graphs of functions 2

CHAPTER 2

GRAPHS OF FUNCTIONS II

2.1 GRAPHS OF FUNCTIONS

• The graph of a function is a set of points on

the Cartesian Plane that satisfy the function

• Information is presented in the form of graphs

• Graph are widely used in science and technology

• Graphs are very useful to researchers, scientists

and economist

The different type of functions and respective power of x

Type of

function

General form Example Highest

power of variable x

Linear

Quadratic

Cubic

Reciprocal

y = ax + c

y = ax2 + bx + c

y = ax3 + bx2 + cx + d

y = a

x

y = 3x

y = -4x + 5

y = 2x2

y = -3x2 + 2x y = 2x2 + 5x + 1

y = 2x3

y = -3x3 + 5x

y = 2x3 - 3x + 6

y = 4

x

y = - 2

x

1

3

-1

2

y = ax3

y = ax3 + bxy = ax3 + bx + c

a ≠ 0

LINEAR FUNCTION

QUADRATIC FUNCTION

QUBIC FUNCTION

RECIPROCAL FUNCTION

• Using calculator to complete the tables

• Using the scale given to mark the points

on the x-axis and y-axis

• Plotting all the points using the scale given

CALC MemoryExample 1Calculate the result

for Y = 3X – 5,

when X = 4, and when X = 6

)3X

- 5

CALC

3X – 5

4 = 7

CALC 6 = 13

ALPHA


for Y = X2 + 3X – 12,

when X = 7, and when X = 8

) 3X

x2 +ALPHA

)

X

- 21

CALC

X2 + 3X – 12

7 = 58CALC 8 = 76

ALPHA


for Y = 2X2 + X – 6,

when X = 3, and when X = -3

)

3

X

x2 +ALPHA

)

X

- 6

CALC

2X2 + X – 6

3 = 15CALC (-) = 9

ALPHA

2


for Y = -X3 + 2X + 5,

when X = 2, and when X = -1

)

1

X

x2

+ 2ALPHA

)

X

+ 5

-X3 + 2X + 5 2 = 1CALC (-) = 4

ALPHA

(-)SHIFT x3

CALC

Example 5Calculate the result

for Y = 6 when X = -3,

X

and when X = 0.5

)

XALPHA

3CALC (-) = -2

6

CALC 0 . 125 =

ab/c 6┘x

Example 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

Y = -2X2 + 40

X 0 0.5 1 1.5 2 3 3.5 4

Y

Y = X3 – 3X + 3

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

Y

Y = -16

X

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

Y

40 39.5 38 35.5 32 22 15.5 8

-15 1 5 3 16.25 1.875 51

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

2(4)2 + 5(4) – 1 = 51

Using Calculator

2 ( )4 x2 + 5 ( 4 )

- 1 =

(-2)3 - 12(-2) + 10 = 26

Using Calculator

2( ) x2

+ 1 0

(-) 1

=

SHIFT - 2

2( )(-)

2( ) 3

+ 1 0

(-) 1

=

V

- 2

2( )(-)

OR

Using Calculator

6

(-3)= -2

3( )6 ÷ (-) =

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

y 15 8 0 -1 0 3 15

y = x2 + 2x

y = (

-3

) 2 + 2 (

-3

) = 3

y = (

2

) 2 + 2 (

2

) = 8

3

8

x -4 -2.5 -1 -0.5 0.5 1 2 3.2 4

y 1 1.6 8 -8 -4 -1.25 -1

4

2

y = -4

( )

=

y = -4

( )

=

-1

4 -2-2

y = -4

x

Completing the table of values

-1 2

2.4 3.0 3.9

x-axis scale : 2 cm to 2 units

Marking the points on the x-axis and y-axis

2 4

-4 -2

-3.6 -3.0 -2.1

x-axis scale : 2 cm to 2 units

10

15

12

14.5

13.5

y-axis scale : 2 cm to 5 units

10.75

-20

-15

-18

-15.5

-16.5

y-axis scale : 2 cm to 5 units

-19.25

The x-coordinate and y-coordinate

xA

B

x

x

x

C

D

(-3,2)

(2,0)

(4,-3)

(0,-4)

Ex

(4,4)

Fx

(-7,-2)

-5

-1 1

x

x

x

x

A

B

C

D

Ex (-0.5,2)

(-1,-3)

(0,3)

(0.3,1.5)

(0.5,-1.5)

-2

-1 1

xx

x

x

A

B

CD (-1,-1.2)

(0,1.2)

(0.3,0.6)

(0.2,-1)

-10

-1 1

x

x

x

x

A

B

C

D(-1,-6)

(0,5)

(0.3,3)

(0.5,-3)

2.1 A Drawing the Graphs

Construct a table for a chosen range of x values, for example

-4 ≤ x ≤ 4

Draw the x-axis and the y-axis and suitable scale for each axis

starting from the origin

Plot the x and y values as coordinate pairs on the Cartesian Plane

Join the points to form a straight line (using ruler) or smooth curve

(using French Curve/flexible ruler) with a sharp pencil

Label the graphs

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

2.1 A Drawing the Graphs

Draw the graph of y = 3x + 2

for -2 ≤ x ≤ 2

solution

x

y

-2 0

-4 2 8

0-2-4 2 4

-2

-4

2

4

6

8

x

y

GRAPH OF A LINEAR FUNCTION

8

3 + 2

22

Draw the graph of y = x2 + 2x for -5 ≤ x ≤ 3

solution

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

y 15 8 0 -1 0 3 15

y = x2 + 2x

3 83

+ 22

-3 -3

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

GRAPH OF A QUADRATIC FUNCTION

Draw the graph of y = x3 - 12x + 3 for -4 ≤ x ≤ 4

solution

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

y -13 12 19 14 3 -8 -13 -6 19

y = x3 - 12x + 3

-13

- 123

-4 -4 + 3

0 1 2 3 4-1-2-3-4

-5

-10

-15

5

10

15

20

25

y

x

x

x

x

x

x

x

x

x

x

GRAPH OF A CUBIC FUNCTION

Draw the graph of y = -4 for -4 ≤ x ≤ 4.

x

solution

y = -4

x

x -4 -2.5 -1 -0.5 0.5 1 2 3.2 4

y 1 1.6 4 8 -8 -4 -2 -1.25 -14

-4

-1

1 2 3 4-1-2-3-4 0

y

x

2

4

6

8

-2

-4

-6

-8

X

X

X

X

X

X

X

X

X

GRAPH OF A RECIPROCAL FUNCTION

xx

USING FRENCH CURVE

F:/User/My Documents/jadual 12 PLOTTING.jpg


x

x

x

x

USING FRENCH CURVE



x

x

x

USING FRENCH CURVE



-10

-15-20

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

2.1 B Finding Values of Variable from a Graph

y = x2 + 2x

Find

(a) the value of

y when

x = -3.5

(b) the value of

x when

y = 11

solution

From the graph;

(a)y = 5

(b) X = -4.4, 2.5

x

x

x

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

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

x =1.5

x

y

0 1 2 3 4

8

6

4

2

-2

-4

-6

-8

-4 -3 -2 -1

xx

x

x

x

x

x

x

-2.2

-1.2

1.8

3.4

( a ) y = -2.2

( b ) x = -1.2

Find

(a) the value of

y when

x = 1.8

(b) the value of

x when

y = 3.4

solution

y = -4

x

Values obtained from

the graphs are

approximations

Notes

2.1 C Identifying the shape of a Graph from

a Given Function

LINEAR a

y

x

y = x

0

b

x

y = -x + 2

0

2

y


a Given Function

QUADRATIC a

y

x

y = x2

0

bx

y = -x2

0

y


a Given Function

CUBIC a

y

x

y = x3

0

b

x

y = -x3 + 2

0

2

y


a Given Function

RECIPROCAL a

y

x0

b

x0

y

y = 1

x

y = -1

x

2.1 D Sketching Graphs of Function

• Sketching a graph means drawing a graph without

the actual data

• When we sketch the graph, we do not use

a graph paper, however we must know the important

characteristics of the graph such as its general form

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

• It helps us to visualise the relationship of the variables

EXAMPLE y = 2x + 4

4

-2 0

y

x

find the x-intercept of

y = 2x + 4.

Substitute y = 0

2x + 4 = 0

2x = -4

x = -2

Thus, x-intercept = -2

find the y-intercept of

y = 2x + 4.

Substitute x = 0

y = 2(0) + 4

y = 4

Thus, y-intercept = 4

draw a straight line that

passes x-intercept and y-intercept

y = 2x + 4

A Sketching The Graph of A Linear Function

B Sketching The Graph of A Quadratic Function

EXAMPLE y = -2x2 + 8

a < 0

the shape of the graph is

y-intercept is 8

find the x-intercept of

y = -2x2 + 8.

Substitute y = 0-2x2 + 8 = 0

-2x2 = -8x2 = 4

Thus, x-intercept = -2 and 2

x0

y

-2 2

8

B Sketching The Graph of A Cubic Function

EXAMPLE y = -3x3 + 5

a < 0

the shape of the graph is

y-intercept is 5

x0

y

5

2.2 The Solution of An Equation By Graphical

Method

Solve the equation x2 = x + 2

Solution

x2 = x + 2

x2 - x – 2 = 0

(x– 2)(x + 1) = 0

x = 2, x = -1

2

y

1

3

4

0-1 1-2 2 x

y = x2

y = x + 2

A

B

• Let y = x + 2

and y = x2

• Draw both

graphs on

the same

axes

• Look at the

points of

intersection:

A and B.

Read the

values of the

coordinates

of x.

x = -1 and

x = 2

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


Method

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

when x = -3 and 2.

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

y 15 8 0 -1 0 3 15

(b) By using scale of 2 cm to 1 unit on the x-axis and 2 cm to 2 units on the y-axis,

draw the graph of y = x2 + 2x for -5 ≤ x ≤ 3.

(c) From your graph, find

(i) the value of y when x = -3.5,

(ii) the value of x when y = 11.

(d) Draw a suitable straight line on your graph to find a value of x which satisfies

the equation of x 2 + x – 4 = 0 for -5 ≤ x ≤ 3.

TABLE 1


Method

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

to be drawn to solve each of the following equations.

a) x2 - 5x - 3 = 4

b) x2 - 5x - 3 = 2x + 4

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

d) x2 - 5x - 10 = 0

e) x2 - 7x - 2 = 0

EXAMPLE 1

solution

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

Therefore, y = 4 is the suitable straight line

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

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


to be drawn to solve the equation: x2 - 5x - 3 = 4

y

solution

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

- 1

-1 on both sides

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

x2 - 5x - 2 = x + 4 - 1

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


to be drawn to solve the equation: x2 - 5x – 2 = x + 4

solution

d) x2 - 5x - 10 = 0 Rearrange the equation

Therefore, y = 7 is the suitable straight line

x2 - 5x = 10

x2 - 5x = 10 - 3- 3

x2 - 5x - 3 = 7 7

-3 on both sides



solution

e) x2 - 7x - 2 = 0 Rearrange the equation

Therefore, y = 2x - 1 is the suitable straight line

x2 = 7x + 2

x2 = 7x + 2 - 5x - 3- 5x - 3

x2 - 5x - 3 = 2x -12x -1

-5x - 3 on both sides



Alternative Method

Since a straight line is needed, we used to eliminate the term, x2.

The following method can be used

y = x2 - 5x - 3 1

0 = x2 - 7x - 2 2

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

y = 2x - 1

e




Method

The graph y = 8 is drawn. Determine the suitable straight line

x

to be drawn to solve each of the following equations.

a) 4 = x + 1

x

b) -8 = -2x - 2

x

EXAMPLE 2

solution

4 = x + 1

xMultiply both sides by 2a

We get 8 = 2x + 2

x


2x + 2


x

to be drawn to solve each the equation: 4 = x + 1

x

solution

-8 = -2x - 2

x

Multiply both sides by -1b

We get 8 = 2x + 2

x


2x + 2


x

to be drawn to solve each the equation: - 8 = -2x - 2

x


Method

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

when x = -3 and 2.

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

y 15 8 0 -1 0 3 15

(b) By using scale of 2 cm to 1 unit on the x-axis and 2 cm to 2 units on the y-axis,

draw the graph of y = x2 + 2x for -5 ≤ x ≤ 3.

(c) From your graph, find

(i) the value of y when x = -3.5,

(ii) the value of x when y = 11.

(d) Draw a suitable straight line on your graph to find a value of x which satisfies

the equation of x 2 + x – 4 = 0 for -5 ≤ x ≤ 3.

TABLE 1

12. (a)

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

y 15 8 0 -1 0 3 15

y = x2 + 2x

y = (-3 ) 2 + 2 (-3 ) = 3

y = ( 2 ) 2 + 2 ( 2 ) = 8

8

solution

3

12. (a)

x

x

x

xx

x

x

x

x

y

x0 1 2

2

4

6

8

10

-1-3-4-5

16

14

12

-2 3

12. (c)

x

x

x

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)

x

x

x

xx

x

x

x

x

y

x0 1 2

2

4

6

8

10

-1-3-4-5

16

14

12

-2 3

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

y = x + 4

x 0 -4

y 4 0

x

x1.5-2.5

Answer:

(d) x = 1.5

x = -2.5

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

=

2.3 REGION REPRESENTING INEQUALITIES IN TWO

VARIABLES2.3 A Determining Whether a Given Point Satisfies y = ax + b,

y > ax + b or y < ax + b

How can we determine whether a given point satisfies

y = 3x + 1, y < 3x + 1or y > 3x + 1 ?




Let us consider the point (3,5). The point can only satisfies one of the

following relations:

(a) y = 3x + 1 (b) y < 3x + 1 (c) y > 3x + 1

y 3x + 1

5 3(3) + 1

5 10

=

<

>

<

Since the y-coordinate of the point (3,5) is 5, which is less than 10,

we conclude that y < 3x + 1 . Therefore, the point (3,5) satisfies the relation

y < 3x + 1

<




Determine whether the following points satisfy y = 3x - 1, y < 3x - 1 or

y > 3x - 1.

(a) (1,-1) (b) (3,10) (c) (2,9)

EXAMPLE




For point (1,-1)

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

Since the y-coordinate of the point (1,-1) is -1, which is less than 2,

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

y < 3x - 1

solution a

2.3REGION REPRESENTING INEQUALITIES IN TWO

VARIABLES

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


For point (3,10)

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

Since the y-coordinate of the point (3,10) is 10, which is greater than 8,

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

y > 3x - 1

solution b


VARIABLES2.3 A

For point (-1,-4)

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

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

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

y = 3x - 1

solution c

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



VARIABLES2.3 B

Determining The Position of A Given Point Relative to

y = ax + b

All the points satisfying y < ax + b are below the graph

All the points satisfying y = ax + b are on the graph

All the points satisfying y > ax + b are above the graph


VARIABLES2.3 B


y = ax + b

-2-4

2

4

6

8

x

y

20

-2

-4

4-6

-8

6

P(4,8)

Q(4,2)

y < xy > x

The point P(4,8) lies

above the line y = x.

This region is represented

by y > x

The point Q(4,2) lies

below the line y = x.


by y < x

Q(4,4)


on the line y = x.


by y = x


VARIABLES2.3 B


y = ax + b

-2-4

2

4

6

8

x

y

20

-2

-4

4-8

-8

8

P(-8,6)

Q(4,4)

y < 3x + 2y > 3x + 2

The point P(-8,6) lies

above the line y = 3x + 2.


by y > 3x + 2


below the line y = 3x + 2.


by y < 3x + 2

Q(2,8)


VARIABLES2.3 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


VARIABLES2.3 D Shading The Regions Representing Given Inequalities

Symbol Type of

Line

< or > Dashed

Line

≤ or ≥ Solid line

The type of line to be drawn depends

on inequality symbol

The table above shows the

symbols of inequality and the

corresponding type of line

to be drawn

HoT TiPs

The dashed line indicates that all points

are not included in the region. The solid

line indicates that all points on the line

are included



0x

y

b

0x

y

b

0x

y

b

0x

y

b

y > ax + b

a > 0 y < ax + b

a > 0

y ≥ ax + b

a > 0 y ≤ ax + b

a > 0



0x

y

by >ax + b

a < 0

0x

y

b y ≥ ax + b

a < 0

x

y

x

y

y ≤ ax + b

a < 0 b

0

y < ax + b

a < 0

0

b



0

y

a

x >a

a > 0

x

y

a

x > a

a < 0

x

y

x

y

x ≤ a

a < 0 x ≤ a

a > 0

0a

x0

0 a


VARIABLES

2.3 EDetermine The Region which Satisfies Two or More

Simultaneous Linear Inequalities

y

x

2

0 2 3-3

1

EXAMPLEShade the region that satisfies

3y < 2x + 6, 2y ≥ -x + 2 and x ≤ 3.

X = 3


VARIABLES

y

x

2

0 2 3-3

1

X = 3

Shade the region that satisfies

3y < 2x + 6, 2y ≥ -x + 2 and x ≤ 3.


VARIABLES

y

x

2

0 2 3-3

1

X = 3A

Region A satisfies 2y ≥ -x + 2, 3y < 2x + 6, and x ≤ 3


VARIABLES2.3 E Determine The Region which Satisfies Two or More


y

x

2

0 2 3-3

1

X = 3A

Region A satisfies

2y ≥ -x + 2,

3y < 2x + 6, and x ≤ 3




y

x

2

0 2 3-3

1

X = 3A

Region A satisfies

2y > -x + 2,

3y ≤ 2x + 6, and x < 3




y

x

3

0

x = 3

Region B satisfies

y ≥ -x + 3,

y < x , and x ≤ 3

B




y

x

3

0

x = 3

Region B satisfies

y > -x + 3,

y ≤ x , and x < 3

B




y

x

3

0

x = -3

Region C satisfies

y > -x + 3,

y ≤ -2x , and x >-3

-3

C

x

x

x

xx

x

x

y

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

xx

x

Region A satisfies

y ≥ x2 + 2x,

y ≤ x + 4,

and x ≥ 0

A

x

x

x

xx

x

x

y

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

xx

x

Region A satisfies

y ≥ x2 + 2x,

y ≤ x < 4,

and x ≥ 0

A


y ≤ 2x + 8, y ≥ x, and y < 8

y

x0

y = 8

y = x

y = 2x + 8

SPM Clone

y ≤ 2x + 8

y ≥ x

y < 8


y ≤ 2x + 8, y ≥ x, and y < 8

y

x0

y = 8

y = x

K1

y = 2x + 8 3

SPM Clone

P2

3.y

x0

y = 8

y = x

K2

y = 2x + 8

2

3.y

x0

y = x

K1

y = 2x + 8 1

y

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

y

xO

y = 2x6

y = 6

Solution:

x = 3

K3

x-intercept = -(-6 ÷2) = 3

x < 3, y ≤ 2x – 6 and y ≥ -6

On the graphs provided, shade the region which satisfies

the three inequalities y ≤ x - 4, y ≤ -3x + 12 and y > -4

[3 marks]

y

x

y = 3x+12

O

y = x4

y

x

y = 3x+12

O

y = x4

y = 4

Solution:

K3

y-intercept =-4

y ≤ x - 4, y ≤ -3x + 12 and y > -4

SPM 2003 PAPER 2

REFER TO QUESTION

NO. 12

12. ( a )

x -4 -2.5 -1 -0.5 0.5 1 2 3.2 4

y 1 1.6 8 -8 -4 -1.25 -1

4

2

K1K1

y = -4

( )

=

y = -4

( )

=

-1

4 -2-2

1 2 3 4-1-2-3-4 0

y

x

2

4

6

8

-2

-4

-6

-8

X

X

X

X

X

X

X

X

X

K1

K1N1

K1N1

12(b)

12. ( a )

x -4 -2.5 -1 -0.5 0.5 1 2 3.2 4

y 1 1.6 8 -8 -4 -1.25 -14 2 K1K0

1 2 3 4-1-2-3-4 0

y

x

2

4

6

8

-2

-4

-6

-8

X

X

XX

X

X

X

X

X

K1

K1N1

K1 N0

12(b)

x

y

0 1 2 3 4

8

6

4

2

-2

-4

-6

-8

-4 -3 -2 -1

xx

x

x

x

x

x

12. ( c )

x

-2.2

-1.2

1.8

3.4

( i ) y = -2.2

( ii ) x = -1.2

P1

P1

12. (d)

x

y

0 1 2 3 4

8

6

4

2

-2

-4

-6

-8

-4 -3 -2 -1

xx

x

x

x

x

x

x

y = -2x - 3

-2.4

0.8

K1K1

x = - 2.4

x = 0.8

N1

4 = 2x + 3

x

- 4 = -2x - 3

x

graphs of functions 2

Documents

Transcript of graphs of functions 2