Graph functions

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GRAPHSOF

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

x0 1 2 3 4 5 6

1

2

3

4

5

6

7

9

8

10

11y = 2x + 3y = 2x + 3

y

Example 1Example 1

Draw the graph of the function Draw the graph of the function y = 2x + 3 for 0 y = 2x + 3 for 0 x x 4 4

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 the Draw the graph of the functionfunction y = 2 – 3x for -1 y = 2 – 3x for -1 x x 4 4

xx -1-1 00 11 22 33 44

yy 55 22 -1-1 -4-4 -7-7x

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

Draw the graph of the functionDraw the graph of the function y = 2xy = 2x22 + 6x + 2 for -4 + 6x + 2 for -4 x x 2 2

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

Draw the graph of the functionDraw the graph of the function y = 4x - 4x2 for -2 y = 4x - 4x2 for -2 x x 4 4



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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

Draw the graph of the functionDraw the graph of the function y = 2xy = 2x33 for -3 for -3 x x 3 3



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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

Draw the graph of the functionDraw the graph of the function y = 4 - 2xy = 4 - 2x33 for -3 for -3 x x 3 3



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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

-3xx 0.50.5 11 22 33 44

yy 44 22 11 0.670.67 0.50.5-4-5

x

2y

x

2y

Example 6Example 6

Draw the graph of the functionDraw the graph of the function

for -5 for -5 x x 4 4



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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

6y

12

-12

x

6y

Try thisTry this

Draw the graph of the functionDraw the graph of the function

for -5 for -5 x x 4 4



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Example 7Example 7

a)a) Complete the table of values Complete 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 = 1.5

x = 1.5

x = -0.5

x = -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 = -0.20x = 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, a y = 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 = ax y = 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 = ax y = 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 a y = 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 = ax2 2 + c, b = 0.+ c, b = 0.

Features of the graph :Features of the graph :

a)a) - shaped- shaped if a > 0 and if 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 the The 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 or By 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 is a = 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 2 4 = -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 is a = -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 1 1 = -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 :

a)a) - shaped- shaped if a > 0 and if 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 the The 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 are The 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 + 4Solution:Solution:

i)i) a = 1 > 0, the graph is a = 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) = 2xx

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 - 8Solution:Solution:

i)i) a = -1 < 0, the graph is a = -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 = ax or 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 - 3x When y = 0, 3 - 3x 33= 1= 1

- 3x- 3x3 3 = - 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 = ax Functions 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

8y )a

x

4y )b

x

y

x

y

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ActivityActivity

1.1. In group, solve the simultaneous equations y = x – 4 In 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 – 4 Then 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 1 What 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 x x 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 - 7y

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 unit x-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 - 7y

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 - 1xx 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 –4 axis, draw the graph of for the values of x in the range –4 x x 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 which Draw 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

4y

x

4y

4. x 4range -the inx of valuesthe for , 3x2x

4

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x

4y

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

4y

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

4y

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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 rectangular Pak 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 – x of 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 the for the values of x in the range 0 ≤ x ≤ 12. From the graph find the width of the enclosure such range 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 the From 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 and and 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 = length Find the area using the formula, area = length width width

• Draw the graph of the area and find the point of intersection and the Draw 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

x8 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 unit x-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 + b b

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 + 4 Determine 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 + bGraph on the right shows the linear Graph on the right shows the linear function y = x + 1.function y = x + 1.

1.1. Complete the following table to determine Complete 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 if Plot the points on the graph to determine if the points fall on the straight line y = x + 1 the 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 the Make 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 points Make a conclusion whether all the points in the same region satisfy the same in 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 which All the points which satisfy the equation satisfy the equation y = ax + b fall on the y = ax + b fall on the line y = ax + bline y = ax + b

PointPoint y-coordinatey-coordinate x + 1x + 1Satisfy 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 or in 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 a All 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 the If 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 region The 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) satisfies Since 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 region The 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) satisfies Since 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 by A 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) y y ax + b or y ax + 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 given If 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 is If 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 - 3Select the point (0, 0)Select the point (0, 0)

y = 0y = 0

2x - 3 = 2(0) - 3 = -32x - 3 = 2(0) - 3 = -3

Since 0 > -3, the point (0, 0) is in Since 0 > -3, the point (0, 0) is in the region which represent the the region which represent the inequality y > -3x + 3.inequality y > -3x + 3.

Shade the region where the point Shade 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

Select the point (0, 0)Select the point (0, 0)

y = 0y = 0

5 – 3x = 5 – 3(0) = 55 – 3x = 5 – 3(0) = 5

Shade the region which represent the inequality y Shade the region which represent the inequality y 5 – 5 – 3x3x

Since 0 < 5, the point (0, 0) is not Since 0 < 5, the point (0, 0) is not in the region which represent the in the region which represent the inequality y inequality y 5 – 3x. 5 – 3x.

Shade the region where the point Shade 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 linear The 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 simultaneous In each diagram, identify the region which satisfies the given simultaneous inequalitiesinequalities

a) y a) y 2x and y 2x and y 5 – x 5 – x b) y ≤ x + 1 and x < 4b) y ≤ x + 1 and x < 4

II

5 x0

y

I

5

y =2x

IV

III

y = 5 - x

The region which satisfy y The region which satisfy y 2x are regions I and II. 2x are regions I and II.

The region which satisfies y The region which satisfies y 5 – x are the region II and 5 – x are the region II and III III

Thus, the region which satisfies both the inequalities y Thus, the region which satisfies both the inequalities y 2x 2x and y and 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 simultaneous In each diagram, identify the region which satisfies the given simultaneous inequalitiesinequalities

a) y a) y 2x and y 2x and y 5 – x 5 – x b) y ≤ x + 1 and x < 4b) y ≤ x + 1 and x < 4

II

4 x0

y

I

1

y = x + 1

IV

III

x = 4

The region which satisfy y The 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 III The region which satisfies x < 4 are the region I and III

Thus, the region which satisfies both the inequalities y Thus, 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 given In each diagram below, shade the region which satisfies the given simultaneous inequalitiessimultaneous inequalities

a) y < 2x, y a) y < 2x, y ½ x, and x + y ≤ 4 ½ x, and x + y ≤ 4

4

x0

y y = 2x

y = ½ x

4

x + y = 4

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a) y < 2x, y a) y < 2x, y ½ x, and x + y ≤ 4 ½ x, and x + y ≤ 4

4

x0

y y = 2x

y = ½ x

4

x + y = 4

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b) y ≤ x + 3, y b) y ≤ x + 3, y 3 - x, and x < 3 3 - x, and x < 3

3

x0

y y = x + 3

y = 3 – x

3

x = 3

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In the diagram below, state three inequalities which define the shaded In the diagram below, state three inequalities which define the shaded region.region.

5

x0

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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