2.0 Graphs of Functions 2

8/3/2019 2.0 Graphs of Functions 2

1/123


2/123


3/123


4/123

CHAPTER 2GRAPHS OF FUNCTIONS II

2.1 GRAPHS OF FUNCTIONS

The graph of a function is a set of points on

the Cartesian Plane that satisfy the function

Information is presented in the form of graphs

Graph are widely used in science and technology

Graphs are very useful to researchers, scientists

and economist


5/123

The different type of functions and respective power of x

Type of

function

General form Example Highest

power of variable x

Linear

Quadratic

Cubic

Reciprocal

y = ax + c

y = ax2 + bx + c

y = ax3 + bx2 + cx + d

y = a

x

y = 3x

y = -4x + 5

y = 2x2

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

y = 2x3

y = -3x3 + 5x

y = 2x

3

- 3x + 6

y = 4

x

y = - 2

x

1

3

-1

2

y = ax3

y = ax3 + bxy = ax3 + bx + c

a 0


6/123

LINEAR FUNCTION


7/123

LINEAR FUNCTION


8/123

QUADRATIC FUNCTION


9/123

QUADRATIC FUNCTION


10/123

QUBIC FUNCTION


11/123

QUBIC FUNCTION


12/123

RECIPROCAL FUNCTION


13/123

RECIPROCAL FUNCTION


14/123

Using calculator to complete the tables

Using the scale given to mark the pointson the x-axis and y-axis

Plotting all the points using the scale given


15/123

COMP


16/123

CALC MemoryCALC MemoryExample 1Example 1Calculate the resultforY = 3X 5,

when X = 4, and when X = 6

)3X

-5

CALC

3X 5

4 = 7CALC 6 = 13

ALPHA


17/123

CALC MemoryCALC MemoryExample 2Example 2Calculate the resultforY = X2 + 3X 12,when X = 7, and when X = 8

) 3X

x2 +ALPHA

)X

- 21CALC X2 + 3X 12

7 = 58CALC

8 = 76

ALPHA


18/123

CALC MemoryCALC MemoryExample 3Example 3Calculate the resultforY = 2X2 + X 6,when X = 3, and when X = -3

)

3

X

x2 +ALPHA

)X

- 6CALC 2X2

+ X 63 = 15

CALC (-) =9

ALPHA

2


19/123

CALC MemoryCALC MemoryExample 4Example 4Calculate the resultforY = -X3 + 2X + 5,when X = 2, and when X = -1

)

1

X

x2

+ 2

ALPHA

)

X

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

(-) = 4

ALPHA

(-)SHIFT x3

CALC


20/123

Example 5Example 5Calculate the resultforY = 6 when X = -3,

X

and when X = 0.5

)XALPHA

3CALC (-) = -2

6

CALC 0 . 125 =

ab/c 6x


21/123

Example 6Example 6Calculate the resultforY = 6 when X = -3,

X

and when X = 0.5

)XALPHA

3CALC (-) = -2

6

CALC 0 . 125 =

x-1 6x-1


22/123

Y = -2X2 + 40

X 0 0.5 1 1.5 2 3 3.5 4

Y

Y = X3 3X + 3

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

Y = -16XX -4 -3 -2 -1 1 2 3 4

Y

40 39.5 38 35.5 32 22 15.5 8

-15 1 5 3 16.25 1.875 51

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


23/123

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

2 ( )4 x2 + 5 ( 4 )

- 1 =


24/123

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

Using Calculator

2( ) x2

+ 1 0

(-) 1

=

SHIFT - 2

2( )(-)

2( ) 3

+ 1 0

(-) 1

=

V- 2

2( )(-)

OR


25/123

Using Calculator6

(-3)= -2

3( )6 (-) =


26/123

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

y 15 8 0 -1 0 3 15

y = x2 + 2x

y = (

-3

) 2 + 2 (

-3

) = 3

y = (

2

)2

+ 2 (

2

) = 8

3

8


27/123

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

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

4

2

y = -4( )

=

y = -4( )

=

-1

4 -2-2

y = -4

x

Completing the table of values

-1 2


28/123

2.4 3.0 3.9

x-axis scale : 2 cm to 2 units

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

2 4


29/123

-4 -2

-3.6 -3.0 -2.1

x-axis scale : 2 cm to 2 units


30/123

10

15

12

14.5

13.5

y-axis scale : 2 cm to 5 units

10.75


31/123

-20

-15

-18

-15.5

-16.5

y-axis scale : 2 cm to 5 units

-19.25


32/123

The x-coordinate and y-coordinate


33/123

xA

B

x

x

x

C

D

(-3,2)

(2,0)

(4,-3)

(0,-4)

Ex

(4,4)

F

x

(-7,-2)


34/123

-5

-1 1

x

x

xx

A

B

C

D

Ex (-0.5,2)

(-1,-3)

(0,3)

(0.3,1.5)

(0.5,-1.5)


35/123

-2

-1 1

x x

xx

A

B

CD (-1,-1.2)

(0,1.2)

(0.3,0.6)

(0.2,-1)


36/123

-10

-1 1

x

x

xx

A

B

C

D(-1,-6)

(0,5)

(0.3,3)

(0.5,-3)


37/123

2.1 A Drawing the Graphs

Construct a table for a chosen range of x values, for example-4 x 4

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

starting from the origin

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

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

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

Label the graphs

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


38/123

2.1 A Drawing the Graphs

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

solution

x

y

-2 0

-4 2 8

0-2-4 2 4

-2

-4

2

4

6

8

x

y

GRAPH OF A LINEAR FUNCTION

8

3 + 2

22


39/123

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

solution

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

y 15 8 0 -1 0 3 15

y = x2 + 2x

3 83

+ 22

-3 -3

y


40/123

x

x

x

x

x

x

x

x

x

y

x0 1 2

2

4

6

8

10

-1-3-4-5

16

14

12

-2 3

y = x2 + 2x

GRAPH OF A QUADRATIC FUNCTION


41/123

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

solution

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

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

y = x3 - 12x + 3

-13

- 123

-4 -4 + 3


42/123

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

-5

-10

-15

5

10

15

20

25

y

x

x

x

x

x

x

x

x

xx

GRAPH OF A CUBIC FUNCTION


43/123

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

x

solution

y = -4

x

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

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

-4

-1


44/123

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

y

x

2

4

6

8

-2

-4

-6

-8

X

X

X

X

X

X

X

X

X

GRAPH OF A RECIPROCAL FUNCTION


45/123

xx

USING FRENCH CURVE


46/123

x

x

x

x

USING FRENCH CURVE


47/123

x

x

x

USING FRENCH CURVE


48/123

-10

-15

-20


49/123


50/123


51/123


52/123

x

x

x

x

x

x

x

x

x

y

x0 1 2

2

4

6

8

10

-1-3-4-5

16

14

12

-2 3-3.5

5

y = 11

-4.4 2.5

2.1 B Finding Values of Variable from a Graph

y = x2 + 2x

Find

(a)the value of

y whenx = -3.5

(b) the value of

x when

y = 11

solution

From the graph;

(a)y = 5

(b) X = -4.4, 2.5


53/123

x

x

x

x

x

x

x

x

x

y

x0 1 2

2

4

6

8

10

-1-3-4-5

16

14

12

-2 3-3.5

5


y = x2 + 2x


54/123

x

x

x

x

x

x

x

x

x

y

x0 1 2

2

4

6

8

10

-1-3-4-5

16

14

12

-2 3

5

11


y = x2 + 2x-4.4 2.5


55/123

x =1.5

y


56/123

x

y

0 1 2 3 4

8

6

4

2

-2

-4

-6

-8

-4 -3 -2 -1

xx

x

x

x

x

x

x

-2.2

-1.2 1.8

3.4

( a ) y = -2.2

( b ) x = -1.2

Find

(a)the value of

y when

x = 1.8

(b) the value of

x when

y = 3.4

solution

y = -4

x

Values obtained fromthe graphs are

approximations

Notes

Id tif i th h f G h f


57/123

2.1 C Identifying the shape of a Graph from

a Given Function

LINEAR a

y

x

y = x

0

b

x

y = -x + 2

0

2

y

Id tif i g th h f G h f


58/123


a Given Function

QUADRATIC a

y

x

y = x2

0

bx

y = -x2

0

y



59/123


a Given Function

CUBIC a

y

x

y = x3

0

b

x

y = -x3 + 2

0

2

y



60/123


a Given Function

RECIPROCAL a

y

x0

b

x0

y

y = 1x

y = -1

x


61/123

2.1 D Sketching Graphs ofFunction

Sketching a graph means drawing a graph withoutthe actual data

When we sketch the graph, we do not use

a graph paper, however we must know the important

characteristics of the graph such as its general form

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

It helps us to visualise the relationship of the variables


62/123

EXAMPLE y = 2x + 4

4

-2 0

y

x

find the x-intercept ofy = 2x + 4.

Substitute y = 0

2x + 4 = 0

2x = -4

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

find the y-intercept of

y = 2x + 4.

Substitute x = 0

y = 2(0) + 4y = 4

Thus, y-intercept = 4

draw a straight line that

passes x-intercept and y-intercept

y = 2x + 4

A Sketching The Graph of A Linear Function


63/123

B Sketching The Graph of A Quadratic Function

EXAMPLE y = -2x2 + 8

a < 0

the shape of the graph is

y-intercept is 8

find the x-intercept of

y = -2x2 + 8.

Substitute y = 0-2x2 + 8 = 0

-2x2 = -8

x2 = 4Thus, x-intercept = -2 and 2

x0

y

-2 2

8


64/123

B Sketching The Graph of A Cubic Function

EXAMPLE y = -3x3 + 5

a < 0

the shape of the graph is

y-intercept is 5

x0

y

5


65/123

2.2 The Solution of An Equation By Graphical

Method

Solve the equation x2 = x + 2

Solution

x2

= x + 2

x2 - x 2 = 0

(x 2)(x + 1) = 0

x = 2, x = -1


66/123

2

y

1

3

4

0-1 1-2 2 x

y = x2

y = x + 2

A

B

Let y = x + 2

and y = x2

Draw both

graphs on

the same

axes

Look at thepoints of

intersection:

A and B.

Read the

values of the

coordinates

of x.

x = -1 and

x = 2

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

The Solution of An Equation By Graphical


67/123


Method

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

when x = -3 and 2.

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

y 15 8 0 -1 0 3 15

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

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

(c) From your graph, find(i) the value of y when x = -3.5,

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

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

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

TABLE 1

2 2 The Solution of An Equation By Graphical


68/123


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


69/123

solution

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

Therefore, y = 4 is the suitable straight line

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

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


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

y


70/123

solution

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

- 1

-1 on both sides

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

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

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


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


71/123

solution

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

Therefore, y = 7 is the suitable straight line

x2 - 5x = 10

x2 - 5x = 10 - 3- 3

x2 - 5x - 3 = 77

-3 on both sides




72/123

solution

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

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

x2 = 7x + 2

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

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

-5x - 3 on both sides




73/123

Alternative Method

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

The following method can be used

y = x2 - 5x - 3 1

0 = x2 - 7x - 2 2

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

y = 2x - 1

e





74/123


Method

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

x

to be drawn to solve each of the following equations.

a) 4 = x + 1

x

b) -8 = -2x - 2x

EXAMPLE 2


75/123

solution

4 = x + 1xMultiply 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


76/123

solution

-8 = -2x - 2

x

Multiply both sides by -1

b

We get 8 = 2x + 2

x


2x + 2


x

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

x



77/123


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


78/123

12. (a)

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

y 15 8 0 -1 0 3 15

y = x2 + 2x

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

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

8

solution

3

12. (a) y


79/123

12. (a)

x

x

x

xx

x

x

x

x

x0 1 2

2

4

6

8

10

-1-3-4-5

16

14

12

-2 3

12. (c)


80/123

( )

x

x

x

xx

x

x

x

x

y

x0 1 2

2

4

6

8

10

-1-3-4-5

16

14

12

-2 3-3.5

5

y = 11

-4.4 2.5

Answer:

(i) y = 5.0

(ii) x = -4.4

x = 2.5

12. (d)


81/123

( )

x

x

x

xx

x

x

x

x

y

x0 1 2

2

4

6

8

10

-1-3-4-5

16

14

12

-2 3

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

y = x +4

x 0 -4

y 4 0

x

x1.5-2.5

Answer:

(d) x = 1.5

x = -2.5


82/123

a 2 + b + c 0


83/123

ax2 + bx + c = 0

x2+ x 4 = 0

a = 1 b = 1 c = -4

MODE EQN

1 1Unknowns ?

2 3Degree?

2 3

2 a ? 1 = b ? 1 = c ?

(-) 4 x1 = 1.561552813 = x2 = -2.561552813

Press 3x

=


84/123

2.3 REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES

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

y > ax + b or y < ax + b

How can we determine whether a given point satisfiesy = 3x + 1, y < 3x + 1or y > 3x + 1 ?

2 3 REGION REPRESENTING INEQUALITIES IN TWO VARIABLES


85/123




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

following relations:

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

y 3x + 1

5 3(3) + 1

5 10

=

3x - 1.

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

EXAMPLE

2 3 REGION REPRESENTING INEQUALITIES IN TWO VARIABLES


87/123




For point (1,-1)

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

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

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

solution a

REGION REPRESENTING INEQUALITIES IN TWO VARIABLES


88/123

2.3REGIONREPRESENTING INEQUALITIESIN TWO VARIABLES



For point (3,10)

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

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

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

solution b


89/123


2.3 A

For point (-1,-4)

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

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

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

solution c

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



90/123


2.3 BDetermining The Position of A Given Point Relative to

y = ax + b

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

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

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



91/123

Q


y = ax + b

-2-4

2

4

6

8

x

y

20

-2

-4

4-6

-8

6

P(4,8)

Q(4,2)

y < xy > x

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

This region is represented

by y > x

The point Q(4,2) liesbelow the line y = x.


by y < x

Q(4,4)

The point Q(4,4) lies

on the line y = x.This region is represented

by y = x



92/123


y = ax + b

-2-4

2

4

6

8

x

y

20

-2

-4

4-8

-8

8

P(-8,6)

Q(4,4)

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

The point P(-8,6) lies

above the line y = 3x + 2.

This region is representedby y > 3x + 2

The point Q(4,4) lies

below the line y = 3x + 2.


by y < 3x + 2

Q(2,8)



93/123

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

-2-4

2

4

6

8

x

y

20

-2

-4

4-8

-8

8

Determine whether the

shaded region in the graph

satisfies y < 3x + 2 ory > 3x + 2

EXAMPLE

solution

The shaded region is

below the graph, y = 3x + 2.

Hence, this shaded region

satisfies y< 3x + 2



94/123

2.3 D Shading The Regions Representing Given Inequalities

Symbol Type ofLine

< or > Dashed

Line

or Solid line

The type of line to be drawn depends

on inequality symbol

The table above shows thesymbols of inequality and thecorresponding type of line

to be drawn

HoT TiPsThe dashed line indicates that all points

are not included in the region. The solid

line indicates that all points on the line

are included



95/123


0

x

y

b

0x

y

b

0x

y

b

0x

y

b

y > ax + b

a > 0y < ax + b

a > 0

y ax + b

a > 0y ax + b

a > 0



96/123


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



97/123


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



98/123

2.3 EDetermine The Region which Satisfies Two or More

Simultaneous Linear Inequalities

y

x

2

0 2 3-3

1

EXAMPLEShade the region that satisfies

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

X = 3



99/123

y

x

2

0 2 3-3

1

X = 3

Shade the region that satisfies

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



100/123

y

x

2

0 2 3-3

1

X = 3A

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



101/123

2.3 E Determine The Region which Satisfies Two or More


y

x

2

0 2 3-3

1

X = 3A

Region A satisfies2y -x + 2,

3y < 2x + 6, and x 3



102/123



y

x

2

0 2 3-3

1

X = 3A

Region A satisfies

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



103/123



y

x

3

0

x = 3

Region B satisfies

y -x + 3,

y < x , and x 3

B



104/123



y

x

3

0

x = 3

Region B satisfies

y > -x + 3,

y x , and x < 3

B



105/123



y

x

3

0

x = -3Region C satisfies

y > -x + 3,

y -2x , and x >-3

-3

C

y

Region A satisfies


106/123

x

x

x

xx

x

x

x0 1 2

2

4

6

8

10

-1-3-4-5

16

14

12

-2 3x

1.5-2.5

y = x2 + 2x

y = x + 4

x

x

x

Region A satisfies

y x2 + 2x,

y x + 4,

and x 0

A

y

Region A satisfies


107/123

x

x

x

xx

x

x

x0 1 2

2

4

6

8

10

-1-3-4-5

16

14

12

-2 3x

1.5-2.5

y = x2 + 2x

y = x + 4

x

x

x

Region A satisfies

y x2 + 2x,

y x < 4,

and x 0

A

ySPM Clone


108/123


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


109/123


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

x0

y = 8

y = x

K1

y = 2x +8 3

SPM Clone

P2

3y


110/123

3.

x0

y = 8

y = x

K2

y = 2x +82


111/123


112/123


113/123

y

xO

y = 2x6

y = 6

Solution:

x = 3

K3

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

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


114/123

On the graphs provided, shade the region which satisfies

the three inequalities y x - 4, y -3x + 12 and y > -4[3 marks]

y

x

y = 3x+12

O

y = x4


115/123

y

x

y = 3x+12

O

y = x4

y = 4

Solution:

K3

y-intercept =-4

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

SPM 2003 PAPER 2


116/123

SPM 2003 PAPER 2

REFER TO QUESTIONNO. 12

12 ( a )


117/123

12.( a )

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

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

4

2

K1K1

y = -4

( )

=

y = -4

( )

=

-1

4 -2-2


118/123

12. ( a )


119/123

12.( a )

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

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

y

8X 12(b)


120/123

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

2

4

6

8

-2

-4

-6

-8

X

X

XX

X

X

X

X

X

K1

K1N1

K1 N0

( )

y

8x12.( c )


121/123

x0 1 2 3 4

6

4

2

-2

-4

-6

-8

-4 -3 -2 -1

xx

x

x

x

x

x

-2.2

-1.2

1.8

3.4

( i ) y = -2.2

( ii ) x = -1.2

P1

P1

y

8x

4 = 2x +3

x


122/123

12.(d)

x0 1 2 3 4

6

4

2

-2

-4

-6

-8

-4 -3 -2 -1

xx

x

x

x

x

x

y = -2x - 3

-2.4

0.8

K1K1

x = - 2.4

x = 0.8

N1

x

- 4 = -2x - 3

x


123/123

Documents

2.0 Graphs of Functions 2