Graph Transformations Higher Unit 1 What is a set Recognising a Function in various formats Composite Functions Exponential and Log Graphs Graph Transformations Trig Graphs Connection between Radians and degrees & Exact values Solving Trig Equations Basic Trig Identities Exam Type Questions www.mathsrevision.com

Sets & Functions Notation & Terminology SETS: A set is a collection of items which have some common property. These items are called the members or elements of the set. Sets can be described or listed using “curly bracket” notation.

Sets & Functions eg {colours in traffic lights} = {red, amber, green} DESCRIPTION LIST eg {square nos. less than 30} = { 0, 1, 4, 9, 16, 25} NB: Each of the above sets is finite because we can list every member

We can describe numbers by the following sets: Sets & Functions We can describe numbers by the following sets: N = {natural numbers} = {1, 2, 3, 4, ……….} W = {whole numbers} = {0, 1, 2, 3, ………..} Z = {integers} = {….-2, -1, 0, 1, 2, …..} Q = {rational numbers} This is the set of all numbers which can be written as fractions or ratios. eg 5 = 5/1 -7 = -7/1 0.6 = 6/10 = 3/5 55% = 55/100 = 11/20 etc

We should also note that Sets & Functions R = {real numbers} This is all possible numbers. If we plotted values on a number line then each of the previous sets would leave gaps but the set of real numbers would give us a solid line. We should also note that N “fits inside” W W “fits inside” Z Z “fits inside” Q Q “fits inside” R

Sets & Functions N W Z Q R When one set can fit inside another we say that it is a subset of the other. The members of R which are not inside Q are called irrational numbers. These cannot be expressed as fractions and include  , 2, 35 etc

This set has no elements and is called the empty set. Sets & Functions To show that a particular element/number belongs to a particular set we use the symbol . eg 3  W but 0.9  Z Examples { x  W: x < 5 } = { 0, 1, 2, 3, 4 } { x  Z: x  -6 } = { -6, -5, -4, -3, -2, …….. } { x  R: x2 = -4 } = { } or  This set has no elements and is called the empty set.

Functions & Mappings 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. If the first set is A and the second B then we often write f: A  B The members of set A are usually referred to as the domain of the function (basically the starting values or even x-values) while the corresponding values or images come from set B and are called the range of the function (these are like y-values).

Functions & Mapping Example Suppose that f: A  B is defined by 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) = 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!

Functions & Mapping f(x) ARROW DIAGRAM A B f(-3) = 0 f(-2) = -2 -2 -2 -1 -2 1 4

NB: This graph consists of 5 separate points. It is not a solid curve. Functions & Graphs In a GRAPH we get : NB: This graph consists of 5 separate points. It is not a solid curve.

Recognising Functions Functions & Graphs Recognising Functions A B a b c d e f g A B e f g a bc d YES Not a function two arrows leaving b!

Not a function - d unused! Functions & Graphs A B Not a function - d unused! a b c d e f g A B a bc d e f g h YES

Recognising Functions from Graphs Functions & Graphs Recognising Functions from Graphs If we have a function f: R  R (R - real nos.) then every vertical line we could draw would cut the graph exactly once! This basically means that every x-value has one, and only one, corresponding y-value!

Function & Graphs Y Function !! x

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

Functions & Graphs Not a function !! Y Cuts graph more than once! X

Functions & Graphs Y Function !! X

COMPOSITION OF FUNCTIONS ( or functions of functions ) Composite Functions 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 .

We can say that h(x) = g(f(x)) “function of a function” Composite Functions ie A B C g f x y z h We can say that h(x) = g(f(x)) “function of a function”

Suppose that f(x) = 3x - 2 and g(x) = x2 +1 Composite Functions g(2)=22 + 1 =5 f(5)=5x3-2 =13 Example 1 f(1)=3x1 - 2 =1 f(1)=3x1 - 2 =1 Suppose that f(x) = 3x - 2 and g(x) = x2 +1 (a) g( f(2) ) = g(4) = 17 g(26)=262 + 1 =677 g(5)=52 + 1 =26 (b) f( g (2) ) = f(5) = 13 (c) f( f(1) ) = f(1) = 1 (d) g( g(5) ) = g(26) = 677

Composite Functions 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)) = g(3x-2) = (3x-2)2 + 1 = 9x2 - 12x + 5 (b) f(g(x)) = f(x2 + 1) = 3(x2 + 1) - 2 = 3x2 + 1 NB: g(f(x))  f(g(x)) in general. CHECK g(f(2)) = 9 x 22 - 12 x 2 + 5 = 36 - 24 + 5 = 17 f(g(2)) = 3 x 22 + 1 = 13

Composite Functions 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)) Put x2 - 6x + 13 = 8 = g(x - 3) then x2 - 6x + 5 = 0 = (x - 3)2 + 4 or (x - 5)(x - 1) = 0 = x2 - 6x + 13 So x = 1 or x = 5 CHECK g(h(5)) = g(2) = 22 + 4 = 8

Composite Functions 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 = {xR: x  -2 or 2 }

Composite Functions x = 0 (0 + 4)(0 - 2) = negative (ii) Suppose that g(x) = (x2 + 2x - 8) x = 3 (3 + 4)(3 - 2) = positive We need (x2 + 2x - 8)  0 x = -5 (-5 + 4)(-5 - 2) = positive Suppose (x2 + 2x - 8) = 0 -4 2 Then (x + 4)(x - 2) = 0 So x = -4 or x = 2 Check values below -4 , between -4 and 2, then above 2 So domain = { xR: x  -4 or x  2 }

Exponential (to the power of) Graphs Exponential Functions A function in the form f(x) = ax where a > 0, a ≠ 1 is called an exponential function to base a . Consider f(x) = 2x x -3 -2 -1 0 1 2 3 f(x) 1 1/8 ¼ ½ 1 2 4 8

Graph The graph of y = 2x (1,2) (0,1) 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.

Log Graphs y = log2x ie x 1/8 ¼ ½ 1 2 4 8 y -3 -2 -1 0 1 2 3 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 “the logarithm to base 2 of x” y = log2x or “log base 2 of x”

Graph y = log2x (2,1) (1,0) The graph of NB: x > 0 Major Points (i) y = log2x passes through the points (1,0) & (2,1) . As x ∞ y ∞ but at a very slow rate and as x  0 y  -∞ .

y = ax Exponential (to the power of) Graphs The graph of y = ax always passes through (0,1) & (1,a) It looks like .. Y y = ax (1,a) (0,1) x

The graph of y = logax always passes through (1,0) & (a,1) Log Graphs The graph of y = logax always passes through (1,0) & (a,1) Y It looks like .. (a,1) (1,0) x y = logax

Graph Transformations We will investigate f(x) graphs of the form 1. f(x) ± k f(x ± k) -f(x) f(-x) kf(x) f(kx) Each moves the Graph of f(x) in a certain way !

Graph of f(x) ± k Transformations 3 y = f(x) y = x2 2 y = f(x) ± k y = x2-3 1 y = x2+ 1 Mathematically y = f(x) ± k moves f(x) up or down Depending on the value of k + k move up - k move down -2 -1 1 2 -1 -2 -3

Graph of -f(x ± k) Transformations Mathematically y = f(x ± k) moves f(x) to the left or right depending on the value of k -k move right + k move left Graph of -f(x ± k) Transformations y = f(x) y = x2 y = f(x ± k) y = (x-1)2 y = (x+2)2 3 2 1 -4 -3 -2 -1 1 2 3

Graph of -f(x) Transformations y = f(x) y = x2 y = -f(x) y = -x2 Mathematically y = –f(x) reflected f(x) in the x - axis

Graph of -f(x) Transformations y = f(x) y = 2x + 3 y = -f(x) y = -(2x + 3) Mathematically y = –f(x) reflected f(x) in the x - axis

Graph of -f(x) Transformations y = f(x) y = x3 y = -f(x) y = -x3 Mathematically y = –f(x) reflected f(x) in the x - axis

Graph of f(-x) Transformations y = f(x) y = x + 2 y = f(-x) y = -x + 2 Mathematically y = f(-x) reflected f(x) in the y - axis

Graph of f(-x) Transformations y = f(x) y = (x+2)2 y = f(-x) y = (-x+2)2 Mathematically y = f(-x) reflected f(x) in the y - axis

Graph of k f(x) Transformations y = f(x) y = x2-1 3 y = k f(x) y = 4(x2-1) 2 y = 0.25(x2-1) 1 Mathematically y = k f(x) Multiply y coordinate by a factor of k k > 1  (stretch in y-axis direction) 0 < k < 1  (squash in y-axis direction) -2 -1 1 2 -1 -4

Graph of -k f(x) Transformations y = f(x) y = x2-1 y = k f(x) y = -4(x2-1) 4 y = -0.25(x2-1) Mathematically y = -k f(x) k = -1 reflect graph in x-axis k < -1  reflect f(x) in x-axis & multiply by a factor k (stretch in y-axis direction) 0 < k < -1  reflect f(x) in x-axis multiply by a factor k (squash in y-axis direction) 1 -2 -1 1 2 -1 -4

Graph of f(kx) Transformations 4 y = f(x) y = (x-2)2 3 y = f(kx) y = (2x-2)2 2 y = (0.5x-2)2 1 1 2 3 4 5 6 Mathematically y = f(kx) Multiply x – coordinates by 1/k k > 1  squashes by a factor of 1/k in the x-axis direction k < 1  stretches by a factor of 1/k in the x-axis direction

Graph of f(-kx) Transformations 4 y = f(x) y = (x-2)2 3 y = f(-kx) y = (-2x-2)2 2 1 y = (-0.5x-2)2 -4 -3 -2 -1 1 2 3 4 Mathematically y = f(-kx) k = -1 reflect in y-axis k < -1  reflect & squashes by factor of 1/k in x direction -1 < k > 0  reflect & stretches factor of 1/k in x direction

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 !

Trig Graphs y = sinx - 2 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. 1 90o 180o 270o 360o -1 -2 y = sinx - 2 -3

Trig Graphs y = cos(x - 50)o 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. 50o 1 90o 180o 270o 360o -1 -2 y = cos(x - 50)o -3

Trig Graphs y = 3sinx Sketch the graph of y = 3sinx If sinx = f(x) then 3sinx = 3f(x) So stretch the sinx graph 3 times vertically. 3 2 1 90o 180o 270o 360o -1 -2 y = 3sinx -3

Trig Graphs y = cos4x Sketch the graph of y = cos4x If cosx = f(x) then cos4x = f(4x) So squash the cosx graph to 1/4 size horizontally 1 90o 180o 270o 360o -1 y = cos4x

Trig Graphs y = 2sin3x 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. 3 2 1 90o 180o 270o 360o -1 -2 y = 2sin3x -3

Radians Radian measure is an alternative to degrees and is based upon the ratio of arc length radius a θ θ(radians) = r θ- theta

Radians θ (radians) = r/r = 1 So θ (radians) = πr /r = π a a = r If the arc length = the radius θ r θ (radians) = r/r = 1 If we now take a semi-circle a Here a = ½ of circumference = ½ of πd θ (180o) r = πr So θ (radians) = πr /r = π

Converting then X π ÷180 degrees radians ÷ π then x 180

Radians Since we have a semi-circle the angle must be 180o. We now get a simple connection between degrees and radians. π (radians) = 180o This now gives us 2π = 360o π /2 = 90o 3π /2 = 270o π /3 = 60o 2π /3 = 120o π /4 = 45o 3π /4 = 135o π /6 = 30o 5π /6 = 150o NB: Radians are usually expressed as fractional multiples of π.

Converting Ex1 72o = 72/180 X π = 2π /5 Ex2 330o = 330/180 X π = 11 π /6 Ex3 2π /9 = 2π /9 ÷ π x 180o = 2/9 X 180o = 40o Ex4 23π/18 = 23π /18 ÷ π x 180o = 23/18 X 180o = 230o

This triangle will provide exact values for Some special values of Sin, Cos and Tan are useful left as fractions, We call these exact values 60º 2 2 60º 30º 3 1 This triangle will provide exact values for sin, cos and tan 30º and 60º

Exact Values ½  ½ 1 1 x 0º 30º 45º 60º 90º 3 Sin xº Cos xº Tan xº 3 2 ½ 1 3 2 1 ½ 3

Exact Values 2 45º 1 1 45º 1 1 Consider the square with sides 1 unit We are now in a position to calculate exact values for sin, cos and tan of 45o

Exact Values ½ ½ 1 1 1 x 0º 30º 45º 60º 90º 1 2 1 2 3 Sin xº Cos xº Tan xº Undefined ½ 1 2 3 2 1 3 2 1 2 1 ½ 1 3

Exact value table and quadrant rules. tan150o = - tan(180 - 150) o = -1/√3 (Q2 so neg) cos300o = cos(360 - 300) o = cos60o = 1/2 (Q4 so pos) sin120o = sin(180 - 120) o = sin60o = √ 3/2 (Q2 so pos) tan300o = - tan(360-300)o = - tan60o = - √ 3 (Q4 so neg)

Find the exact value of cos2(5π/6) – sin2(π/6) Exact value table and quadrant rules. Find the exact value of cos2(5π/6) – sin2(π/6) cos(5π/6) = cos150o = cos(180 - 150)o = - cos30o = - √3 /2 (Q2 so neg) = 1/2 sin(π/6) = sin30o cos2(5π/6) – sin2(π/6) = (- √3 /2)2 – (1/2)2 = ¾ - 1/4 = 1/2

Prove that sin(2 π /3) = tan (2 π /3) Exact value table and quadrant rules. Prove that sin(2 π /3) = tan (2 π /3) cos (2 π /3) sin(2π/3) = sin120o = sin(180 – 120)o = sin60o = √3/2 cos(2 π /3) = cos120o = cos(180 – 120)o = - cos60o = -1/2 tan(2 π /3) = tan120o = tan(180 – 120)o = -tan60o = - √3 sin(2 π /3) cos (2 π /3) LHS = = √3/2 ÷ -1/2 = √3/2 X -2 = - √3 = tan(2π/3) = RHS

Solving Trig Equations 1 2 3 4 Sin +ve All +ve 180o - xo 180o + xo 360o - xo Tan +ve Cos +ve created by Mr. Lafferty

Solving Trig Equations C A S T 0o 180o 270o 90o Solving Trig Equations Graphically what are we trying to solve Example 1 Type 1: Solving the equation sin xo = 0.5 in the range 0o to 360o 1 2 3 4 sin xo = (0.5) xo = sin-1(0.5) xo = 30o There is another solution xo = 150o (180o – 30o = 150o) created by Mr. Lafferty

Solving Trig Equations C A S T 0o 180o 270o 90o Solving Trig Equations Graphically what are we trying to solve Example 2 : Solving the equation cos xo - 0.625 = 0 in the range 0o to 360o 1 2 3 4 cos xo = 0.625 xo = cos -1 (0.625) xo = 51.3o There is another solution xo = 308.7o (360o - 53.1o = 308.7o) created by Mr. Lafferty

Solving Trig Equations C A S T 0o 180o 270o 90o Solving Trig Equations Graphically what are we trying to solve Example 3 : Solving the equation tan xo – 2 = 0 in the range 0o to 360o 1 2 3 4 tan xo = 2 xo = tan -1(2) xo = 63.4o There is another solution x = 180o + 63.4o = 243.4o created by Mr. Lafferty

Solving Trig Equations C A S T 0o 180o 270o 90o Solving Trig Equations Graphically what are we trying to solve Example 4 Type 2 : Solving the equation sin 2xo + 0.6 = 0 in the range 0o to 360o sin 2xo = (-0.6) 2xo = sin-1(0.6) 2xo = 37o ( always 1st Q First) 2xo = 217o , 323o 577o , 683o ...... ÷2 xo = 108.5o , 161.5o 288.5o , 341.5o created by Mr. Lafferty

Solving Trig Equations C A S T 0o 180o 270o 90o Solving Trig Equations Graphically what are we trying to solve Example 5 Type 3 : Solving the equation 2sin (2xo - 30o) - √3 = 0 in the range 0o to 360o 2sin (2x - 30o) = √3 sin (2x - 30o) = √3 ÷ 2 2x - 30o = sin-1(√3 ÷ 2) 2xo - 30o = 60o , 120o ,420o , 480o ......... 2xo = 90o , 150o ,450o , 530o ......... ÷2 xo = 45o , 75o 225o , 265o created by Mr. Lafferty

Solving Trig Equations C A S T 0o 180o 270o 90o Solving Trig Equations Graphically what are we trying to solve Example 5 Type 4 : Solving the equation cos2x = 1 in the range 0o to 360o cos2 xo = 1 cos xo = ± 1 cos xo = 1 xo = 0o and 360o cos xo = -1 xo = 180o created by Mr. Lafferty

Solving Trig Equations C A S T 0o 180o 270o 90o C A S T 0o 180o 270o 90o Solving Trig Equations Example 6 Type 5 : Solving the equation 3sin2x + 2sin x - 1 = 0 in the range 0o to 360o Let p = sin x We have 3p2 + 2p - 1 = 0 Factorise (3p – 1)(p + 1) = 0 3p – 1 = 0 p + 1 = 0 p = 1/3 p = - 1 sin x = 1/3 sin x = -1 xo = 19.5o and 160.5o xo = 270o

Trig Identities An identity is a statement which is true for all values. eg 3x(x + 4) = 3x2 + 12x eg (a + b)(a – b) = a2 – b2 Trig Identities (1) sin2θ + cos2 θ = 1 (2) sin θ = tan θ cos θ θ ≠ an odd multiple of π/2 or 90°.

Trig Identities (1) sin2θo + cos2 θo = a2 +b2 = c2 c a sinθo = a/c b Reason a2 +b2 = c2 c a sinθo = a/c θo b cosθo = b/c (1) sin2θo + cos2 θo =

Simply rearranging we get two other forms Trig Identities Simply rearranging we get two other forms sin2θ + cos2 θ = 1 sin2 θ = 1 - cos2 θ cos2 θ = 1 - sin2 θ

Since θ is between 0 < θ < π/2 Trig Identities Example1 sin θ = 5/13 where 0 < θ < π/2 Find the exact values of cos θ and tan θ . cos2 θ = 1 - sin2 θ Since θ is between 0 < θ < π/2 then cos θ > 0 = 1 – (5/13)2 So cos θ = 12/13 = 1 – 25/169 tan θ = sinθ cos θ = 5/13 ÷ 12/13 = 144/169 cos θ = √(144/169) = 5/13 X 13/12 = 12/13 or -12/13 tan θ = 5/12

Trig Identities Given that cos θ = -2/ √ 5 where π< θ < 3 π /2 Find sin θ and tan θ. Since θ is between π< θ < 3 π /2 sinθ < 0 sin2 θ = 1 - cos2 θ = 1 – (-2/ √ 5 )2 Hence sinθ = - 1/√5 = 1 – 4/5 tan θ = sinθ cos θ = - 1/ √ 5 ÷ -2/ √ 5 = 1/5 = - 1/ √ 5 X - √5 /2 sin θ = √(1/5) = 1/ √ 5 or - 1/ √ 5 Hence tan θ = 1/2

Higher Maths Graphs & Functions Strategies Click to start

Graphs & Functions Higher The following questions are on Graphs & Functons Non-calculator questions will be indicated You will need a pencil, paper, ruler and rubber. Click to continue

Graphs & Functions Higher 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) a) Reflect across the y axis b) Now scale by 2 in the y direction Hint Previous Quit Quit Next

Graphs & Functions Higher The diagram shows a sketch of part of the graph of a trigonometric function whose equation is of the form Determine the values of a, b and c 2a 1 a = 4 1 in  2 in 2  a is the amplitude: b = 2 b is the number of waves in 2 c = 1 c is where the wave is centred vertically Hint Previous Quit Quit Next

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) Hint Previous Quit Quit Next

Graphs & Functions Higher a) Express in the form b) On the same diagram sketch i) the graph of ii) the graph of c) Find the range of values of x for which is positive b) a) c) Solve: Hint 10 - f(x) is positive for -1 < x < 5 Previous Quit Quit Next

Graphs & Functions Higher The graph of a function f intersects the x-axis at (–a, 0) and (e, 0) as shown. There is a point of inflexion at (0, b) and a maximum turning point at (c, d). Sketch the graph of the derived function m is + m is + m is - Hint f(x) Previous Quit Quit Next

Graphs & Functions Higher Functions f and g are defined on suitable domains by and a) Find expressions for: i) ii) b) Solve a) b) Hint Previous Quit Quit Next

Graphs & Functions Higher The diagram shows the graphs of two quadratic functions Both graphs have a minimum turning point at (3, 2). Sketch the graph of and on the same diagram sketch the graph of y=f(x) y=g(x) Hint Previous Quit Quit Next

Graphs & Functions Higher are defined on a suitable set of real numbers. a) Find expressions for b) i) Show that ii) Find a similar expression for and hence solve the equation a) b) Now use exact values Repeat for ii) Hint equation reduces to Previous Quit Quit Next

Graphs & Functions Higher A sketch of the graph of y = f(x) where is shown. The graph has a maximum at A and a minimum at B(3, 0) a) Find the co-ordinates of the turning point at A. b) Hence, 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. c) Write down the range of values of k for which g(x) = k has 3 real roots. a) Differentiate for SP, f(x) = 0 when x = 1 t.p. at A is: b) Graph is moved 2 units to the left, and 4 units up t.p.’s are: Hint c) For 3 real roots, line y = k has to cut graph at 3 points from the graph, k  4 Previous Quit Quit Next

Graphs & Functions Higher a) Find b) If find in its simplest form. a) b) Hint Previous Quit Quit Next

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) graph moves to the left 1 unit b) graph is reflected in the x axis graph is then scaled 2 units in the y direction Hint Previous Quit Quit Next

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. c) Find the area enclosed between this graph and the x-axis. a) b) c) Graph cuts x axis at 0 and 1 Now evaluate Hint Area Previous Quit Quit Next

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

Graphs & Functions Higher You have completed all 13 questions in this section Previous Quit Quit Back to start

Graphs & Functions Higher Table of exact values 30° 45° 60° sin cos tan 1 Previous