Graph functions

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GRAPHS
OF
FUNCTIONS II

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2.12.1 Graphs of functions.Graphs of functions.
a)a) Graphs of linear functionsGraphs of linear functions
Scale :Scale : x-axis 2 cm to 1 unitx-axis 2 cm to 1 unit
y-axis 2 cm to 1 unity-axis 2 cm to 1 unit
xx 00 11 22 33 44
yy 33 55 77 99 1111
x
0 1 2 3 4 5 6
1
2
3
4
5
6
7
9
8
10
11
y = 2x + 3y = 2x + 3
y
Example 1Example 1
Draw the graph of the functionDraw the graph of the function
y = 2x + 3 for 0y = 2x + 3 for 0 ≤≤ xx ≤≤ 44
By plotting the ordered pairs
from the table, the graph as
drawn on the right is obtained.

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Draw the graph of theDraw the graph of the
functionfunction y = 2 – 3x for -1y = 2 – 3x for -1 ≤≤ xx
≤≤ 44
xx -1-1 00 11 22 33 44
yy 55 22 -1-1 -4-4 -7-7
x
0 1 2 3 4 5-1
-12
-10
-8
-6
-4
-2
2
4
6
y = 2 – 3xy = 2 – 3x
y
-10-10
Try this

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b)b) Graphs of quadratic functionsGraphs of quadratic functions
Example 2Example 2
y = 2xy = 2x22
+ 6x + 2 for -4+ 6x + 2 for -4 ≤≤ xx ≤≤ 22
x
-3 -2 -1 0 1 2-4
2
4
6
8
10
12
14
16
18
y = 2xy = 2x22
+ 6x + 2+ 6x + 2
y
xx -4-4 -3-3 -2-2 -1-1 00 11 22
yy 1010 22 -2-2 -2-2 22 1010 2222
20
22
-2

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x
-1 0 1 2 3 4-2
-45
-40
-35
-30
-25
-20
-15
-10
-5
y = 4x - 4xy = 4x - 4x22
y
xx -2-2 -1-1 00 11 22 33 44
yy -24-24 -8-8 00 00 -8-8 -24-24 - 48- 48
5
-50
Example 3Example 3
y = 4x - 4x2 for -2y = 4x - 4x2 for -2 ≤≤ xx ≤≤ 44

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c)c) Graphs of cubic functionsGraphs of cubic functions
xx -3-3 -2-2 -1-1 00 11 22 33
yy -54-54 -16-16 -2-2 00 22 1616 5454
x
-1 0 1 2 3 4-2
-40
-30
-20
-10
10
20
30
40
50
y = 2xy = 2x33
y
60
-50
-3
-60
Example 4Example 4
y = 2xy = 2x33
for -3for -3 ≤≤ xx ≤≤ 33

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xx -3-3 -2-2 -1-1 00 11 22 33
yy 5858 2020 66 44 22 -12-12 -50-50
x
-1 0 1 2 3 4-2
-40
-30
-20
-10
10
20
30
40
50
y = 4 - 2xy = 4 - 2x33
y
60
-50
-3
-60
Example 5Example 5
y = 4 - 2xy = 4 - 2x33
for -3for -3 ≤≤ xx ≤≤ 33

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d)d) Graphs of reciprocal functionsGraphs of reciprocal functions
xx -5-5 -4-4 -2-2 -1-1 -0.5-0.5
yy -0.4-0.4 -0.5-0.5 -1-1 -2-2 -4-4
x-1 0 1 2 3 4-2
-4
-3
-2
-1
1
2
3
4
5
y
-5
-3
xx 0.50.5 11 22 33 44
yy 44 22 11 0.670.67 0.50.5
-4-5
x
2
y =
x
2
y =
Example 6Example 6
for -5for -5 ≤≤ xx ≤≤ 44

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xx -5-5 -4-4 -2-2 -1-1 -0.5-0.5
yy 1.21.2 1.51.5 33 66 1212
x-1 0 1 2 3 4-2
-8
-6
-4
-2
2
4
6
8
10
y
-10
-3
xx 0.50.5 11 22 33 44
yy -12-12 -6-6 -3-3 -2-2 -1.5-1.5 -4-5
x
6
y −=
12
-12
x
6
y −=
Try thisTry this
for -5for -5 ≤≤ xx ≤≤ 44

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Example 7Example 7
a)a) Complete the table of valuesComplete the table of values
below an draw the graph forbelow an draw the graph for
y = xy = x33
– 3x - 2– 3x - 2
xx -2-2 -1-1 00 11 22
yy - 4- 4 00 -2-2 - 4- 4 00
x
-1 0 1 2-2
-4
-3
-2
-1
y = xy = x33
– 3x - 2– 3x - 2
y
b)b) From the graph, find.From the graph, find.
i)i) The value of y whenThe value of y when
a) x = 1.5a) x = 1.5 b) x = -0.5b) x = -0.5
ii)ii) The value of x whenThe value of x when
a) y = 0a) y = 0 b) y = -1.5b) y = -1.5
B Find from the graphs, the values of y given the values of x and vice versa

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x
-1 0 1 2-2
-4
-3
-2
-1
y = xy = x33
– 3x - 2– 3x - 2
y
When x = 1.5, y = -3.15When x = 1.5, y = -3.15
x=1.5x=1.5
x=-0.5x=-0.5
When x = -0.5, y = -0.65When x = -0.5, y = -0.65
When y = 0, x = -1, dan 2When y = 0, x = -1, dan 2
y = -1.5y = -1.5
When y = -1.5,When y = -1.5,
x = -1.65x = -1.65
x = -0.20x = -0.20
x = 1.8x = 1.8
b)b) From the graph, find.From the graph, find.
i)i) The value of y whenThe value of y when
a) x = 1.5a) x = 1.5 b) x = -0.5b) x = -0.5
ii)ii) The value of x whenThe value of x when
a) y = 0a) y = 0 b) y = -1.5b) y = -1.5

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1. Graphs of linear functions1. Graphs of linear functions
y = ax + c , a and c are constants, ay = ax + c , a and c are constants, a ≠≠ 0.0.
Shape : straight lineShape : straight line
a > 0 ( positive )a > 0 ( positive ) a < 0 ( negative )a < 0 ( negative )
x
y
x
y
C Shapes of Graphs

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2.2. Graphs of quadratic functions.Graphs of quadratic functions.
y = axy = ax 22
+ bx + c , a, b, and c are constants, a+ bx + c , a, b, and c are constants, a ≠≠ 0.0.
Shape : parabolaShape : parabola
a > 0 ( positive)a > 0 ( positive) a < 0 ( negative )a < 0 ( negative )
x
y
x
y

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3.3. Graphs of cubic functionsGraphs of cubic functions
y = axy = ax 33
, a is a constant and a, a is a constant and a ≠≠ 0.0.
Shape : S-shapeShape : S-shape
x
y
x
y

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4.4. Graphs of reciprocal functionsGraphs of reciprocal functions
y = a /x , a is a constant and ay = a /x , a is a constant and a ≠≠ 0.0.
Shape : HyperbolaShape : Hyperbola
x
y
x
y

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a)a) To sketch the graphs of a linear functions.To sketch the graphs of a linear functions.
DD Sketch the graph of a given functionSketch the graph of a given function
To draw a graph of y = ax + c with a is the gradient and c is the y-To draw a graph of y = ax + c with a is the gradient and c is the y-
intercept,intercept,
i)i) Look for two suitable pointsLook for two suitable points
ii)ii) Then plot the two points, andThen plot the two points, and
iii)iii) Joint the two points to get the sketch of the graph.Joint the two points to get the sketch of the graph.
Example 8Example 8
Sketch the graph for y = 2x - 4Sketch the graph for y = 2x - 4
Solution :Solution :
When x = 0,When x = 0, y = 2(0) – 4 = - 4y = 2(0) – 4 = - 4
When y = 0,When y = 0, 0 = 2x – 40 = 2x – 4
x = 2x = 2
xx
yy
22
- 4- 4
Thus, the two points are (0, -4) and (2, 0)Thus, the two points are (0, -4) and (2, 0)

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b)b) To sketch the graphs of quadratic functionsTo sketch the graphs of quadratic functions
To sketch the graph of y = axTo sketch the graph of y = ax22
+ bx + c, a+ bx + c, a ≠≠ 0.0.
i)i) Determine the shape of the graphDetermine the shape of the graph
ii)ii) Mark the y-intercept or/and the x-interceptMark the y-intercept or/and the x-intercept
iii)iii) Determine the axis of symmetryDetermine the axis of symmetry
(a > 0 :(a > 0 : ∪∪ - shape, a < 0 :- shape, a < 0 : ∩∩ - shape)- shape)
iv)iv) Determine the minimum or maximum pointDetermine the minimum or maximum point

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Case 1 : Function of type y = axCase 1 : Function of type y = ax22
+ c, b = 0.+ c, b = 0.
Features of the graph :Features of the graph :
α)α) ∪∪ - shaped- shaped if a > 0 andif a > 0 and ∩∩ - shaped- shaped if a < 0 :a < 0 :
b)b) The y-intercept is cThe y-intercept is c
c)c) The x-intercept is obtained by substituting y = 0 to solve theThe x-intercept is obtained by substituting y = 0 to solve the
equation axequation ax22
+ c = 0.+ c = 0.
d)d) The axis of symmetry is y-axis.The axis of symmetry is y-axis.
e)e) By substituting x = 0, the coordinate of the minimum orBy substituting x = 0, the coordinate of the minimum or
maximum point are (0, c)maximum point are (0, c)

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Example 9Example 9
Sketch the graph of the function y = 2xSketch the graph of the function y = 2x22
- 8- 8
i)i) a = 2 > 0, the graph isa = 2 > 0, the graph is ∪∪ - shaped.- shaped.
ii)ii) c = -8, then the y-intercept is -8c = -8, then the y-intercept is -8
iii)iii) When y = 0, 2xWhen y = 0, 2x22
– 8 = 0– 8 = 0
x =x = √√ 4 = -2 or 24 = -2 or 2
Hence, x-intercept are -2 and 2.Hence, x-intercept are -2 and 2.
xx
yy
00
iv)iv) The axis of symmetry is the y-axisThe axis of symmetry is the y-axis
22-2-2
-8-8
v)v) When x = 0, y = -8.When x = 0, y = -8.
Hence, the minimum point is (0, -8)Hence, the minimum point is (0, -8)

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11-1-1
33
xx
yy
00
Example 10Example 10
Sketch the graph of the function y = -3xSketch the graph of the function y = -3x22
+ 3+ 3
i)i) a = -3 < 0, the graph isa = -3 < 0, the graph is ∩∩ - shaped.- shaped.
ii)ii) c = 3, then the y-intercept is 3c = 3, then the y-intercept is 3
iii)iii) When y = 0, -3xWhen y = 0, -3x22
+ 3 = 0+ 3 = 0
x =x = √√ 1 = -1 or 11 = -1 or 1
Hence, the x-intercept are -1 and 1.Hence, the x-intercept are -1 and 1.
iv)iv) The axis of symmetry is the y-axisThe axis of symmetry is the y-axis
v)v) When x = 0, y = 3.When x = 0, y = 3.
Hence, the minimum point is (0, 3)Hence, the minimum point is (0, 3)

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Case 2 : Function of type y = axCase 2 : Function of type y = ax22
+ bx + c.+ bx + c.
Features of the graph :Features of the graph :
α)α) ∪∪ - shaped- shaped if a > 0 andif a > 0 and ∩∩ - shaped- shaped if a < 0 :a < 0 :
b)b) The y-intercept is cThe y-intercept is c
c)c) The x-intercept is obtained by substituting y = 0 to solve theThe x-intercept is obtained by substituting y = 0 to solve the
equation axequation ax22
+ c = 0.+ c = 0.
d)d) The axis of symmetry is y = -(b/2a)The axis of symmetry is y = -(b/2a)
e)e) The coordinate of the minimum or maximum point areThe coordinate of the minimum or maximum point are
obtained by substituting x = -(b/2a) into the equation.obtained by substituting x = -(b/2a) into the equation.

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Sketch the graph of the function y = xSketch the graph of the function y = x22
– 4x + 4– 4x + 4
Solution:Solution:
i)i) a = 1 > 0, the graph isa = 1 > 0, the graph is ∪∪ - shaped.- shaped.
ii)ii) c = 4, then the y-intercept is 4.c = 4, then the y-intercept is 4.
iv)iv) The axis of symmetryThe axis of symmetry
x = -b/2a, where a = 1, b = -4x = -b/2a, where a = 1, b = -4
x = -(-4)/2(1) = 2x = -(-4)/2(1) = 2
xx
yy
00v)v) When x = 2When x = 2
y = (2)y = (2)22
– 4(2) + 4 = 0– 4(2) + 4 = 0
Hence, the minimum point is (2, 0)Hence, the minimum point is (2, 0)
22
44
iii)iii) When y = 0, xWhen y = 0, x22
- 4x + 4 = 0- 4x + 4 = 0
x = 0x = 0
Hence, the x-intercept is 0Hence, the x-intercept is 0

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xx
yy
00 22
- 4- 4
- 8- 8
Try thisTry this
Sketch the graph of the function y = - xSketch the graph of the function y = - x22
+ 4x - 8+ 4x - 8
Solution:Solution:
i)i) a = -1 < 0, the graph isa = -1 < 0, the graph is ∩∩ - shaped.- shaped.
ii)ii) c = -8, then the y-intercept is -8.c = -8, then the y-intercept is -8.
iv)iv) The axis of symmetryThe axis of symmetry
x = -b/2a, where a = -1, b = 4x = -b/2a, where a = -1, b = 4
x = -(4)/2(-1) = 2x = -(4)/2(-1) = 2
v)v) When x = 2When x = 2
y = -(2)y = -(2)22
+ 4(2) - 8 = - 4+ 4(2) - 8 = - 4
Hence, the minimum point is (2, -4)Hence, the minimum point is (2, -4)
iii)iii) When y = 0, -xWhen y = 0, -x22
+ 4x - 8 = 0+ 4x - 8 = 0
x is not possiblex is not possible

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c)c) To sketch the graphs of cubic funtionsTo sketch the graphs of cubic funtions
Functions of type y = axFunctions of type y = ax33
or y = axor y = ax33
+ c , a+ c , a ≠≠ 0.0.
Features,Features,
ii)ii) The y-intercept is c.The y-intercept is c.
iii)iii) The x-intercept is obtained by finding the value of x when y = 0.The x-intercept is obtained by finding the value of x when y = 0.
i)i) Determine the s-shaped of the graph.Determine the s-shaped of the graph.
(a > 0 : s-shaped , a < 0 : s-shaped )(a > 0 : s-shaped , a < 0 : s-shaped )

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Sketch the graphs of the functions belowSketch the graphs of the functions below
a) y = 4xa) y = 4x33
b) y = 4 - 2xb) y = 4 - 2x33
a)a) Solution;Solution;
ii)ii) When x = 0, y = 0When x = 0, y = 0
The graph passes through the origin (0, 0)The graph passes through the origin (0, 0)
xx
yy
00
i)i) a = 4 > 0, the graph is of the shapea = 4 > 0, the graph is of the shape
iii)iii) When x = 1, y = 4When x = 1, y = 4
The graph passes through the point (1, 4)The graph passes through the point (1, 4) • (1, 4)

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xx
yy
00
33
Sketch the graphs of the functions belowSketch the graphs of the functions below
a) y = 4xa) y = 4x33
b) y = 3 - 3xb) y = 3 - 3x33
b)b) Solution;Solution;
ii)ii) When x = 0, y = 3When x = 0, y = 3
The graph passes through the origin (0, 3)The graph passes through the origin (0, 3)
iii)iii) When y = 0, 3 - 3xWhen y = 0, 3 - 3x 33
= 1= 1
- 3x- 3x33
= - 3= - 3
x = 1x = 1
The graph passes through the point (1, 0)The graph passes through the point (1, 0)
i)i) a = -3 < 0, the graph is of the shapea = -3 < 0, the graph is of the shape
1

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d)d) To sketch graphs of reciprocal functionsTo sketch graphs of reciprocal functions
Functions of type y = a / x or y = axFunctions of type y = a / x or y = ax –1–1
..
Features:Features:
a)a) a > 0 ( positive )a > 0 ( positive ) a < 0 ( negative )a < 0 ( negative )
x
y
x
y
b) As the values of x tend towards positive or negative infinity, the
graph approaches the x-axis but does not intersect it. The same
applies as the values of y tend towards positive or negative infinity.

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Sketch the graph of the following functions.Sketch the graph of the following functions.
x
8
y)a =
x
4
y)b −=
x
y
x
y

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ActivityActivity
1.1. In group, solve the simultaneous equations y = x – 4In group, solve the simultaneous equations y = x – 4
and y = -2x + 5.and y = -2x + 5.
2.2. Then on the same axes, draw graphs of y = x – 4Then on the same axes, draw graphs of y = x – 4
and y = -2x + 5.and y = -2x + 5.
3.3. State the point of intersection of the graph.State the point of intersection of the graph.
4.4. What is the relation between the value obtained in 1What is the relation between the value obtained in 1
and 2 above?and 2 above?
2.22.2 Solution of an equation by the graph methodSolution of an equation by the graph method
A Point(s) of intersection of two graphs.

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y = x – 4y = x – 4 --------------------------(1)--------------------------(1)
y = -2x + 5y = -2x + 5 --------------------------(2)--------------------------(2)
(1) = (2)(1) = (2) x – 4 = -2x + 5x – 4 = -2x + 5
3x = 93x = 9
x = 3x = 3
y = -1y = -1
y = x – 4y = x – 4 xx 00 44
yy - 4- 4 00
y = -2x + 5y = -2x + 5 xx 00 22
yy 55 11 0 1 2 3-3 -2 -1
-2
-1
1
2
x
y
-4
-3
3
4
5
y = x - 4
y = -2x + 5
Based on the graph ,Based on the graph ,
Point of intersection is (3, -1)Point of intersection is (3, -1)

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a)a) Complete the table below with the values of y for the function y = xComplete the table below with the values of y for the function y = x22
+ 3x –7.+ 3x –7.
b)b) Using a scale of 2 cm to 1 unit on the x-axis and y-axis,Using a scale of 2 cm to 1 unit on the x-axis and y-axis,
Draw the graph of y = xDraw the graph of y = x22
+ 3x –7 for the values of x in the range –5+ 3x –7 for the values of x in the range –5 ≤≤ xx ≤≤ 2.2.
c)c) From the graph, findFrom the graph, find
i)i) values of x when y = -2values of x when y = -2
ii)ii) value of y when x = -0.5value of y when x = -0.5
d)d) Use the graph to solve the following equations.Use the graph to solve the following equations.
i)i) xx22
+ 3x –7 = 0+ 3x –7 = 0
ii)ii) xx22
+ 2x – 6 = 0+ 2x – 6 = 0
xx -5-5 -4-4 -3-3 -2-2 -1-1 00 11 22
yy 33 -7-7 -9-9 -7-7 -3-3

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a)a) y = xy = x22
+ 3x -7+ 3x -7
x
-1 0 1 2-2
-8
-7
-6
-5
-4
-3
-2
-1
y= x2
+ 3x - 7
y
xx -5-5 -4-4 -3-3 -2-2 -1-1 00 11 22
yy 33 -7-7 -9-9 -7-7 -3-3-3-3 -9-9 33
1
-9
-3-4-5
2
3
c) Based on the graph,c) Based on the graph,
y = -2,y = -2,
y = -2
x = -4.2,x = -4.2, dan x = 1.15.dan x = 1.15.
Scale ;Scale ; x-axis 2 cm to 1 unitx-axis 2 cm to 1 unit
x = -0.5,x = -0.5, y = -8.3y = -8.3

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x
-1 0 1 2-2
-8
-7
-6
-5
-4
-3
-2
-1
y= x2
+ 3x - 7
y
1
-9
-3-4-5
2
3d)d) y = xy = x22
+ 3x -7+ 3x -7
0 = x0 = x22
+ 3x -7+ 3x -7
y = 0y = 0
Based on the graph,Based on the graph,
x = -4.5 and 1.5x = -4.5 and 1.5
y = xy = x22
+ 3x - 7+ 3x - 7
0 = x0 = x22
+ 2x - 6+ 2x - 6
y = x - 1y = x - 1
y = x - 1y = x - 1
xx 00 22
yy -1-1 11
x = -3.6 and 1.6x = -3.6 and 1.6

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a)a) Complete the table below with the values of y for the functionComplete the table below with the values of y for the function
b)b) Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 2 unit on the y-Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 2 unit on the y-
axis, draw the graph of for the values of x in the range –4axis, draw the graph of for the values of x in the range –4 ≤≤ xx ≤≤ 4.4.
c)c) From the graph, findFrom the graph, find
i)i) value of y when x = 1.8value of y when x = 1.8
ii)ii) value of x when y = 3.4value of x when y = 3.4
d)d) Draw a suitable straight line on the graph to find all the values of x whichDraw a suitable straight line on the graph to find all the values of x which
satisfy the equationsatisfy the equation
State the values of x.State the values of x.
xx -4-4 -2.5-2.5 -1-1 -0.5-0.5 0.50.5 11 22 3.23.2 44
yy 11 1.61.6 88 -8-8 -4-4 -1.25-1.25 -1-1
x
4
y −=
x
4
y −=
4.x4range -theinxofvaluesthefor, ≤≤+= 3x2
x
4

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x
4
y −=
xx -4-4 -2.5-2.5 -1-1 -0.5-0.5
yy 11 1.61.6 44 88
xx 0.50.5 11 22 3.23.2 44
yy -8-8 -4-4 -2-2 -1.25-1.25 -1-1
x-1 0 1 2 3 4-2
-8
-6
-4
-2
2
4
6
8
10
y
-10
-3-4
x
4
y −=
12
-12
c)c) i)i) y = -2.3y = -2.3
ii)ii) x = -1.25x = -1.25
d)d)
-- 0 = + 2x + 30 = + 2x + 3
----------------------------------------------
y = -2x – 3y = -2x – 3
x = -2.35, x = 0.85x = -2.35, x = 0.85
x
4
y −=

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C Solve problems involving the solution of an equation by
the graphical method
Pak Samad has a 24 m length of fencing to build a rectangularPak Samad has a 24 m length of fencing to build a rectangular
enclosure to keep goats. The width of the enclosure is x m.enclosure to keep goats. The width of the enclosure is x m.
a)a) Show that the area A mShow that the area A m22
of the enclosure is given by A = 12x – xof the enclosure is given by A = 12x – x22
b)b) Draw the graph of the function A = 12x – xDraw the graph of the function A = 12x – x22
for the values of x in thefor the values of x in the
range 0 ≤ x ≤ 12. From the graph find the width of the enclosure suchrange 0 ≤ x ≤ 12. From the graph find the width of the enclosure such
that the area enclose isthat the area enclose is
i)i) 25 m25 m22
b)b) 30 m30 m22
c)c) From the graph, find the dimensions which give the enclosure theFrom the graph, find the dimensions which give the enclosure the
maximum area.maximum area.

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Understand the problemUnderstand the problem
• The perimeter of the rectangular enclosure is 24 m and its width is x mThe perimeter of the rectangular enclosure is 24 m and its width is x m
• Find the value of x such that the area of the rectangles are 25 mFind the value of x such that the area of the rectangles are 25 m22
andand
30 m30 m22
respectively.respectively.
• Find the dimensions that will give the maximum area.Find the dimensions that will give the maximum area.
Plan a strategyPlan a strategy
• Given the width, find the length in terms of x.Given the width, find the length in terms of x.
• Find the area using the formula, area = lengthFind the area using the formula, area = length ×× widthwidth
• Draw the graph of the area and find the point of intersection and theDraw the graph of the area and find the point of intersection and the
maximum pointmaximum point

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Carry out the strategyCarry out the strategy
• Given the perimeter = 24 m and width = x mGiven the perimeter = 24 m and width = x m
Hence, length = (24 – 2x ) / 2 = 12 - xHence, length = (24 – 2x ) / 2 = 12 - x
x m
(12 – x) m
Thus, the area
A = x(12 – x)
= 12x – x2

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b) y = 12 x - xb) y = 12 x - x22
x
8 10 126
5
10
15
20
25
30
35
40
A
xx 00 22 44 66 88 1010 1212
yy 00 2020 3232 3636 3232 2020 00
420
Based on the graphBased on the graph
Scale ;Scale ; x-axis 2 cm to 2 unitx-axis 2 cm to 2 unit
A = 25
When the area is 25 m2
, the
width is 2.7 m or 9.3 m
A = 30
When the area is 30 m2
, the
width is 3.6 m or 8.4 m
Maximum area = 36
Based on the graph, the
maximum area is 36m2
.
The width is 6 m.
The length is 12 – 6 = 6 m

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2.32.3 Region representing inequalities in two variablesRegion representing inequalities in two variables
AA Determine whether a given point satisfies y = ax +b, y > ax + b or y < ax +Determine whether a given point satisfies y = ax +b, y > ax + b or y < ax +
bb
To determine whether a point (x, y) satisfies y = ax +b, y > ax + b or y <To determine whether a point (x, y) satisfies y = ax +b, y > ax + b or y <
ax + bax + b , substitute the value of x into ax + b and compare with the y-
coordinate.
a) When the y-coordinate of the point (x, y) is equal to the value ax + b, we
say that the point (x, y) satisfies the equation y = ax + b.
b) When the y-coordinate of the point (x, y) is greater than the value ax + b,
we say that the point (x, y) satisfies the inequality y > ax + b.
c) When the y-coordinate of the point (x, y) is less than the value ax + b, we
say that the point (x, y) satisfies the inequality y < ax + b.

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Determine whether the following points satisfy y = 3x + 4 or y > 3x + 4Determine whether the following points satisfy y = 3x + 4 or y > 3x + 4
or y < 3x + 4.or y < 3x + 4.
a)a) A(3, 15)A(3, 15)
y = 15y = 15 3x + 4 = 3(3) + 4 = 133x + 4 = 3(3) + 4 = 13
Since 15 > 13, the point A(3, 15)satisfies y > 3x + 4Since 15 > 13, the point A(3, 15)satisfies y > 3x + 4
Substitute y = 15 and x = 3 into 3x + 4Substitute y = 15 and x = 3 into 3x + 4

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Substitute y = -2 and x = -2 into 3x + 4Substitute y = -2 and x = -2 into 3x + 4
y = -2y = -2 3x + 4 = 3(-2) + 4 = -23x + 4 = 3(-2) + 4 = -2
The point (-2, -2) satisfies y = 3x + 4The point (-2, -2) satisfies y = 3x + 4
c)c) C(1, -1)C(1, -1)
y = -1y = -1 3x + 4 = 3(1) + 4 = 73x + 4 = 3(1) + 4 = 7
Since -1 < 7, the point (1, -1) satisfies y < 3x + 4Since -1 < 7, the point (1, -1) satisfies y < 3x + 4
Substitute y = -1 and x = 1 in 3x + 4Substitute y = -1 and x = 1 in 3x + 4
b)b) B(-2, -2)B(-2, -2)

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BB Position of a point relative to the graph y = ax + bPosition of a point relative to the graph y = ax + b
Graph on the right shows the linearGraph on the right shows the linear
function y = x + 1.function y = x + 1.
1.1. Complete the following table to determineComplete the following table to determine
if the given points satisfy y = x + 1, y > x +if the given points satisfy y = x + 1, y > x +
1 or y < x + 1.1 or y < x + 1.
2.2. Plot the points on the graph to determine ifPlot the points on the graph to determine if
the points fall on the straight line y = x + 1the points fall on the straight line y = x + 1
or in a region above or below the line.or in a region above or below the line.
3.3. Make a conclusion on the positions of theMake a conclusion on the positions of the
points which satisfy y = ax + b, or y > ax +points which satisfy y = ax + b, or y > ax +
b or y < ax + b relative to the line y = ax +b or y < ax + b relative to the line y = ax +
bb
4.4. Make a conclusion whether all the pointsMake a conclusion whether all the points
in the same region satisfy the samein the same region satisfy the same
inequality.inequality.
x
-4
C
2-2 4
y
0
-2
4
2
-4

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a)a) All the points whichAll the points which
satisfy the equationsatisfy the equation
y = ax + b fall on they = ax + b fall on the
line y = ax + bline y = ax + b
PointPoint y-coordinatey-coordinate x + 1x + 1
Satisfy y = x +Satisfy y = x +
1, y > x + 1 or1, y > x + 1 or
y < x + 1y < x + 1
Situated on the line,Situated on the line,
in a region above orin a region above or
below the linebelow the line
A(0, 1)A(0, 1) 11 11 y = x + 1y = x + 1 on the lineon the line
B(2, 3)B(2, 3) 33 33
C(-4, -3)C(-4, -3)
D(-2, 3)D(-2, 3)
E(-4, 0)E(-4, 0)
F(1, 5)F(1, 5)
G(1, -2)G(1, -2)
H(3, 0)H(3, 0)
I(4, 2)I(4, 2)
x
-4
C
2-2 4
y
0
-2
4
2
-4
y = x + 1 On the line
-3 -2 y < x + 1 Below the line
• A
• B
• C
• D
3 -2 y > x + 1 Above the line
• E
0 -4 y > x + 1 Above the line
• F
5 2 y > x + 1 Above the line
• G
-2 1 y < x + 1 Below the line
• H
0 4 y < x + 1 Below the line
• I
2 5 y < x + 1 Below the line
Findings :
b)b) All the points which satisfy the inequality y > ax + b or y < ax + b are situated in aAll the points which satisfy the inequality y > ax + b or y < ax + b are situated in a
region above or below the line y = ax + b respectively.region above or below the line y = ax + b respectively.
c)c) If one point in a region satisfies y > ax + b or y < ax + b, then all the points in theIf one point in a region satisfies y > ax + b or y < ax + b, then all the points in the
region satisfy the same inequality.region satisfy the same inequality.

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C The region satisfying v > ax + b or y < ax + b
To identify whether a region satisfies inequality y > ax + b or y < ax +
b,
Step 1
Substitute a point in the region to determine if it satisfies the inequality
y > ax + b or y < ax + b.
Step 2
If the point satisfies y > ax + b, then the region where the point lies
satisfies inequality y > ax + b.
Step 3
If the point satisfies y < ax + b, then the region where the point lies
satisfies inequality y < ax + b.

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The diagram shows the graph of y = 5x + 2. Determine if the shaded regionThe diagram shows the graph of y = 5x + 2. Determine if the shaded region
satisfies the inequality y > 5x + 2 or y < 5x + 2.satisfies the inequality y > 5x + 2 or y < 5x + 2.
x
-4 2-2 4
6
0
-2
y
4
2
Select the origin (0, 0)Select the origin (0, 0)
Substitute y = 0Substitute y = 0
Substitute x = 0 into 5x + 2Substitute x = 0 into 5x + 2
5(0) + 2 = 25(0) + 2 = 2
Since 0 < 4, the point (0, 0) satisfiesSince 0 < 4, the point (0, 0) satisfies
inequality y < 5x + 2inequality y < 5x + 2
Thus he shaded region satisfies y < 5x + 2Thus he shaded region satisfies y < 5x + 2

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Try ThisTry This
x
-4 2-2 4
6
0
-2
y
4
2
The diagram shows the graph of y = -3x + 3. Determine if the shaded regionThe diagram shows the graph of y = -3x + 3. Determine if the shaded region
satisfies the inequality y > -3x + 3 or y < -3x + 3.satisfies the inequality y > -3x + 3 or y < -3x + 3.
Select the point (2, 0)Select the point (2, 0)
Substitute y = 0Substitute y = 0
Substitute x = 2 into -3x + 3Substitute x = 2 into -3x + 3
-3(2) + 3 = -3-3(2) + 3 = -3
Since 0 > -3, the point (2, 0) satisfiesSince 0 > -3, the point (2, 0) satisfies
inequality y > -3x + 3inequality y > -3x + 3
Thus he shaded region satisfies y > -3x + 3Thus he shaded region satisfies y > -3x + 3

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DD Shade the region of the inequalitiesShade the region of the inequalities
A given linear inequality can be represented by a region which can be determine byA given linear inequality can be represented by a region which can be determine by
shading.shading.
i)i) y > ax + b or y < ax + b.y > ax + b or y < ax + b.
ii)ii) yy ≥≥ ax + b or yax + b or y ≤≤ ax + b.ax + b.
Shading can be done by following the steps below.Shading can be done by following the steps below.
a)a) Draw the graph of the linear equation y = ax + b.Draw the graph of the linear equation y = ax + b.
i)i) Use a dashed line for the inequality “ > ” or “ < ” lUse a dashed line for the inequality “ > ” or “ < ” l
ii)ii) Use a solid line for the inequality “Use a solid line for the inequality “≥≥ ” or “” or “≤≤ ””
b)b) Select a suitable point such as the origin (0, 0)Select a suitable point such as the origin (0, 0)
c)c) If the point satisfies the given inequality, then the required region is where the givenIf the point satisfies the given inequality, then the required region is where the given
point lies. Shade the region.point lies. Shade the region.
d)d) If the point does not satisfies the given inequality, then the required region isIf the point does not satisfies the given inequality, then the required region is
opposite the region where the given point lies. Shade the opposite region.opposite the region where the given point lies. Shade the opposite region.

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Shade the region which represent the inequalities y < 2x – 3.Shade the region which represent the inequalities y < 2x – 3.
x
y
1 2 3 4 5-1 0
-2
-1
-3
1
2
-4
y = 2x - 3
y = 0y = 0
2x - 3 = 2(0) - 3 = -32x - 3 = 2(0) - 3 = -3
Since 0 > -3, the point (0, 0) is inSince 0 > -3, the point (0, 0) is in
the region which represent thethe region which represent the
inequality y > -3x + 3.inequality y > -3x + 3.
Shade the region where the pointShade the region where the point
(0, 0) is situated.(0, 0) is situated.

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Try ThisTry This
x
y
2 4 6 8-2 0
4
6
2
y = 0y = 0
5 – 3x = 5 – 3(0) = 55 – 3x = 5 – 3(0) = 5
Shade the region which represent the inequality yShade the region which represent the inequality y ≥≥ 5 – 3x5 – 3x
Since 0 < 5, the point (0, 0) is notSince 0 < 5, the point (0, 0) is not
in the region which represent thein the region which represent the
inequality yinequality y ≥≥ 5 – 3x.5 – 3x.
Shade the region where the pointShade the region where the point
(0, 0) is not situated.(0, 0) is not situated.

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EE The region which satisfies two or more simultaneous linearThe region which satisfies two or more simultaneous linear
inequalitiesinequalities
To determine the common region by shading, use the
method below
a) For inequalities y > ax + b or y ≥ ax + b, shade the region
above the line y = ax + b.
b) For inequalities y < ax + b or y ≤ ax + b, shade the region
below the line y = ax + b.
c) Find the intersection of the region which represent the
inequalities.
d) The intersection is the common region which satisfies the
simultaneous inequalities.

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In each diagram, identify the region which satisfies the given simultaneousIn each diagram, identify the region which satisfies the given simultaneous
a) ya) y ≥≥ 2x and y2x and y ≥≥ 5 – x5 – x b) y ≤ x + 1 and x < 4b) y ≤ x + 1 and x < 4
II
5
x
0
y
I
5
y =2x
IV
III
y = 5 - x
The region which satisfy yThe region which satisfy y ≥≥ 2x are regions I and II.2x are regions I and II.
The region which satisfies yThe region which satisfies y ≥≥ 5 – x are the region II and5 – x are the region II and
IIIIII
Thus, the region which satisfies both the inequalities yThus, the region which satisfies both the inequalities y ≥≥ 2x2x
and yand y ≥≥ 5 – x is region II.5 – x is region II.

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In each diagram, identify the region which satisfies the given simultaneousIn each diagram, identify the region which satisfies the given simultaneous
a) ya) y ≥≥ 2x and y2x and y ≥≥ 5 – x5 – x b) y ≤ x + 1 and x < 4b) y ≤ x + 1 and x < 4
II
4 x
0
y
I
1
y = x + 1
IV
III
x = 4
The region which satisfy yThe region which satisfy y ≤ x + 1 are regions I and II.≤ x + 1 are regions I and II.
The region which satisfies x < 4 are the region I and IIIThe region which satisfies x < 4 are the region I and III
Thus, the region which satisfies both the inequalities yThus, the region which satisfies both the inequalities y ≤ x≤ x
+ 1 and x < 4 is region I.+ 1 and x < 4 is region I.

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In each diagram below, shade the region which satisfies the givenIn each diagram below, shade the region which satisfies the given
simultaneous inequalitiessimultaneous inequalities
a) y < 2x, ya) y < 2x, y ≥≥ ½ x, and x + y ≤ 4½ x, and x + y ≤ 4
4
x
0
y
y = 2x
y = ½ x
4
x + y = 4

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In each diagram below, shade the region which satisfies the givenIn each diagram below, shade the region which satisfies the given
simultaneous inequalitiessimultaneous inequalities
b) y ≤ x + 3, yb) y ≤ x + 3, y ≥≥ 3 - x, and x < 33 - x, and x < 3
3
x
0
y
y = x + 3
y = 3 – x
3
x = 3

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In the diagram below, state three inequalities which define the shadedIn the diagram below, state three inequalities which define the shaded
region.region.
5
x
0
y y = x + 1
x + y = 5
5
(4, 5)
The region lies above the line y = x + 1.
The inequality is y ≥ x + 1
The region lies above the line x + y = 5
The inequality is x + y ≥ 5.
The y-intercept of line x + y = 5 is y = 5.
The dashed line is y = 5.
The region lies below the line y = 5.
The inequality is y < 5.
Thus, the three inequalities which
define the shaded region are
y ≥ x + 1, x + y ≥ 5 and y < 5.

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