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Unit 2 EF 1.3
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Higher
Higher MathsHigher Maths
www.mathsrevision.comwww.mathsrevision.com
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
NOTE: Always state any coordinates given on f(x)
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
NOTE: Always state any coordinates given on f(x)
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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Higher
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
NOTE: Always state any coordinates given on f(x)
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
NOTE: Always state any coordinates given on f(x)
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
NOTE: Always state any coordinates given on f(x)
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
Explain the effect the following have
(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
Explain the effect the following have
(a)-f(x) flip in x-axis
(b)f(-x) flip in y-axis
(c)f(x) ± k move up or down
Explain the effect the following have
(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
Graphs & Functions Higher
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
Graphs & Functions Higher
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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Higher
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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Higher
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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Higher
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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Higher
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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Higher
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
Completing the Square
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.
Completing the Square
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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Higher
Completing the Square
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 !
Completing the Square
Complete the square for 2x2 - 8x + 9
and hence sketch function.
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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Higher
Completing the Square
sketch function.f(x) = a(x + b)2 + c
= 2(x - 2)2 + 1
Mini. Pt. ( 2, 1)
(2,1)
(0,9)
Unit 2 EF 1.3
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Higher
Half the x term and square the
coefficient
Take out coefficient of x2
compensate
Completing the Square
Complete the square for 7 + 6x – x2
and hence sketch function.
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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Higher
Completing the Square
sketch function.f(x) = a(x + b)2 + c
= -(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
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
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
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
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
Difference 2 squares
Take out common
factor
25(4s2 - 1) = 0
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Solving Quadratic Equations
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
Difference 2 squares
2(x – 2)(x + 2) = 0(x – 2)(x + 2) = 0
Difference 2 squares
5(4 – 5e)(4 + 5e) = 0(4 – 5e)(4 + 5e) = 0
(x – 2) = 0 and(x + 2) = 0
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Solving Quadratic Equations
Nat 5 Examples
Solve ( find the roots ) for the following
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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Solving Quadratic Equations
Nat 5 Examples
Solve ( find the roots ) for the following
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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Higher
created by Mr. Laffertycreated by Mr. Lafferty
Example : Solve x2 + 3x – 3 = 0
2-(3) ± ((3) -4(1)(-3))x =
2(1)
ax2 + bx + c
1 3 -3
-3 ± 21=
2
Quadratic FormulaQuadratic Formula
Unit 2 EF 1.3
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created by Mr. Laffertycreated by Mr. Lafferty
-3 + 21x =
2-3 - 21
x = 2
and
x = 0.8 x = -3.8
Quadratic FormulaQuadratic Formula
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Nat 5 Examples
Solve ( find the roots ) for the following
√ 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
Solving Quadratic Equations
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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Higher
Solving Quadratic Inequations
Solve the inequation x2 + 5x – 6 > 0
1. Solve the equation = 0 2. Sketch graph
3. Read off solution
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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Higher
Extra Practice
HHM Ex 8F and Ex8K
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Solving Quadratic Inequations
Solve the inequation 3 - 2x – x2 < 0
1. Solve the equation = 0 2. Sketch graph
3. Read off solution
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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Higher
Example
-2 1 5
30
y = f(x)
Finding a Polynomial From Its Zeros
Unit 2 EF 1.3
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Higher
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
Finding a Polynomial From Its Zeros
Unit 2 EF 1.3
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Higher
Formula is f(x) = 3(x + 2)(x – 1)(x – 5)
f(x) = (3x + 6)(x2 – 6x + 5)
f(x) = 3x3 – 12x2 – 21x + 30
Finding a Polynomial From Its Zeros
Quad Demo Cubic Demo
Unit 2 EF 1.3
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Extra Practice
HHM Ex 7H
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Nat 5
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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Nat 5
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) .
Unit 2 EF 1.3
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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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Higher
Functions & GraphsFunctions & Graphs
X
Y Not a function !!
Cuts graph
more than once!
Unit 2 EF 1.3
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Higher
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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Nat 5
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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Nat 5
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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Nat 5
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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Nat 5
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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Higher
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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Composite FunctionsComposite Functions
A B C
x y zf g
h
We can say that h(x) = g(f(x))
“function of a function”
DEMO
Unit 2 EF 1.3
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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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Higher
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.
Composite FunctionsComposite Functions
3x - 2 x2 +1
Unit 2 EF 1.3
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Higher
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
Composite FunctionsComposite Functions
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 }
Composite FunctionsComposite Functions
Unit 2 EF 1.3
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Higher
(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
Graphs & Functions Higher
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)
Graphs & Functions Higher
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
Graphs & Functions Higher
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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Higher
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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Higher
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
Unit 2 EF 1.3
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Higher
Inverse Function
Find the inverse function given
f(x) = x2
Example
Remember
f(x) is simply the
y-coordinate
y = x2
Using Changing the subject
rearrange into
x =
x = √y
Rewrite replacing y with
x.
This is the inverse function
f-1(x) = √x
Unit 2 EF 1.3
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Higher
Inverse Function
Find the inverse function given
f(x) = 4x - 1
Example
Remember
f(x) is simply the
y-coordinate
y = 4x - 1
Using Changing the subject
rearrange into
x =
x =
Rewrite replacing y with
x.
This is the inverse function
f-1(x) =
y + 1
4
x + 1
4
Unit 2 EF 1.3
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Higher
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