Unit 2 EF Higher Higher Maths Composite Functions Exponential and Log Graphs Graph Transformations Trig Graphs Inverse function Mindmap Exam Question Type Derivative Graphs f’(x) Completing the Square Solving equations / /Inequations

Unit 2 EF 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) x 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(90 o,0) A(45 o,0.5) C(135 o,-0.5) B(90 o,1) A(45 o,1.5) C(135 o,0.5)

Unit 2 EF Higher Extra Practice HHM Ex 3C

f(x) x 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) x Transformation -f(x) (x, y)  (x, -y) Mapping f(x) Flip in x-axis Flip in x-axis

Unit 2 EF 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(90 o,0) A(45 o,0.5) C(135 o,-0.5) A(45 o,-0.5) C(135 o,0.5)

Unit 2 EF Higher Extra Practice HHM Ex 3G

f(x) x Transformation f(-x) (x, y)  (-x, y) Mapping f(x) Flip in y-axis Flip in y-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 Higher Extra Practice HHM Ex 3I

f(x) x Transformation kf(x) (x, y)  (x, ky) Mapping f(x) Stretch in y-axis 2f(x) 0.5f(x) Compress in y-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) x Transformation f(kx) (x, y)  (1/kx, y) Mapping f(x) Compress in x-axis f(2x) f(0.5x) Stretch in x-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 Higher Extra Practice HHM Ex 3K & 3M

Unit 2 EF 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) f(x + 1) + 2 -f(x) - 2 (1,3) (-1,-3) (1,3) (-1,-3)

(1,-2) (-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 / compress in y direction (e)f(kx) stretch / compress in x direction Name : (-1,-3) (1,3) (-2,-1) f(x + 1) + 2 -f(x) - 2 (-2,0) (1,3) (0,-6) (-1,-3) (1,3) (1,7) (-1,-5) (2,2) (-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 = 2 f (- x ) Graphs & Functions Higher a) Reflect across the y axis b) Now scale by 2 in the y direction

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. 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

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 a) b) Solve: 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. Graph is moved 2 units to the left, and 4 units up t.p.’s are: (1,4) (-1,8)

Outcome 3 Higher Trig Graphs The same transformation rules apply to the basic trig graphs. NB: If f(x) =sinx  then3f(x) = 3sinx  andf(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 !

Outcome 3 Higher 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 o 180 o 270 o 360 o DEMO

Outcome 3 Higher 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 o 90 o 180 o 270 o 360 o DEMO

Outcome 3 Higher 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  o 180 o 270 o 360 o DEMO

Outcome 3 Higher 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  o 180 o 270 o 360 o DEMO

Outcome 3 Higher Trig Graphs Sketch the graph of y = 2sin3x  If sinx  = f(x) then 2sin3x  = 2f(3x) So squash the sinx  graph to 1 / 3 size horizontally and also double its height. y = 2sin3x  90 o o 180 o 270 o DEMO

created by Mr. Lafferty Trig Graph o 180 o 270 o 360 o Write down equations for graphs shown ? Combinations Higher y = 0.5sin2x o y = 2sin4x o - 1 Write down the equations in the form f(x) for the graphs shown? y = 0.5f(2x) y = 2f(4x) - 1

DEMO created by Mr. Lafferty Trig Graphs o 180 o 270 o 360 o Combinations y = cos2x o + 1 y = -2cos2x o - 1 Higher Write down the equations for the graphs shown? Write down the equations in the form f(x) for the graphs shown? y = f(2x) + 1 y = -2f(2x) - 1

Unit 2 EF Higher Extra Practice HHM Ex 4A & 4B Show-me boards

Unit 2 EF Higher A function in the form f(x) = a x where a > 0, a ≠ 1 is called an exponential function to base a. Exponential (to the power of) Graphs Exponential Functions Consider f(x) = 2 x x f(x) 11 / 8 ¼ ½

Unit 2 EF Higher The graph of y = 2 x (0,1) (1,2) Major Points (i) y = 2 x 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 Higher ie y x 1 / 8 ¼ ½ 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 = log 2 x “the logarithm to base 2 of x” or “log base 2 of x” Log Graphs

Unit 2 EF Higher The graph of y = log 2 x (1,0) (2,1) Major Points (i) y = log 2 x 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 Higher The graph of y = a x always passes through (0,1) & (1,a) It looks like.. x Y y = a x (0,1) (1,a) Exponential (to the power of) Graphs

Unit 2 EF Higher The graph of y = log a x always passes through (1,0) & (a,1) It looks like.. x Y y = log a x (1,0) (a,1) Log Graphs

Unit 2 EF Higher x Y f -1 (x) = log a x (1,0) (a,1) Connection (0,1) (1,a) f(x) = a x

Unit 2 EF Higher Extra Practice HHM Ex 2H HHM Ex 3N, 3O and 15K HHM Ex 3P

f(x) x Derivative f’(x) f(x) All to do with GRADIENT ! f’(x) Demo

Unit 2 EF 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 Higher Half the x term and square the coefficient. Completing the Square Complete the square for x 2 + 2x + 3 and hence sketch function. f(x) = a(x + b) 2 + c x 2 + 2x + 3 x 2 + 2x + 3 (x 2 + 2x + 1) + 3 Compensate (x + 1) a = 1 b = 1 c = 2 -1 Tidy up !

Unit 2 EF Higher Completing the Square sketch function. f(x) = a(x + b) 2 + c = (x + 1) Mini. Pt. ( -1, 2) (-1,2) (0,3)

Unit 2 EF Higher 2(x 2 - 4x) + 9 Half the x term and square the coefficient. Take out coefficient of x 2 term. Compensate ! Completing the Square Complete the square for 2x 2 - 8x + 9 and hence sketch function. f(x) = a(x + b) 2 + c 2x 2 - 8x + 9 2x 2 - 8x + 9 2(x 2 – 4x + 4) + 9 Tidy up 2(x - 2) a = 2 b = 2 c = 1 - 8

Unit 2 EF Higher Completing the Square sketch function. f(x) = a(x + b) 2 + c = 2(x - 2) Mini. Pt. ( 2, 1) (2,1) (0,9)

Unit 2 EF Higher Half the x term and square the coefficient Take out coefficient of x 2 compensate Completing the Square Complete the square for 7 + 6x – x 2 and hence sketch function. f(x) = a(x + b) 2 + c -x 2 + 6x + 7 -x 2 + 6x + 7 -(x 2 – 6x + 9) + 7 Tidy up -(x - 3) a = -1 b = 3 c = (x 2 - 6x) + 7

Unit 2 EF Higher Completing the Square sketch function. f(x) = a(x + b) 2 + c = -(x - 3) Mini. Pt. ( 3, 16) (3,16) (0,7)

Given, express in the form Hence sketch function. Quadratic Theory Higher (-1,9) (0,-8)

Quadratic Theory Higher a)Write in the form b)Hence or otherwise sketch the graph of a) b) For the graph of 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 Higher Extra Practice HHM Ex 8D

8-Oct-15 Created by Mr. Solving Quadratic Equations Nat 5 Examples Solve ( find the roots ) for the following x(x – 2) = 0 x = 0and x - 2 = 0 x = 2 4t(3t + 15) = 0 4t = 0and3t + 15 = 0 t = -5t = 0and

8-Oct-15 Created by Mr. Solving Quadratic Equations Nat 5 Examples Solve ( find the roots ) for the following x 2 – 4x = 0 x(x – 4) = 0 x = 0and x - 4 = 0 x = 4 16t – 6t 2 = 0 2t(8 – 3t) = 0 2t = 0and8 – 3t = 0 t = 8/3t = 0and Common Factor Common Factor

8-Oct-15 Created by Mr. Solving Quadratic Equations Nat 5 Examples Solve ( find the roots ) for the following x 2 – 9 = 0 (x – 3)(x + 3) = 0 x = 3and x = s 2 – 25 = 0 25(2s – 1)(2s + 1) = 0 2s – 1 = 0and2s + 1 = 0 s = - 0.5s = 0.5and Difference 2 squares Difference 2 squares Take out common factor 25(4s 2 - 1) = 0

Solving Quadratic Equations Nat 5 Examples 2x 2 – 8 = 0 2(x 2 – 4) = 0 x = 2andx = – 125e 2 = 0 5(16 – 25e 2 ) = 0 4 – 5e = 0and4 + 5t = 0 e = - 4/5e = 4/5and 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) = 0and(x + 2) = 0

Solving Quadratic Equations Nat 5 Examples Solve ( find the roots ) for the following x 2 + 3x + 2 = 0 (x + 2)(x + 1) = 0 x = - 2andx = - 1 SAC Method x x 2 1 x + 2 = 0x + 1 = 0and 3x 2 – 11x - 4 = 0 (3x + 1)(x - 4) = 0 x = - 1/3andx = 4 SAC Method 3x x x + 1 = 0andx - 4 = 0

Solving Quadratic Equations Nat 5 Examples Solve ( find the roots ) for the following x 2 + 5x + 4 = 0 (x + 4)(x + 1) = 0 x = - 4andx = - 1 SAC Method x x 4 1 x + 4 = 0x + 1 = 0and 1 + x - 6x 2 = 0 (1 + 3x)(1 – 2x) = 0 x = - 1/3andx = 0.5 SAC Method x -2x 1 + 3x = 0and1 - 2x = 0

Unit 2 EF Higher created by Mr. Lafferty When we cannot factorise or solve graphically quadratic equations we need to use the quadratic formula. ax 2 + bx + c Quadratic Formula

Unit 2 EF Higher created by Mr. Lafferty Example : Solve x 2 + 3x – 3 = 0 ax 2 + bx + c 13-3 Quadratic Formula

Unit 2 EF Higher created by Mr. Lafferty and Quadratic Formula

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 = 5x = 1and √ both sides √of 7 = ± √7 (x + 2) 2 – 7 = 0 (x + 2) 2 = 7 x + 2 = ± √7 x = -2 ± √7 x = -2 + √7x = -2 - √7and Solving Quadratic Equations

Unit 2 EF Higher Extra Practice HHM Ex 8E and Ex8G

Higher Solving Quadratic Inequations To solve inequations ( inequalities) the steps are 1.Solve the equation = 0 2.Sketch graph 3.Read off solution

Higher Solving Quadratic Inequations Solve the inequation x 2 + 5x – 6 > 0 1. Solve the equation = 0 2. Sketch graph 3. Read off solution x 2 + 5x – 6 = 0 (x - 1)(x + 6) = 0 x = 1 and x = x < -6 and x > 1

Unit 2 EF Higher Extra Practice HHM Ex 8F and Ex8K

Higher Solving Quadratic Inequations Solve the inequation 3 - 2x – x 2 < 0 1. Solve the equation = 0 2. Sketch graph 3. Read off solution 3 - 2x – x 2 = 0 (x + 3)(x - 1) = 0 x = - 3 and x = x < -3 and x > 1 x 2 + 2x – 3 = 0 Demo

Unit 2 EF 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 Higher Example y = f(x) Finding a Polynomial From Its Zeros

Unit 2 EF Higher f(x) has zeros at x = -2, x = 1 and x = 5, so it has factors (x +2), (x – 1) and (x – 5) sof(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 Higher Formula isf(x) = 3(x + 2)(x – 1)(x – 5) f(x) = (3x + 6)(x 2 – 6x + 5) f(x) = 3x 3 – 12x 2 – 21x + 30 Finding a Polynomial From Its Zeros Quad Demo Cubic Demo

Unit 2 EF Higher Extra Practice HHMEx 7H

Nat 5 What are Functions ? Functions describe how one quantity relates to another Car Parts 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.

Nat 5 What are Functions ? Functions describe how one quantity relates to another Dirty Washing Machine Clean OutputInput yx Function f(x) y = f(x)

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 Higher X Y Function !! Functions & Graphs

Unit 2 EF Higher Function & Graphs x Y Function !!

Unit 2 EF Higher x Y Not a function !! Cuts graph more than once ! Function & Graphs x must map to one value of y

Unit 2 EF Higher Functions & Graphs X Y Not a function !! Cuts graph more than once!

Unit 2 EF Higher Functions & 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 - Coordinate f(x) = x 2 + 3x - 1

Nat 5 Finding the Function Find the output or input values for the functions below : f(x) = x 2 f: 0 f: 1 f: f(x) = 4x f(x) = 3x Examples

Unit 2 EF Higher Functions & Mapping Functions can be illustrated in three ways: 1) by a formula. 2) by arrow diagram. 3) by a graph (ie co-ordinate diagram). Example Suppose that f: A  B is defined by f(x) = x 2 + 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!

Nat 5 The standard way to represent a function is by a formula. Function Notation Example f(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 =5 a + 4

Nat 5 For the function h(x) = 10 – x 2. Calculate h(1), h(-3) and h(5) h(1) = Examples h(-3) = h(5) = h(x) = 10 – x 2  Function Notation 10 – 1 2 = 9 10 – (-3) 2 = 10 – 9 = 1 10 – 5 2 =10 – 25 = -15

Nat 5 For the function g(x) = x 2 + x Calculate g(0), g(3) and g(2a) g(0) = Examples g(3) = g(2a) = g(x) = x 2 + x  Function Notation = = 12 (2a) 2 +2a =4a 2 + 2a

Unit 2 EF 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 Functions

Unit 2 EF Higher Composite Functions ABCABC x y z f g h We can say that h(x) = g(f(x)) “function of a function” DEMO

Unit 2 EF Higher Composite Functions f(2)=3 x 2 – 2 =4 g(4)= =17 f(5)=5x3-2 =13 Example 1 Suppose that f(x) = 3x - 2 and g(x) = x 2 +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)= =677 g(2)= =5 f(1)=3x1 - 2 =1 g(5)= =26

Unit 2 EF Higher Suppose that f(x) = 3x - 2 and g(x) = x 2 +1 Find formulae for (a) g(f(x)) (b) f(g(x)). (a) g(f(x)) =( ) 2 + 1= 9x x + 5 (b) f(g(x)) =3( ) - 2= 3x CHECK g(f(2)) =9 x x 2 + 5= = 17 f(g(2)) =3 x = 13 NB: g(f(x))  f(g(x)) in general. Composite Functions 3x - 2x 2 +1

Unit 2 EF Higher Let h(x) = x - 3, g(x) = x and k(x) = g(h(x)). If k(x) = 8 then find the value(s) of x. k(x) = g(h(x)) = ( ) = x 2 - 6x + 13 Put x 2 - 6x + 13 = 8 then x 2 - 6x + 5 = 0 or (x - 5)(x - 1) = 0 So x = 1 or x = 5 Composite Functions x - 3

Unit 2 EF Higher Choosing a Suitable Domain (i) Suppose f(x) = 1. x Clearly x  0 So x 2  4 So x  -2 or 2 Hence domain = {x  R: x  -2 or 2 } Composite Functions

Unit 2 EF Higher (ii) Suppose that g(x) =  (x 2 + 2x - 8) We need (x 2 + 2x - 8)  0 Suppose (x 2 + 2x - 8) = 0 Then (x + 4)(x - 2) = 0 So x = -4 or x = 2 So domain = { x  R: x  -4 or x  2 } Composite Functions Sketch graph -42

Graphs & Functions Higher The functions f and g are defined on a suitable domain by a) Find an expression for b) Factorise a) Difference of 2 squares Simplify 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. a) b)

Graphs & Functions Higher a) Find b) If find in its simplest form. a) b)

Graphs & Functions Higher Functions 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. a) b)

Unit 2 EF Higher Inverse Functions A Inverse function is simply a function in reverse Input Function Output f(x) = x 2 + 3x - 1 InputOutput f -1 (x) = ?

Unit 2 EF 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 Higher Inverse Function Find the inverse function given f(x) = x 2 Example Remember f(x) is simply the y-coordinate y = x 2 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 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 x + 1 4

Unit 2 EF Higher Are you on Target ! Update you log book Make sure you complete and correct MOST of the Composite FunctionComposite Function questions in the past paper booklet.

f(x) Graphs & Functions y = -f(x) y = f(-x) y = f(x) ± k y = f(kx) Move vertically up or downs depending on k flip in y-axis flip in x-axis + - Stretch or compress vertically depending on k y = kf(x) Stretch or compress horizontally depending on k f(x) y = f(x ± k) Move horizontally left 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) = x g(x) = 1 x x Domain x-axis values Input Range y-axis values Output x x Restrictionx ≠ 0 (x – 2)(x + 2) ≠ 0 x ≠ 2x ≠ -2 g(f(x)) g(f(x)) = f(x) = x g(x) = 1 x x Domain x-axis values Input Range y-axis values Output f(g(x)) Restriction x 2 ≠ 0 1 x = Similar to composite Area Write down g(x) with brackets for x g(x) = 1 ( ) inside bracket put f(x) g(f(x)) = 1 x x x2x2 f(g(x)) = Write down f(x) with brackets for x f(x) = ( ) inside bracket put g(x) f(g(x)) = 1 x2x2 - 4

Functions & Graphs TYPE questions (Sometimes Quadratics) Sketching Graphs Composite Functions Steps : 1.Outside function stays the same EXCEPT replace x terms with a ( ) 2.Put inner function in bracket You need to learn basic movements Exam questions normally involve two movements Remember order BODMAS Restrictions : 1.Denominator NOT ALLOWED to be zero 2.CANNOT take the square root of a negative number