Introduction This Chapter focuses on sketching Graphs We will also be looking at using them to solve...

Introduction

• This Chapter focuses on sketching Graphs

• We will also be looking at using them to solve Equations

• There will also be some work on Graph transformations

Sketching CurvesSketching Cubics

You need to be able to sketch equations of the form:

This involves finding the places where the graph crosses the axes, in the same way you do when sketching a Quadratic.

4A

3 2y ax bx cx d

( )( )( )y x a x b x c

or

A cubic equation will take one of the following shapes

For any x3

For any -x3




4A

3 2y ax bx cx d

( )( )( )y x a x b x c

or

ExampleSketch the graph of the function:

( 2)( 1)( 1)y x x x

If y = 0

0 ( 2)( 1)( 1)x x x

So x = 2, 1 or -1(-1,0) (1,0) and (2,0)

If x = 0

(0 2)(0 1)(0 1)y

So y = 2

(0,2)




4A

3 2y ax bx cx d

( )( )( )y x a x b x c

or


( 2)( 1)( 1)y x x x

(-1,0) (1,0) (2,0) (0,2)

x

y

2

2

-1

1

If we substitute in x = 3, we get a value of y = 8. The curve must be increasing after this

point…




4A

3 2y ax bx cx d

( )( )( )y x a x b x c

or


( 2)(1 )(1 )y x x x

If y = 0

0 ( 2)(1 )(1 )x x x

So x = 2, 1 or -1(-1,0) (1,0) and (2,0)

If x = 0

(0 2)(1 0)(1 0)y

So y = -2

(0,-2)




4A

3 2y ax bx cx d

( )( )( )y x a x b x c

or


( 2)(1 )(1 )y x x x

(-1,0) (1,0) (2,0) (0,-2)

x

y

2

-2

-1

1

If we substitute in x = 3, we get a value of y = -8. The curve must be decreasing after this

point…




4A

3 2y ax bx cx d

( )( )( )y x a x b x c

or


2( 1) ( 1)y x x

If y = 0

20 ( 1) ( 1)x x So x = 1 or -1

(-1,0) and (1,0)

If x = 0

2(0 1) (0 1)y So y = 1

(0,1)




4A

3 2y ax bx cx d

( )( )( )y x a x b x c

or


2( 1) ( 1)y x x

(-1,0) (1,0) (0,1)

x

y

1

-1

1


point…

‘repeated root’




4A

3 2y ax bx cx d

( )( )( )y x a x b x c

or


3 22 3y x x x

If y = 0

0 ( 3)( 1)x x x So x = 0, 3 or -1

(0,0) (3,0) and (-1,0)

If x = 0

0(0 3)(0 1)y So y = 0

(0,0)

2( 2 3)y x x x

( 3)( 1)y x x x

Factorise

Factorise fully




4A

3 2y ax bx cx d

( )( )( )y x a x b x c

or


3 22 3y x x x

(0,0) (3,0) (-1,0)

x

y

0-1

3


point…

Sketching Curves

Sketching CubicsYou need to be able to sketch and interpret cubics that are variations of y = x3

This will be covered in more detail in C2. You can still plot the graphs in the same way we have seen before. This topic is offering a ‘shortcut’ if you can understand it.

4B


3y x

x

yy = x3

Sketching Curves



4B


3y x

x

yy = x3

y = -x3

A cubic with a negative ‘x3’ will be reflected in the x-axis

‘Whatever you get for x3, you now have the negative of

that..’

5

-5

Sketching Curves



4B


3( 1)y x

x

yy = x3

When a value ‘a’ is added to a cubic, inside a bracket, it is a

horizontal shift of ‘-a’

‘I will now get the same values for y, but with values of x that

are 1 less than before’

y = (x + 1)3

1

When x = 0:

3(0 1)y 1y

y-intercept

-1

Sketching Curves



4B


3(3 )y x

x

yy = x3

y = (3 - x)3

27

When x = 0:

3(3 0)y 27y

y-intercept

3(3 )y x 3( 3)y x

Reflected in the x-

axis

Horizontal shift, 3 to the right

3

Sketching CurvesThe Reciprocal Function

You need to be able to sketch the ‘reciprocal’ function. This takes the form:

Where ‘k’ is a constant.

ky

x

ExampleSketch the graph of the function 1

yx

and its asymptotes.

124-4-2-1y

10.50.25-0.25-0.5-1x x

y

y = 1/x

4C

You cannot divide by 0, so you get no value at this point

These are where the

graph ‘never reaches’, in this case the

axes…




ky

x


yx

and its asymptotes.

x

y

y = 1/x

4C

y = 3/x

The curve will be the same, but further out…




ky

x


yx

and its asymptotes.

x

y

y = 1/x

4C

y = -1/x

The curve will be the same, but reflected in

the x-axis

Sketching CurvesSolving Equations and Sketching

You need to be able to sketch 2 equations on a set of axes, as well as solve equations based on graphs.

ExampleOn the same diagram, sketch the

following curves:

4D

( 3)y x x 2 (1 )y x x and

x

y

( 3)y x x Quadratic ‘U’ shapeCrosses through 0 and 3

0 3

( 3)y x x

2 (1 )y x x Cubic ‘negative’ shapeCrosses through 0

and 1. The ‘0’ is repeated so just

‘touched’

1

2 (1 )y x x




following curves:

4D

( 3)y x x 2 (1 )y x x andFind the co-ordinates of the points of

intersection

These will be where the graphs are equal…

x

y

0 3

( 3)y x x

1

2 (1 )y x x

2( 3) (1 )x x x x 2 2 33x x x x 3 3 0x x 2( 3) 0x x

Expand bracketsGroup

together

Factorise

0x 2 3 0x 2 3x

3x




following curves:

4D

( 3)y x x 2 (1 )y x x andFind the co-ordinates of the points of

intersection

These will be where the graphs are equal…2( 3) (1 )x x x x

2 2 33x x x x 3 3 0x x 2( 3) 0x x

Expand bracketsGroup

together

Factorise

0x 2 3 0x 2 3x

3x

( 3)y x x

x=-√3 x=0 x=√3

( 3)y x x ( 3)y x x ( 3)y x x

3( 3 3)y 0(0 3)y

0y 3 3 3y

3( 3 3)y

3 3 3y

(0,0)(-√3 , 3+3√3) (√3 , 3-3√3)




following curves:

4D

2 ( 1)y x x 2

yx

and

x

y

2 ( 1)y x x Cubic ‘positive’ shapeCrosses through 0

and 1. The ‘0’ is repeated.

02

yx

Reciprocal ‘positive’ shape

Does not cross any axes

1

y = 2/x



How does the graph show there are 2 solutions to the equation..


following curves:

4D

2 ( 1)y x x 2

yx

and

x

y

0 1

y = 2/x2 2( 1) 0x x

x

2 2( 1)x x

x

2 2( 1) 0x x

x

Set equations equal, and re-

arrange

And they cross in 2 places…

Sketching CurvesMore Transformations

You have seen that a curve with the following function:

Will be transformed horizontally ‘-a’ units.

A curve with this function:

Will be transformed vertically ‘a’ units

4E

( )f x a

( )f x a

f(x)f(x + 2)

f(x) + 2

x

y

2 units left

2 units up

f(x + 2) The x values reduce by 2 for the same y values

f(x) + 2 The y values from the original function increase by 2


Sketch the following functions:

f(x) = x2

Standard curve Label known points

g(x) = (x + 3)2

Moved 3 units left Work out new ‘key points’

h(x) = x2 + 3 Moved 3 units up Work out new ‘key points’

4E

x

y f(x)

0

x

y g(x)

-3x

y h(x)

3

9


Given that:

i) f(x) = x3

Sketch the curve where y = f(x - 1). State any locations where the graphs crosses the axes.

f(x) = x3

f(x – 1) = (x – 1)3

So for this curve, when x = 0, y = -1It therefore crosses at y = -1

4E

f(x)

0x

y

f(x – 1)

1-1


Given that:

i) g(x) = x(x – 2)

Sketch the curve where y = g(x + 1). State any locations where the graphs crosses the axes.

g(x) is a positive quadratic crossing at 0 and 2.

g(x) = x(x – 2)

g(x + 1) = (x + 1)(x + 1 – 2) g(x + 1) = (x + 1)(x – 1)

So for this curve, when x = 0, y = -1It therefore crosses at y = -1

4E

g(x)

0x

y

1-1 2

g(x + 1)

x’s replaced with ‘x + 1’

-1


Given that:i) h(x) = 1/x

Sketch the curve where y = h(x) + 1. State any locations where the graphs crosses the axes and the equations of any asymptotes.

h(x) is a positive reciprocal graph

h(x) = 1/x

h(x) + 1 = 1/x + 1

The asymptotes are: x = 0 (the y-axis)

y = 1

It will cross the x-axis at -1 since this value will make the equation = 0

4E

h(x)

x

y

1

-1

h(x) + 1

Sketching CurvesEven more Transformations

You also need to be able to perform transformations of the form:

this is a horizontal stretch of 1/a.

You also need to know:

this is a vertical stretch by factor ‘a’

4F

( )f ax

( )af x

(2 )y f x( )y f x‘We will get the same y values, using half the x

values’

This is because the x values get multiplied by 2

before the y values are worked out

2 ( )y f x( )y f x‘We will get y values twice as big, using the same x

values’

This is because when we work out the y values,

they are doubled after


Given that f(x) = 9 – x2, sketch the curve with equation;

a) y = f(2x)

Sketch the original curve, working out key points.

If x = 0

If y = 0

4F

x

y f(x)

-3 3

9

29y x

(3 )(3 )y x x

9y

29y x

0 (3 )(3 )x x

(0,9)

(3,0) (-3,0)



a) y = f(2x)

Substitute ‘2x’ in place of ‘x’

If x = 0

If y = 0

4F

x

y f(x)

-3 3

9

x

y f(2x)

-1.5 1.5

929 (2 )y x

(3 2 )(3 2 )y x x

9y

29 4y x

0 (3 2 )(3 2 )x x

(0,9)

(-1.5,0)

(1.5,0)

29 4y x



a) y = 2f(x)

f(x), the original equation, is doubled..

If x = 0

If y = 0

4F

x

y f(x)

-3 3

9

x

y 2f(x)

-3 3

18

29y x

2(3 )(3 )y x x

18y

22(9 )y x

0 2(3 )(3 )x x

(0,18)

(3,0) (-3,0)

22(9 )y x

Summary

• We have learnt the shapes of several different curves

• We have learnt how to apply transformations to those curves

• We have also looked at how to work out the ‘key points’

Introduction This Chapter focuses on sketching Graphs We will also be looking at using them to solve...

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Transcript of Introduction This Chapter focuses on sketching Graphs We will also be looking at using them to solve...