Unit 2 EF 1.3 Higher Higher Maths Composite Functions Exponential and Log Graphs Graph Transformations Trig

Unit 2 EF 1.3

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Higher

Higher MathsHigher Maths

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Composite Functions

Exponential and Log Graphs

Graph Transformations

Trig Graphs

Inverse functionMindmap

Exam Question Type

Derivative Graphs f’(x)

Completing the Square

Solving equations / /Inequations

Unit 2 EF 1.3

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Higher

Graph Transformations

We will investigate f(x) graphs of the form

1. f(x) ± k

2. f(x ± k)

3. -f(x)

4. f(-x)

5. kf(x)

6. f(kx)

Each affect the

Graph of f(x) in a certain

way !

f(x)

0 2 4 6 8x

-2-4-6

2

4

6

-2

-4

-6

Transformation f(x) ± k

(x , y) (x , y ± k)

Mapping

f(x) + 5 f(x) - 3

f(x)

Transformation f(x) ± k

Keypoints

y = f(x) ± k

moves original f(x) graph vertically up or down

+ k move up

- k move down

Only y-coordinate changes

NOTE: Always state any coordinates given on f(x)

on f(x) ± k graph

Demo

f(x) - 2

A(-1,-2) B(1,-2)C(0,-3)

f(x) + 1

B(90o,0)

A(45o,0.5)

C(135o,-0.5)

B(90o,1)

A(45o,1.5)

C(135o,0.5)

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 3C

f(x)

0 2 4 6 8x

-2-4-6

2

4

6

-2

-4

-6

Transformation f(x ± k)

(x, y) (x ± k , y)

Mapping

f(x - 2) f(x + 4)

f(x)

Transformation f(x ± k)

Keypoints

y = f(x ± k)

moves original f(x) graph horizontally left or right

+ k move left

- k move right

Only x-coordinate changes


on f(x ± k) graph

Demo

f(x)

0 2 4 6 8x

-2-4-6

2

4

6

-2

-4

-6

Transformation -f(x)

(x, y) (x , -y)

Mapping

f(x)

Flip inx-axis

Flip inx-axis

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 3E

Transformation -f(x)

Keypoints

y = -f(x)

Flips original f(x) graph in the x-axis

y-coordinate changes sign


on -f(x) graph

Demo

- f(x)

A(-1,0) B(1,0)

C(0,1)

- f(x)

B(90o,0)

A(45o,0.5)

C(135o,-0.5)A(45o,-0.5)

C(135o,0.5)

Unit 2 EF 1.3

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Extra Practice

HHM Ex 3G

f(x)

0 2 4 6 8x

-2-4-6

2

4

6

-2

-4

-6

Transformation f(-x)

(x, y) (-x , y)

Mapping

f(x)

Flip iny-axis

Flip iny-axis

Transformation f(-x)

Keypoints

y = f(-x)

Flips original f(x) graph in the y-axis

x-coordinate changes sign


on f(-x) graph

Demo

f(-x)

B(0,0)

C’(-1,1)

A’(1,-1)A(-1,-1)

C (1,1)

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 3I

f(x)

0 2 4 6 8x

-2-4-6

2

4

6

-2

-4

-6

Transformation kf(x)

(x, y) (x , ky)

Mapping

f(x)

Stretch iny-axis

2f(x) 0.5f(x)

Compress iny-axis

Transformation kf(x)

Keypoints

y = kf(x)

Stretch / Compress original f(x) graph in the

y-axis direction

y-coordinate changes by a factor of k


on kf(x) graph

Demo

f(x)

0 2 4 6 8x

-2-4-6

2

4

6

-2

-4

-6

Transformation f(kx)

(x, y) (1/kx , y)

Mapping

f(x)

Compress inx-axis

f(2x) f(0.5x)

Stretch inx-axis

Transformation f(kx)

Keypoints

y = f(kx)

Stretch / Compress original f(x) graph in the

x-axis direction

x-coordinate changes by a factor of 1/k


on f(kx) graph

Demo

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 3K & 3M

Unit 2 EF 1.3

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Higher

You need to be able to work with combinations

Combining Transformations

Demo

(1,3)

(-1,-3)

(1,3)

(-1,-3)

2f(x) + 1

f(0.5x) - 1

f(-x) + 1

-f(x + 1) - 3

Explain the effect the following have

(a)-f(x)

(b)f(-x)

(c)f(x) ± k


(d)f(x ± k)

(e)kf(x)

(f)f(kx)

Name :

(-1,-3)

(1,3)

(-1,-3)

(1,3)

f(x + 1) + 2

-f(x) - 2

(1,3)

(-1,-3)

(-1,-3)

(1,3)

(1,3)

(-1,-3)

(1,-2)

(1,3)(-1,4)

(-1,-3)(1,-5)

(1,3)

(-1,1)

(-1,-3)

(0,5)

2f(x) + 1

f(0.5x) - 1

f(-x) + 1

-f(x + 1) - 3


(a)-f(x) flip in x-axis

(b)f(-x) flip in y-axis

(c)f(x) ± k move up or down


(d)f(x ± k) move left or right

(e)kf(x) stretch / compressin y

direction

(e)f(kx) stretch / compress

in x direction

Name :

(-1,-3)

(1,3)

(-1,-3)

(1,3)

(-2,-1)

f(x + 1) + 2

-f(x) - 2

(-2,0)

(1,3)

(0,-6)

(-1,-3)

(-1,-3)

(1,3)

(1,7)

(-1,-5)

(2,2)

(1,3)

(-2,-4)(-1,-3)

The diagram shows the graph of a function f.

f has a minimum turning point at (0, -3) and a

point of inflexion at (-4, 2).

a) sketch the graph of y = f(-x).

b) On the same diagram, sketch the graph of y = 2f(-x)

Graphs & Functions Higher

a) Reflect across the y axis

b) Now scale by 2 in the y direction-1 3 4

2

y = f(-x)

-3

y

x

4

y = 2f(-x)

-6


Part of the graph of is shown in the diagram.

On separate diagrams sketch the graph of

a) b)

Indicate on each graph the images of O, A, B, C, and D.

( )y f x

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

a)

b)

graph moves to the left 1 unit

graph is reflected in the x axis

graph is then scaled 2 units in the y direction

(2, 1)

(2, -1)

(2, 1)

5

y=f(x)

y= -f(x)

y= 10 - f(x)

Graphs & Functions Higher =

a) On the same diagram sketch

i) the graph of

ii) the graph of

b) Find the range of values of x for

which is positive

2( ) 4 5f x x x

( )y f x

10 ( )y f x

10 ( )f x

2( 2) 1x

a)

b) Solve:210 ( 2) 1 0x

2( 2) 9x ( 2) 3x 1 or 5x

10 - f(x) is positive for -1 < x < 5


A sketch of the graph of y = f(x) where is shown.

The graph has a maximum at A (1,4) and a minimum at B(3, 0)

.

Sketch the graph of

Indicate the co-ordinates of the turning points. There is no need to

calculate the co-ordinates of the points of intersection with the axes.

3 2( ) 6 9f x x x x

( ) ( 2) 4g x f x

Graph is moved 2 units to the left, and 4 units up(3, 0) (1, 4)

(1, 4) ( 1, 8)

t.p.’s are:

(1,4)

(-1,8)

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Outcome 3Higher

Trig Graphs

The same transformation rules

apply to the basic trig graphs.

NB: If f(x) =sinx then 3f(x) = 3sinx

and f(5x) = sin5x

Think about sin replacing f !

Also if g(x) = cosx then g(x) – 4 = cosx – 4

and g(x + 90) = cos(x + 90) Think about cos replacing g !

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Outcome 3Higher

Sketch the graph of y = sinx - 2

If sinx = f(x) then sinx - 2 = f(x) - 2

So move the sinx graph 2 units down.

y = sinx - 2

Trig Graphs

1

-1

-2

-3

090o 180o 270o 360o

DEMO

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Outcome 3Higher

Sketch the graph of y = cos(x - 50)

If cosx = f(x) then cos(x - 50) = f(x - 50)So move the cosx graph 50 units right.

Trig Graphs

y = cos(x - 50)o

1

-1

-2

-3

0

50o

90o 180o 270o 360o

DEMO

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Outcome 3Higher

Trig Graphs

Sketch the graph of y = 3sinx

If sinx = f(x) then 3sinx = 3f(x)

So stretch the sinx graph 3 times vertically.

y = 3sinx

1

-1

-2

-3

0

2

3

90o 180o 270o 360o

DEMO

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Outcome 3Higher

Trig Graphs

Sketch the graph of y = cos4x

If cosx = f(x) then cos4x = f(4x)

So squash the cosx graph to 1/4 size horizontally

y = cos4x

1

-1

090o 180o 270o 360o

DEMO

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Outcome 3Higher

Trig Graphs

Sketch the graph of y = 2sin3xIf sinx = f(x) then 2sin3x = 2f(3x)So squash the sinx graph to 1/3 size horizontally and also double its height.

y = 2sin3x

90o

1

-1

-2

-3

0

2

3

360o180o 270o

DEMO

DEMO

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Trig Graph Trig Graph

1

2

3

-3

-2

-1

090o 180o 270o 360o

Write down equations for

graphs shown ?

CombinationsHigher

y = 0.5sin2xo + 0.5

y = 2sin4xo- 1

Write down the equations in the form f(x) for the graphs shown?

y = 0.5f(2x) + 0.5

y = 2f(4x) - 1

DEMO

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Trig GraphsTrig Graphs

1

2

3

-3

-2

-1

090o 180o 270o 360o

Combinationsy = cos2xo + 1

y = -2cos2xo - 1Higher

Write down the equations for the graphs shown?

Write down the equations in the form f(x) for the graphs shown?

y = f(2x) + 1y = -2f(2x) - 1

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 4A & 4B

Show-me boards

Unit 2 EF 1.3

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A function in the form f(x) = ax where a > 0, a ≠ 1

is called an exponential function to base a .

Exponential (to the power of) Graphs

Exponential Functions

Consider f(x) = 2x

x -3 -2 -1 0 1 2 3

f(x) 1 1/8 ¼ ½ 1 2 4 8

Unit 2 EF 1.3

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The graph of

y = 2x

(0,1)(1,2)

Major Points

(i) y = 2x passes through the points (0,1) & (1,2)

(ii) As x ∞ y ∞ however as x -∞ y 0 .(iii) The graph shows a GROWTH function.

Graph

Unit 2 EF 1.3

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ie

y -3 -2 -1 0 1 2 3

x 1/8 ¼ ½ 1 2 4 8

To obtain y from x we must ask the question

“What power of 2 gives us…?”

This is not practical to write in a formula so we say

y = log2x“the logarithm to base 2 of x”

or “log base 2 of x”

Log Graphs

Unit 2 EF 1.3

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Higher

The graph of

y = log2x (1,0)

(2,1)

Major Points

(i) y = log2x passes through the points (1,0) & (2,1) .(ii) As x ∞ y ∞ but at a very slow rate

and as x 0 y -∞ .

NB: x > 0

Graph

Unit 2 EF 1.3

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Higher

The graph of y = ax always passes through (0,1) & (1,a)

It looks like ..

x

Y

y = ax

(0,1)

(1,a)

Exponential (to the power of) Graphs

Unit 2 EF 1.3

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Higher

The graph of y = logax always passes through (1,0) & (a,1)

It looks like ..

x

Y

y = logax

(1,0)

(a,1)

Log Graphs

Unit 2 EF 1.3

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x

Y

f-1(x) = logax

(1,0)

(a,1)

Connection

(0,1)

(1,a)

f(x) = ax

Unit 2 EF 1.3

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Extra Practice

HHM Ex 2H

HHM Ex 3N , 3O and 15K

HHM Ex 3P

f(x)

0 2 4 6 8x

-2-4-6

2

4

6

-2

-4

-6

Derivative f’(x)

f(x)

All to do with

GRADIENT !

+

+0

-

-

-0

+

+

f’(x)

Demo

Unit 2 EF 1.3

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Higher


This is a method for changing the format

of a quadratic equation

so we can easily sketch or read off key information

Completing the square format looks like

f(x) = a(x + b)2 + c

Warning ! The a,b and c values are different

from the a ,b and c in the general quadratic function

Unit 2 EF 1.3

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Higher

Half the x term and square the

coefficient.


Complete the square for x2 + 2x + 3

and hence sketch function.

f(x) = a(x + b)2 + c

x2 + 2x + 3

x2 + 2x + 3

(x2 + 2x + 1) + 3 Compensate

(x + 1)2 + 2

a = 1

b = 1

c = 2

-1Tidy up !

Unit 2 EF 1.3

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sketch function.f(x) = a(x + b)2 + c

= (x + 1)2 + 2

Mini. Pt. ( -1, 2) (-1,2)

(0,3)

Unit 2 EF 1.3

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Higher

2(x2 - 4x) + 9 Half the x term and square the

coefficient.

Take out coefficient of

x2 term.Compensate !


Complete the square for 2x2 - 8x + 9


f(x) = a(x + b)2 + c

2x2 - 8x + 9

2x2 - 8x + 9

2(x2 – 4x + 4) + 9 Tidy up

2(x - 2)2 + 1

a = 2

b = 2

c = 1

- 8

Unit 2 EF 1.3

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= 2(x - 2)2 + 1

Mini. Pt. ( 2, 1)

(2,1)

(0,9)

Unit 2 EF 1.3

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Half the x term and square the

coefficient

Take out coefficient of x2

compensate


Complete the square for 7 + 6x – x2


f(x) = a(x + b)2 + c

-x2 + 6x + 7

-x2 + 6x + 7

-(x2 – 6x + 9) + 7 Tidy up

-(x - 3)2 + 16

a = -1

b = 3

c = 16

+ 9-(x2 - 6x) + 7

Unit 2 EF 1.3

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= -(x - 3)2 + 16

Mini. Pt. ( 3, 16)

(3,16)

(0,7)

Given , express in the form

Hence sketch function.

Quadratic Theory Higher

2( ) 2 8f x x x ( )f x 2x a b

2( ) ( 1) 9f x x

(-1,9)

(0,-8)

Quadratic Theory Higher

a) Write in the form

b) Hence or otherwise sketch the graph of

2( ) 6 11f x x x 2x a b

( )y f x

a) 2( ) ( 3) 2f x x

b) For the graph of 2y x moved 3 places to left and 2 units up.

minimum t.p. at (-3, 2) y-intercept at (0, 11)

(-3,2)

(0,11)

Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 8D

19 Apr 202319 Apr 2023 Created by Mr. [email protected] by Mr. [email protected]

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Solving Quadratic Equations

Nat 5 Examples

Solve ( find the roots ) for the following

x(x – 2) = 0

x = 0 andx - 2 = 0

x = 2

4t(3t + 15) = 0

4t = 0 and 3t + 15 = 0

t = -5 t = 0 and


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Nat 5 Examples


x2 – 4x = 0

x(x – 4) = 0

x = 0 andx - 4 = 0

x = 4

16t – 6t2 = 0

2t(8 – 3t) = 0

2t = 0 and 8 – 3t = 0

t = 8/3 t = 0 and

Common

Factor

Common

Factor


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Nat 5 Examples


x2 – 9 = 0

(x – 3)(x + 3) = 0

x = 3 and x = -3

100s2 – 25 = 0

25(2s – 1)(2s + 1) = 0

2s – 1 = 0 and2s + 1 = 0

s = - 0.5s = 0.5 and

Difference 2 squares


Take out common

factor

25(4s2 - 1) = 0

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Nat 5 Examples

2x2 – 8 = 0

2(x2 – 4) = 0

x = 2 and x = - 2

80 – 125e2 = 0

5(16 – 25e2) = 0

4 – 5e = 0 and4 + 5t = 0

e = - 4/5 e = 4/5 and

Common

Factor

Common

Factor


2(x – 2)(x + 2) = 0(x – 2)(x + 2) = 0


5(4 – 5e)(4 + 5e) = 0(4 – 5e)(4 + 5e) = 0

(x – 2) = 0 and(x + 2) = 0

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Nat 5 Examples


x2 + 3x + 2 = 0

(x + 2)(x + 1) = 0

x = - 2 andx = - 1

SAC Method

x

x

2

1

x + 2 = 0 x + 1 = 0 and

3x2 – 11x - 4 = 0

(3x + 1)(x - 4) = 0

x = - 1/3 and x = 4

SAC Method

3x

x

+ 1

- 4

3x + 1 = 0 andx - 4 = 0

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Nat 5 Examples


x2 + 5x + 4 = 0

(x + 4)(x + 1) = 0

x = - 4 andx = - 1

SAC Method

x

x

4

1

x + 4 = 0 x + 1 = 0 and

1 + x - 6x2 = 0

(1 + 3x)(1 – 2x) = 0

x = - 1/3 and x = 0.5

SAC Method

1

1

+3x

-2x

1 + 3x = 0 and1 - 2x = 0

Unit 2 EF 1.3

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Higher

created by Mr. Laffertycreated by Mr. Lafferty

When we cannot factorise or solve graphically quadratic equations we need to use the

quadratic formula.

2-b ± (b -4ac)x =

2a

ax2 + bx + c

Quadratic FormulaQuadratic Formula

Unit 2 EF 1.3

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Example : Solve x2 + 3x – 3 = 0

2-(3) ± ((3) -4(1)(-3))x =

2(1)

ax2 + bx + c

1 3 -3

-3 ± 21=

2


Unit 2 EF 1.3

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-3 + 21x =

2-3 - 21

x = 2

and

x = 0.8 x = -3.8


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Nat 5 Examples


√ both sides

√4 = ± 2

(x – 3)2 – 4 = 0

(x – 3)2 = 4

x – 3 = ± 2

x = 3 ± 2

x = 5 x = 1and

√ both sides √of 7 = ±

√7

(x + 2)2 – 7 = 0

(x + 2)2 = 7

x + 2 = ± √7

x = -2 ± √7

x = -2 + √7 x = -2 - √7and


Unit 2 EF 1.3

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Higher

Extra Practice

HHM Ex 8E and Ex8G

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Solving Quadratic Inequations

To solve inequations ( inequalities) the steps are

1. Solve the equation = 0

2. Sketch graph

3. Read off solution

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Solve the inequation x2 + 5x – 6 > 0

1. Solve the equation = 0 2. Sketch graph


x2 + 5x – 6 = 0

(x - 1)(x + 6) = 0

x = 1 and x = - 6

-6 1

x < -6 and x > 1

Unit 2 EF 1.3

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Extra Practice

HHM Ex 8F and Ex8K

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Solve the inequation 3 - 2x – x2 < 0

1. Solve the equation = 0 2. Sketch graph


3 - 2x – x2 = 0

(x + 3)(x - 1) = 0

x = - 3 and x = 1

-3 1

x < -3 and x > 1

x2 + 2x – 3 = 0

Demo

Unit 2 EF 1.3

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Higher

If a function f(x) has roots/zeros at a, b and c

then it has factors (x – a), (x – b) and (x – c)

And can be written as f(x) = k(x – a)(x – b)(x – c).

Functions from Graphs

Unit 2 EF 1.3

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Example

-2 1 5

30

y = f(x)

Finding a Polynomial From Its Zeros

Unit 2 EF 1.3

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f(x) has zeros at x = -2, x = 1 and x = 5,

so it has factors (x +2), (x – 1) and (x – 5)

so f(x) = k (x +2)(x – 1)(x – 5)

f(x) also passes through (0,30) so replacing x by 0

and f(x) by 30 the equation becomes

30 = k X 2 X (-1) X (-5)

ie 10k = 30

ie k = 3


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Formula is f(x) = 3(x + 2)(x – 1)(x – 5)

f(x) = (3x + 6)(x2 – 6x + 5)

f(x) = 3x3 – 12x2 – 21x + 30


Quad Demo Cubic Demo

Unit 2 EF 1.3

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Extra Practice

HHM Ex 7H

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What are Functions ?

Functions describe how one quantity

relates to another

Car Part

s

Assembly line

Cars

Defn: A function or mapping is a relationship between two sets in which each member of the first set is connected to exactly one member in the

second set.

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Nat 5

What are Functions ?

Functions describe how one quantity

relates to another

Dirty

Washing Machine

Clean

OutputInputyx

Functionf(x)

y = f(x)

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Defining a Functions

A function can be thought of as the relationship between

Set A (INPUT - the x-coordinate)

and

SET B the y-coordinate (Output) .

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X

YFunction !

!

Functions & GraphsFunctions & Graphs

Unit 2 EF 1.3

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Function & Graphs Function & Graphs

x

YFunction !

!

Unit 2 EF 1.3

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x

YNot a

function !!

Cuts graph

more than once !

Function & GraphsFunction & Graphs

x must map to

one value of y

Unit 2 EF 1.3

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Functions & GraphsFunctions & Graphs

X

Y Not a function !!

Cuts graph

more than once!

Unit 2 EF 1.3

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Functions & MappingsFunctions & Mappings

A function can be though of as a black box

x - Coordinate

Input

Domain

Members (x - axis)Co-Domain

Members (y - axis)

Image

Range

Function

Output

y - Coordinatef(x) = x2+ 3x - 1

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Finding the Function

Find the output or input values for the functions below :

6

7

8

36

49

64

f(x) = x2

f: 0

f: 1

f:2

-1

3

7f(x) = 4x - 1

4 12

f(x) = 3x

5 15

6 18

Examples

Unit 2 EF 1.3

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Functions & MappingFunctions & Mapping

Functions can be illustrated in three ways:

1) by a formula.

2) by arrow diagram.

3) by a graph (ie co-ordinate diagram).

ExampleSuppose that f: A B is defined by

f(x) = x2 + 3x where A = { -3, -2, -1, 0, 1}.FORMULA

then f(-3) = 0 , f(-2) = -2 , f(-1) = -2 , f(0) = 0 ,f(1) = 4

NB: B = {-2, 0, 4} = the range!

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The standard way to represent a function

is by a formula.

Function Notation

Examplef(x) = x + 4

We read this as “f of x equals x + 4”

or

“the function of x is x + 4

f(1) = 5 is the value of f at 1

f(a) = a + 4 is the value of f at a

1 + 4 = 5

a + 4

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For the function h(x) = 10 – x2.

Calculate h(1) , h(-3) and h(5)

h(1) =

Examples

h(-3) = h(5) =

h(x) = 10 – x2

Function Notation

10 – 12 = 9

10 – (-3)2 =

10 – 9 = 1

10 – 52 =

10 – 25 = -15

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For the function g(x) = x2 + x

Calculate g(0) , g(3) and g(2a)

g(0) =

Examples

g(3) = g(2a) =

g(x) = x2 + x

Function Notation

02 + 0 =

0

32 + 3 =

12

(2a)2 +2a =

4a2 + 2a

Unit 2 EF 1.3

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COMPOSITION OF FUNCTIONS

( or functions of functions )

Suppose that f and g are functions where

f:A B and g:B C

with f(x) = y and g(y) = z

where x A, y B and z C.

Suppose that h is a third function where

h:A C with h(x) = z .

Composite FunctionsComposite Functions

Unit 2 EF 1.3

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A B C

x y zf g

h

We can say that h(x) = g(f(x))

“function of a function”

DEMO

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Composite FunctionsComposite Functionsf(2)=3x2 – 2 =4

g(4)=42 + 1 =17

f(5)=5x3-2 =13Example 1

Suppose that f(x) = 3x - 2 and g(x) = x2 +1

(a) g( f(2) ) = g(4) = 17

(b) f( g (2) ) = f(5) = 13

(c) f( f(1) ) = f(1) = 1

(d) g( g(5) ) = g(26)= 677

f(1)=3x1 - 2 =1

g(26)=262

+ 1 =677

g(2)=22 + 1 =5

f(1)=3x1 - 2 =1

g(5)=52 + 1 =26

Unit 2 EF 1.3

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Suppose that f(x) = 3x - 2 and g(x) = x2 +1

Find formulae for (a) g(f(x)) (b) f(g(x)).

(a) g(f(x)) = ( )2 + 1

= 9x2 - 12x + 5

(b) f(g(x)) = 3( ) - 2= 3x2 + 1

CHECK

g(f(2)) = 9 x 22 - 12 x 2 + 5

= 36 - 24 + 5= 17

f(g(2)) = 3 x 22 + 1= 13

NB: g(f(x)) f(g(x)) in general.


3x - 2 x2 +1

Unit 2 EF 1.3

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Let h(x) = x - 3 , g(x) = x2 + 4 and k(x) = g(h(x)). If k(x) = 8 then find the value(s) of x.

k(x) = g(h(x))

= ( )2 + 4

= x2 - 6x + 13

Put x2 - 6x + 13 = 8

then x2 - 6x + 5 = 0

or (x - 5)(x - 1) = 0

So x = 1 or x = 5


x - 3

Unit 2 EF 1.3

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Choosing a Suitable Domain

(i) Suppose f(x) = 1 . x2 - 4

Clearly x2 - 4 0

So x2 4

So x -2 or 2

Hence domain = {xR: x -2 or 2 }


Unit 2 EF 1.3

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(ii) Suppose that g(x) = (x2 + 2x - 8)

We need (x2 + 2x - 8) 0

Suppose (x2 + 2x - 8) = 0

Then (x + 4)(x - 2) = 0

So x = -4 or x = 2

So domain = { xR: x -4 or x 2 }

Composite FunctionsComposite FunctionsSketch graph

-4 2


The functions f and g are defined on a suitable domain by

a) Find an expression for b) Factorise

2 2( ) 1 and ( ) 2f x x g x x

( ( ))f g x ( ( ))f g x

a) 22 2( ( )) ( 2) 2 1f g x f x x

2 22 1 2 1x x Difference of 2 squares

Simplify 2 23 1x x

b)


Functions and are defined on suitable domains.

a) Find an expression for h(x) where h(x) = f(g(x)).

b) Write down any restrictions on the domain of h.

1( )

4f x

x

( ) 2 3g x x

( ( )) (2 3)f g x f x a)1

2 3 4x

1

( )2 1

h xx

b) 2 1 0x 1

2x


3( ) 3 ( ) , 0x

f x x and g x x

a) Find

b) If find in its simplest form.

( ) where ( ) ( ( ))p x p x f g x

3

3( ) , 3

xq x x

( ( ))p q x

3( ) ( ( ))x

p x f g x f

a) 33

x 3 3x

x

3( 1)x

x

b)

33 1

3333

3

( ( )) x

xx

p q x p

9 3

33 3x x

9 3(3 ) 3

3 3

x x

x

3 3

3 3

x x

x

x

Graphs & Functions HigherFunctions f and g are defined on the set of real numbers by

a) Find formulae for

i) ii)

b) The function h is defined by

Show that and sketch the graph of h.

2( ) 1 and ( )f x x g x x

( ( ))f g x ( ( ))g f x

( ) ( ( )) ( ( ))h x f g x g f x

2( ) 2 2h x x x

a)

b)

2 2( ( )) ( ) 1f g x f x x 2( ( )) ( 1) 1g f x g x x

22( ) 1 1h x x x 2 2( ) 1 2 1h x x x x 22 2x x

Unit 2 EF 1.3

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Inverse FunctionsInverse Functions

A Inverse function is simply a function in reverse

Input

Function

Outputf(x) = x2+ 3x - 1

InputOutputf-1(x) = ?

Unit 2 EF 1.3

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Inverse Function

Find the inverse function given

f(x) = 3x

Example

Remember

f(x) is simply the

y-coordinate

y = 3x

Using Changing the subject

rearrange into

x =

x =y

3

Rewrite replacing y with

x.

This is the inverse function

f-1(x) =x

3

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Inverse Function


f(x) = x2

Example

Remember

f(x) is simply the

y-coordinate

y = x2


rearrange into

x =

x = √y


x.


f-1(x) = √x

Unit 2 EF 1.3

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Inverse Function


f(x) = 4x - 1

Example

Remember

f(x) is simply the

y-coordinate

y = 4x - 1


rearrange into

x =

x =


x.


f-1(x) =

y + 1

4

x + 1

4

Unit 2 EF 1.3

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Are you on Target !

• Update you log book

• Make sure you complete and correct MOST of the Composite Functionquestions in the past paper booklet.

f(x)

f(x)

Graphs & Functions

y = -f(x)

y = f(-x)

y = f(x) ± k

y = f(kx)

Move verticallyup or downs

depending on k

flip iny-axis

flip inx-axis

+

- Stretch or compressvertically

depending on k

y = kf(x)

Stretch or compress

horizontally depending on k

f(x)

f(x)f(x)

f(x)y = f(x ± k)

Move horizontallyleft or right

depending on k

+-

Remember we can combine

these together !!

0 < k < 1 stretch

k > 1 compress

0 < k < 1 compress

k > 1 stretch

Composite Functions

A complex function made up of 2 or

more simpler functions

= +

f(x) = x2 - 4 g(x) = 1x

x

Domainx-axis valuesInput

Rangey-axis valuesOutput

x2 - 41

x2 - 4

Restriction x2 - 4 ≠ 0

(x – 2)(x + 2) ≠ 0

x ≠ 2 x ≠ -2

g(f(x)) g(f(x)) =

f(x) = x2 - 4g(x) = 1x

x

Domainx-axis valuesInput

Rangey-axis valuesOutput

f(g(x))

Restriction x2 ≠ 0

1

x

2- 4 =

Similar to composite

Area

Write down g(x) with brackets for x

g(x) =1

( )

inside bracket put f(x)

g(f(x)) =1

x2 - 4

1x

- 41x2

f(g(x)) =Write down f(x) with brackets for x

f(x) = ( )2 - 4

inside bracket put g(x)

f(g(x)) =1

x2- 4

Functions & Graphs

TYPE questions(Sometimes Quadratics)

SketchingGraphs

CompositeFunctions

Steps :

1.Outside function staysthe same EXCEPT replacex terms with a ( )

2. Put inner function in bracket

You need to learn basic movements

Exam questionsnormally involve two movements

Remember orderBODMAS

Restrictions :

1.Denominator NOT ALLOWEDto be zero

2.CANNOT take the square rootof a negative number

Documents

Unit 2 EF 1.3 Higher Higher Maths Composite Functions Exponential and Log Graphs Graph Transformations Trig