX
X
X
X
X
X
K1
K1N1
K1 N0
( )
y
8x12.( c )
-
8/3/2019 2.0 Graphs of Functions 2
121/123
x0 1 2 3 4
6
4
2
-2
-4
-6
-8
-4 -3 -2 -1
xx
x
x
x
x
x
-2.2
-1.2
1.8
3.4
( i ) y = -2.2
( ii ) x = -1.2
P1
P1
y
8x
4 = 2x +3
x
-
8/3/2019 2.0 Graphs of Functions 2
122/123
12.(d)
x0 1 2 3 4
6
4
2
-2
-4
-6
-8
-4 -3 -2 -1
xx
x
x
x
x
x
y = -2x - 3
-2.4
0.8
K1K1
x = - 2.4
x = 0.8
N1
x
- 4 = -2x - 3
x
-
8/3/2019 2.0 Graphs of Functions 2
123/123