Graphs of Functions II 2

8/3/2019 Graphs of Functions II 2

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CHAPTER 2

GRAPHS OF FUNCTIONS II

2.1 GRAPHS OF FUNCTIONS

The graph of a function is a set of points on

the Cartesian Plane that satisfy the function

Information is presented in the form of graphs

Graph are widely used in science and technology Graphs are very useful to researchers, scientists

and economist


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The different type of functions and respective

power of x

Type of

function

General form Example Highest

power of variable x

Linear

Quadratic

Cubic

Reciprocal

y = ax + c

y = ax2 + bx + c

y = ax3

+ bx2

+ cx + d

y = a

x

y = 3x

y = -4x + 5

y = 2x2

y = -3x2 + 2x

y = 2x2 + 5x + 1

y = 2x3

y = -3x3 + 5x

y = 2x3 - 3x + 6

y = 4

x

y = - 2

x

1

3

-1

2

y = ax3

y = ax3 + bxy = ax3 + bx + c

a 0


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LINEAR FUNCTION


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LINEAR FUNCTION


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QUADRATIC FUNCTION


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QUADRATIC FUNCTIONQUADRATIC FUNCTION


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QUBIC FUNCTION


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QUBIC FUNCTION


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RECIPROCAL FUNCTION


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RECIPROCAL FUNCTION


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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


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COMP


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CALC MemoryCALC MemoryExample 1Example 1Calculate the result

for Y = 3X 5,

when X = 4, and when X = 6

)3X

- 5CALC

3X 5

4 = 7

CALC 6 = 13

ALPHA


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CALC MemoryCALC MemoryExampleExample 22Calculate the result

for Y = X2 + 3X 12,

when X = 7, and when X = 8

) 3X

x2 +ALPHA

)

X

- 21CALCX

2

+ 3X 12

7 = 58

CALC 8 = 76

ALPHA


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for Y = 2X2 + X 6,

when X = 3, and when X = -3

)

3

X

x2 +ALPHA

)

X

- 6CALC2X

2

+ X 6

3 = 15CALC (-) = 9

ALPHA

2


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for Y = -X3 + 2X + 5,

when X = 2, and when X = -1

)

1

X

x2

+ 2

ALPHA

)

X

+ 5-X3 + 2X + 5 2 = 1

CALC (-) = 4

ALPHA

(-)SHIFT x3

CALC


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Example 5Example 5Calculate the result

for Y = 6 when X = -3,

X

and when X = 0.5

)

XALPHA

3CALC (-) = -2

6

CALC 0 . 125 =

ab/c 6x


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Example 6Example 6Calculate the result

for Y = 6 when X = -3,

X

and when X = 0.5

)

XALPHA

3CALC (-) = -2

6

CALC 0 . 125 =

x-1 6x-1


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Y = -2X2 + 40

X 0 0.5 1 1.5 2 3 3.5 4

Y

Y = X3 3X + 3

X -3 -2 -1 0 0.5 1 1.5 2

Y

Y = -16X

X -4 -3 -2 -1 1 2 3 4

Y

40 39.5 38 35.5 32 22 15.5 8

-15 1 5 3 1.625 1.875 51

4 5.33 8 16 -16 -8 -5.33 -4


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2(4)2 + 5(4) 1 =51Using Calculator

2 ( )4 x2 + 5 ( 4 )

- 1 =


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(-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


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Using Calculator

6

(-3)= -2

3( )6 (-) =


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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


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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


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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


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-4 -2

-3.6 -3.0 -2.1

x-axis scale : 2 cm to 2 units


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10

15

12

14.5

13.5

y-axis scale : 2 cm to 5 units

10.75


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-20

-15

-18

-15.5

-16.5

y-axis scale : 2 cm to 5 units

-19.25


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The x-coordinate and y-coordinate


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xA

B

x

x

x

C

D

(-3,2)

(2,0)

(4,-3)

(0,-4)

Ex

(4,4)

F

x

(-7,-2)


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-5

-1 1

x

x

x

x

A

B

C

D

Ex (-0.5,2)

(-1,-3)

(0,3)

(0.3,1.5)

(0.5,-1.5)


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-2

-1 1

xx

x

x

A

B

CD (-1,-1.2)

(0,1.2)

(0.3,0.6)

(0.2,-1)


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-10

-1 1

x

x

x

x

A

B

C

D (-1,-6)

(0,5)

(0.3,3)

(0.5,-3)


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2.1 A2.1 A Drawing the Graphs

Construct a table fora chosenrange of x values, forexample-4 x 4

Draw the x-axis andthe y-axis andsuitable scale foreach axis

starting from the origin

Plotthe x and y values ascoordinate pairsonthe CartesianPlane

Jointhe pointsto form a straight line (usingruler) orsmoothcurve

(using FrenchCurve/flexible ruler) with a sharp pencil

Label the graphs

To draw the graph of a function, follow these steps;


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2.1 A2.1 A Drawing the Graphs

Draw the graphof 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 ALINEAR FUNCTION

8

3 + 2

22


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Draw the graphof 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


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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


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Draw the graphof 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


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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 ACUBIC FUNCTION


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Draw the graphof y = -4 for-4 x 4.

x

solutiony = -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


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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


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xx

USING FRENCH CURVE


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x

x

x

x

USING FRENCH CURVE


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x

x

x

USING FRENCH CURVE


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x x

x

xx

x

xx x

x

x


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-10

-15

-20


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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

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


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x

x

x

x

x

x

x

x

x

y

x0 1 2

2

4

6

8

10

-1-3-4-5

16

14

12

-2 3-3.5

5


y = x2 + 2x

i di l f i bl f G h


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x

x

x

x

x

x

x

x

x

y

x0 1 2

2

4

6

8

10

-1-3-4-5

16

14

12

-2 3

5

11


y = x2 + 2x

-4.4 2.5


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x =1.5


y


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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

-1.21.8

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

Valuesobtained fromthe graphs are

approximations

Notes


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2.1 CIdentifying the shape of a Graph from

a Given Function

LINEAR aa

y

x

y = x

0

bb

x

y = -x + 2

0

2

y


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a Given Function

QUADRATIC aa

y

x

y = x2

0

bbx

y = -x2

0

y


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a Given Function

CUBIC aa

y

x

y = x3

0

bb

x

y = -x3 + 2

0

2

y


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a Given Function

RECIPROCAL aa

y

x0

bb

x0

y

y = 1x

y = -1

x

0


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2.1 D Sketching Graphs of Function

Sketching a graph meansdrawing a graph withoutthe actual data

When we sketchthe graph, we donotuse

a graph paper,howeverwe must know the important

characteristicsofthe graphsuch as itsgeneral form

(shape),the y-intercept and x-intercept

Ithelpsusto visualise the relationship ofthe variables


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Example y = 2x + 4

4

-2 0

y

x

findthe x-interceptofy = 2x + 4.

Substitute y = 02x + 4 = 0

2x = -4

x = -2Thus, x-intercept = -2

findthe y-interceptof

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


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B Sketching The Graph of AQuadratic 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


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B Sketching The Graph of A Cubic Function

Example y = -3x3 + 5

a < 0

the shape ofthe graph is

y-intercept is 5

x0

y

5


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2.2The 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


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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

graphson

the same

axes

Look atthepointsof

intersection:

A and B.

Readthe

valuesofthe

coordinates

of x.

x = -1 and

x = 2

Solve the equation x2 = x + 2 by usingthe Graphical Method

The Solution of An Equation By Graphical


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2.2 The Solution of An Equation By Graphical

Method

12. (a) Complete Table 1 forthe equation y = x2 + 2x by writingdownthe 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 usingscale of 2 cm to 1 unitonthe x-axis and 2 cm to 2 unitsonthe

y-axis,

draw the graphof y = x2 + 2x for-5 x 3.

(c) From yourgraph, 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 yourgraphto find a value of x which

satisfiesthe equationof x 2 + x 4 = 0 for-5 x 3.

TABLE 1

2 2 The Solution of An Equation By Graphical


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Method

The graph y = x2 - 5x - 3 is drawn. Determinethe suitable straight line

to be drawn to solveeachofthefollowing 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


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solution

a) x2 - 5x - 3 = 4 x2 - 5x - 3 = y4

Therefore, y = 4 is the suitable straight line

b) x2 - 5x - 3 = 2x + 2 x2 - 5x - 3 = y2x + 2 y

Therefore, y = 2x + 2 is the suitable straight line


to be drawn to solvetheequation: x2 - 5x - 3 = 4

y


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solution

c) x2 - 5x - 2 = x + 4

- 1

-1 on both sides

Therefore, y = x + 3 is the suitable straight line

x2 - 5x 2 = x + 4 - 1

x2 - 5x - 3 = x + 3x + 3


to be drawn to solvetheequation: x2 - 5x 2 = x + 4


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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




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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




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Alternative MethodSince 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 - 2x - 5 2

1 + 2 y+0 = -5x + (-2x) - 3 + ( -5)

y = -7x - 8


to be drawn to solvetheequation: -x2 - 2x - 5 = 0


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Alternative Method


to be drawn to solvetheequation: -x2 - 2x - 5 = 0

X2

= -2x - 5

Substitute x2 = -2x - 5 into y = x2 5x - 3

y = (-2x - 5) 5x - 3

y = -2x - 5 5x - 3

y = -7x - 8


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y = x3 - 5x - 3 1

0 = x3 - 7x - 2 2

1 - 2 y-0 = -5x - (-7x) - 3 - ( -2)

y = 2x - 1




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Alternative Method



X3

= 7x + 2

Substitute x3 = 7x + 2 into y = x3 5x - 3

y = (7x + 2) 5x - 3

y = 7x + 2 5x - 3

y = 2x - 1


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y = x2 - 5x - 3 1

0 = x2 - 2x - 4 2

1 - 2 y-0 = -5x - (-2x) - 3 - ( -4)

y = -3x + 1


to be drawn to solvetheequation: x2 - 2x = 4

x2

2x - 4 = 0

2


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Alternative Method

Substitute x2 = 2x + 4 into y = x2 5x - 3


to be drawn to solvetheequation: x2 - 2x = 4

X2

= 2x + 4

y = (2x + 4) 5x - 3

y = 2x + 4 5x - 3

y = -3x + 1


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solution

4 = x + 1

xMultiply both sides by 2

We get 8 = 2x + 2

xTherefore, y = 2x + 2 is the suitable straight line

2x + 2

The graph y = 8 is drawn. Determinethe suitable straight line

x

to be drawn to solveeachtheequation: 4 = x + 1

x


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solution

-8 = -2x - 2

x

Multiply both sides by -1

We get 8 = 2x + 2

x

Therefore, y = 2x + 2 is the suitable straight line

2x + 2

The graph y = 8 is drawn. Determinethe suitable straight line

xto be drawn to solveeachtheequation: - 8 = -2x - 2

x

2 2 The Solution of An Equation By Graphical


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Method

12. (a) CompleteTable 1 fortheequation y = x2 + 2x by writing down the values ofy

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 scaleof 2 cm to 1 uniton the x-axis and 2 cm to 2 units on the y-axis,

draw the graphofy = x2 + 2x for -5 x 3.

(c) From your graph, find

(i) the valueofy when x = -3.5,

(ii) the valueofx when y = 11.

(d) Draw a suitable straight lineon your graphtofind a valueofx which satisfies

theequation of x 2 + x 4 = 0 for -5 x 3.

TABLE 1

l


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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


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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)


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( )

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)


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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 +0 = 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

ax2 + bx + c = 0


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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

=


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2.3 REGION REPRESENTING INEQUALITIES IN TWO VARIABLES

2.3 A Determining Whether a Given Point Satisfies y = ax + b,

y > ax + b or y < ax + b

How can we determine whethera given pointsatisfiesy = 3x + 1, y < 3x + 1ory > 3x + 1 ?

2 3 REGION REPRESENTING INEQUALITIES IN TWO VARIABLES


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y > ax + b or y < ax + bLetusconsiderthe point (3,5). The pointcanonly satisfiesone ofthe

followingrelations:

(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


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Forpoint (1,-1)

When x = 1, y = 3(1) - 1 = 2

Since the y-coordinate ofthe point (1,-1) is -1, which is lessthan 2,

we conclude that y < 3x - 1 . Therefore,the point (1,-1) satisfiesthe relationy < 3x - 1

Solution a

REGION REPRESENTING INEQUALITIES IN TWO VARIABLES


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2.3REGION REPRESENTING INEQUALITIES IN TWO VARIABLES



Forpoint (3,10)

When x = 3, y = 3(3) - 1 = 8

Since the y-coordinate ofthe point (3,10) is 10, which isgreaterthan 8,

we conclude that y > 3x - 1 . Therefore,the point (3,10) satisfiesthe relationy > 3x - 1

Solution b


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2.3 A

Forpoint (-1,-4)

When x = -1, y = 3(-1) - 1 = -4

Since the y-coordinate ofthe point (-1,-4) is -4, which is equal to -4,

we conclude that y = 3x - 1 . Therefore,the point (-1,-4) satisfiesthe relationy = 3x - 1

Solution c

Determining Whether a Given Point Satisfies y = ax + b,



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2.3 B

Determining The Position of A Given Point Relative to

y = ax + b

All the pointssatisfying y < ax + b are below the graph

All the pointssatisfying y = ax + b are onthe graph

All the pointssatisfying y > ax + b are above the graph



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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 < x

The point P(4,8) liesabove the line y = x.

This region is represented

by y > x

The point Q(4,2) lies

below the line y = x.

This region is represented

by y < x

Q(4,4)


on the line y = x.This region is represented

by y = x y > x



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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 + 2

The pointP(-8,6) lies

above the line y = 3x + 2.

Thisregion isrepresentedby y > 3x + 2


below the line y = 3x + 2.

Thisregion is

represented

by y < 3x + 2

Q(2,8)

y > 3x + 2



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2.3 C Identifying The Region Satisfying y > ax + b or y < ax + b

-2-4

2

4

6

8

x

y

20

-2

-4

4-8

-8

8

Determine whether the

shaded region in the graph

satisfies y < 3x + 2 or

y > 3x + 2

EXAMPLE

solution

The shaded region is

below the graph, y = 3x + 2.

Hence, this shaded region

satisfies y< 3x + 2



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2.3 D Shading The Regions Representing Given Inequalities

Symbol Type ofLine

< or> Dashed

Line

or Solid line

The type of line to be drawndepends

on inequality symbol

The table above showsthesymbolsof inequality andthecorrespondingtype of line

to be drawn

HoT TiPsThe dashed line indicatesthat all points

are not included inthe region. The solid

line indicatesthat all pointsonthe line

are included



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0x

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



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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



100/140


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



101/140

2.3 EDetermine The Region which Satisfies Two or More

Simultaneous Linear Inequalities

y

x

2

0 2 3-3

1

ExampleShade the regionthatsatisfies

3y < 2x + 6, 2y -x + 2 and x 3.

X = 3



102/140

y

x

2

0 2 3-3

1

X = 3

Shade the region that satisfies

3y < 2x + 6, 2y -x + 2 and x 3.



103/140

y

x

2

0 2 3-3

1

X = 3A

RegionA satisfies 2y -x + 2, 3y < 2x + 6, and x 3



104/140

2.3 E Determine The Region which Satisfies Two or More


y

x

2

0 2 3-3

1

X = 3A

RegionA satisfies

2y -x + 2,

3y < 2x + 6, and x 3



105/140



y

x

2

0 2 3-3

1

X = 3A

RegionA satisfies

2y > -x + 2,3y 2x + 6, and x < 3



106/140



y

x

3

0

x = 3

Region B satisfies

y -x + 3,

y < x , and x 3

B



107/140



y

x

3

0

x = 3

Region B satisfies

y > -x + 3,

y x , and x < 3

B



108/140



y

x

3

0

x =-3RegionCsatisfies

y > -x + 3,

y -2x , and x >-3

-3

C

y

RegionA satisfies


109/140

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

RegionA satisfies

y x2 + 2x,

y x + 4,

and x 0

A

y

Region A satisfies


110/140

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


111/140

Shade the regionthatsatisfies

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


112/140

Shadethe region that satisfies

y 2x + 8, y x, and y < 8

x0

y = 8

y = x

K3

y = 2x + 8 3

SPM Clone

3y


113/140

3.

x0

y = 8

y = x

K2

y = 2x + 82

3y


114/140

3.

x0

y = x

K1

y = 2x + 81


115/140

y

xO

y = 2x6

On the graphs provided, shade the region which satisfies

the three inequalities x < 3, y 2x 6 and y -6[3 marks]

y = 6


116/140

y

xO

y = 2x6

y = 6

Solution:

x = 3

K3

x-intercept = -(-6 2) = 3

x < 3, y 2x 6 and y -6


117/140

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


118/140

y

x

y = 3x+12

O

y = x4

y = 4

Solution:

K3

y-intercept =-4

y x - 4, y -3x + 12 and y > -4

Cloned SPM Question


119/140

12. (a) CompleteTable 1 fortheequation y = x2 + 2x by writing down the values ofy

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 scaleof 2 cm to 1 uniton the x-axis and 2 cm to 2 units on the y-axis,

draw the graphofy = x2 + 2x for -5 x 3.


(i) the valueofy when x = -3.5,(ii) the valueofx when y = 11.


theequation of x 2 + x 4 = 0 for -5 x 3.

TABLE 1

Q

12. (a) Complete Table 1 for the equation y = x2 + 2x by writing

down the values of y when x = -3 and 2.


120/140

x -5 -4 -3 -2 -1 0 1 2 3

y 15 8 0 -1 0 3 158

Solution

3

) 2X

x2 +ALPHA

)X

CALC

X2 + 2X

-3 = 3

CALC 2 =8

ALPHA

CALC

-5 = 15

down the values of y when x 3 and 2.

2 cm to 1 unit on the x-axis and 2 cm to 2 units on the y-axis, for -5 x 3.


121/140

12. (b)y


122/140

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

-2

y


12 (c)


123/140

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 = x2 + 2x

12. (c)

(i)

-2

y


12 (c)


124/140

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

5

11

y = x2 + 2x-4.4 2.5

12. (c)

(ii)

-2

(d) Draw a suitable straight lineon your graphtofind a valueofx

hi h i fi h i f 2 f


125/140

which satisfies theequation of x 2 + x 4 = 0 for -5 x 3.

y = x2 + 2x + 0 10 = x2 + x - 4 2

1 - 2 y-0 = 2x - x + 0 - ( -4)

y = x + 4

x 0 -4

y 4 0

Answer:

(d) x =

x =

12. (d)y y = x2 + 2x + 0


126/140

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

y0 = 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

-2


127/140

x -5 -4 -3 -2 -1 0 1 2 3

y 15 8 0 -1 0 3 1583

12. (a)

K1K1

12. (b)y

P1


128/140

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

P1

x

x

K2

N1

K2

-2


129/140

12. (c) (i) y = 5

(ii) x = -4.4, 2.5

12. (d) x = 1.5, -2.5

y = x2 + 2x + 00 = x2 + x - 4-

y = x + 4

x 0 -4

y 4 0

P1

P1P1

N1

Cloned SPM Question


130/140

12. (a) CompleteTable 1 fortheequation by writing down the values of y

when x = -1.5 and 2.

x -4 -3 -1.5 -1 -0.5 0.5 1 1.5 2 4

y 0.75 1 3 6 -6 -3 -2 -0.75

(b) By using scaleof 2 cm to 1 uniton the both axis, draw the graphof

for -4 x 4.


(i) the valueofy when x = 1.6,(ii) the valueofx when y = 3.3.


theequation of for -4 x 4.

TABLE 1

x

y3

!

,3

xy !

423

! x

x

Solution


131/140

x -4 -3 -1.5 -1 -0.5 0.5 1 1.5 2 4

y 0.75 1 3 6 -6 -3 -2 -0.75

Solution

)

XALPHA

1CALC (-) = 23

CALC

2 -1.5=

ab/c -3x(-). 5

2 -1.5

(b)

5

6yx


132/140

1

2

3

4

5

0 x

-1

-2

-3

-4

-5

-

-1-2-3-4 1 2 3 4

xx

x

x

x

x

xx

x

(c)

5

6yx


133/140

1

2

3

4

5

0 x

-1

-2

-3

-4

-5

-

-1-2-3-4 1 2 3 4

xx

x

x

x

x

xx

x

1.6

-1.9

3.3

-0.9


134/140

(d) Graph already drawn: ,3

xy ! 1

Equationto be solved: 423

0 ! xx

2

1 2- 42 ! xy

Solution

Th h i d D t i th it bl t i ht li3

(d)


135/140

Solution

Multiply both sides by -1

We get = -2x - 4

Therefore, y = -2x - 4 is the suitable straight line

-2x - 4

The graph is drawn. Determinethe suitable straight line

to be drawn to solvetheequation:

xy !

42

3! x

x

(d)

423

! x

x

x

3 y

x!

3


136/140

)2X

- 4CALC

-2X 4

2 = 0CALC 0 = -4

ALPHA

(-)

(-)

y = -2X 4

5

6yx


137/140

1

2

3

4

0 x

-1

-2

-3

-4

-5

-

-1-2-3-4 1 2 3 4

xx

x

x

x

x

xx

x

x

x

-2.6

0.6


138/140

x -4 -3 -1.5 -1 -0.5 0.5 1 1.5 2 4

y 0.75 1 3 6 -6 -3 -2 -0.752 -1.5

12. (a)

K1K1

5

6yx

P1


139/140

1

2

3

4

0 x

-1

-2

-3

-4

-5

-

-1-2-3-4 1 2 3 4

xx

x

x

x

x

xx

x

x

x

K2

N1

K2


140/140

12. (c) (i) y = -1.9

(ii) x = -0.9

12. (d) x = -2.6, 0.6

P1

P1

N1N1

Graphs of Functions II 2

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Transcript of Graphs of Functions II 2