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KING KHALID ISLAMIC COLLEGE OF VICTORIA Proudly presents… Mathematical Methods CONTENTS: (Click on topic you want) 1. Factorisation 2. Graphing 3. Differentiation 4. Integration Mathematical Methods CONTENTS: (Click on topic you want) 2. Graphing Quadratics Cubics Quartics Circular functions Exponential Logarithmic Domain and Range RETURN TO MAIN MENU Factorisation (1.01) 12x3 + 8x2 – 4x = 4x (3x2 + 2x – 1) 1. Common factors 2. Difference of Perfect Squares a2 – b2 = (a + b)(a – b) 3. Perfect Squares a2 + 2ab + b2 = (a + b)2 a2 - 2ab + b2 = (a - b)2 4. Sum and Product x2 + (a + b)x + ab = (x + a)(x + b) 5. Completion of Perfect Squares eg x2 – 6x + 3 can’t be factorised by 1, 2, 3, or 4, so = x2 – 6x + 9 - 9 + 3 Completion of square = (x – 3)2 - (6 )2 = (x – 3 + 6)(x –3 - 6) Difference of squares Factorisation (1.01) Cubic expressions 1. Sum of Cubes (a + b)3 = (a + b)(a2 – ab + b2) 2. Difference of Cubes (a - b)3 = (a - b)(a2 + ab + b2) Factorisation (1.01) Factor Theorem For a polynomial in x, P(x) , if (x + a) is a factor, then P(-a) = 0 eg Find the factors of x3 + 4x2 – 7x - 10 Try (x - 1): Test P(1) = (1)3 + 4(1)2 – 7(1) – 10 = -12 (x – 1) is not a factor Try (x + 1): Test P(-1) = (-1)3 + 4(-1)2 – 7(-1) – 10 = 0 (x + 1) is a factor Try (x – 2): Test P(2) = (2)3 + 4(2)2 – 7(2) – 10 = 0 (x - 2) is a factor If (x + 1) and (x – 2) are factors then (x + 5) must be the third factor (1 x –2 x 5 = -10) Remainder Theorem If P(-a) does not equal zero, the number left is the remainder of P(x) divided by (x + a) Division of a polynomial RETURN TO CONTENTS (1.01) If (x – 3) is a factor of 3x3 – 7x2 – 9x + 9, find the other factor. 3x2 + 2x - 3 ) (x – 3) 3x3 – 7x2 – 9x + 9 3x3 – 9x2 +2x2 - 9x 2x2 – 6x - 3x + 9 -3x + 9 No remainder 3x3 – 7x2 – 9x + 9 = (x – 3)(3x2 +2x – 3) Graphing All graphs follow a standard set of transformation rules: 1. Reflections Negative before the function, y = - f(x) reflects vertically (top for bottom). Negative inside the function, y = f(-x) reflects horizontally (left for right). 2. Dilations Number (>1) before the function, y = a.f(x), dilates (stretches) vertically. Number (>1) inside the function, y = f(ax), dilates (compresses) horizontally. 3. Translations Number added to or subtracted from the function, slides up or down. Number added or subtracted inside the function (inside the brackets), slides the graph horizontally in the opposite direction. Parabolas y= y 5 x2 3 1 -4 -2 2 -1 -3 -5 4 x Parabolas y - Vertical translation 5 y = x2 + 2 3 1 -4 -2 2 -1 -3 -5 4 x Parabolas y - Vertical translation 5 y = x2 - 3 3 1 -4 -2 2 -1 -3 -5 4 x Parabolas y - Vertical reflection 5 y = - x2 3 1 -4 -2 2 -1 -3 -5 4 x Parabolas y - Vertical translation - Vertical reflection 5 3 y = 5 - x2 1 -4 -2 2 -1 -3 -5 4 x Parabolas y - Horizontal translation 5 y = (x – 4)2 3 1 -4 -2 2 -1 -3 -5 4 x Parabolas y - Horizontal translation 5 y = (x + 3)2 3 1 -4 -2 2 -1 -3 -5 4 x Parabolas y - Horizontal translation 5 - Vertical reflection 3 y = - (x – 1)2 1 -4 -2 2 -1 -3 -5 4 x RETURN TO CONTENTS Parabolas y Vertical and horizontal translation 5 3 y = (x – 2)2 1 -3 Turning point form -4 -2 2 -1 -3 Turning point at (2, -3) -5 4 x Cubics y 5 y = x3 3 1 Three factors are the same, so Point of inflection occurs at (0, 0) -4 -2 2 -1 -3 -5 4 x Cubics y 5 y = (x – 3)3 + 1 3 Turning point form 1 Three factors are the same, so Point of inflection occurs at (3, 1) -4 -2 2 -1 -3 -5 4 x Cubics y What is the equation to this curve? y = (x + 1)(x - 2)(x - 4) 6 2 -4 -2 2 -2 -6 4 x RETURN TO CONTENTS Cubics y What is the equation to this curve? 5 3 y = -(x + 3)(x - 1)2 1 or -4 y=– x3 - x2 + 5x - 3 -2 2 -1 -3 4 x Quartics y What is the equation to this curve? 6 y = x(x + 3)(x - 3)(x –5) 2 or -4 -2 2 -2 y = x4 – 5x3 - 9x2 + 45x -6 4 x RETURN TO CONTENTS Quartics y What is the equation to this curve? y = x(x + 3)(x - 3)(x –5) 6 2 or y = x4 – 5x3 - 9x2 + 45x -4 -2 2 -2 -6 4 x TRIGONOMETRIC functions (Circular Functions) Circular functions y y = a sin nx a a – amplitude (the height) 0 n Period, T = -a (the length) 2 n 2 n x Circular functions y y = 3sin2x + 2 5 Amplitude, a = 3 Vertical Translation = +2 x 0 -5 Period, T = Circular functions y What is the equation to this curve? 1 y = sin x 0 -1 2 x Circular functions y What is the equation to this curve? 1 0 y = cos x -1 2 x Circular functions y What is the equation to this curve? 5 0 y = tan x -5 2 x Circular functions y What is the equation to this curve? 4 0 y = 4 sin x -4 2 x Circular functions What is the equation to this curve? y 5 y = 5 sin 2x + 2 0 -5 x Circular functions л 2 л-θ л θ–л or θ+л θ Only Sin is positive All functions are 1 positive 2 3 4 Only Only Tan Cos is positive is positive 3л 2 0, 2л 2л - θ Circular functions What is the equation to this curve? RETURN TO CONTENTS y 5 y = 5 sin 2x + 2 0 -5 x Exponential Equations y = ax (1.02) 1 To solve an exponential equation, make the bases the same. eg 25x = 125 Becomes 52x = 53 Now, equate indices: 2x = 3 and so, x = 1.5 Exponential Equations (1.02) To solve an exponential equation when you cannot find common bases. eg 73x = 66 No common base, so take log10 of both sides: log10 73x = log10 66 and so, 3x.log10 7 = log10 66 By rearrangement, 3x = log10 66 log10 7 and, x = 0.718 Exponential Equations eg becomes or (1.02) RETURN TO CONTENTS 9x - 4 3x + 3 = 0 (32)x + 4(3x) + 3 = 0 (3x)2 + 4(3x) + 3 = 0 Notice the quadratic? To make solving easier, let 3x = a a2 + 4a + 3 = 0 and (a - 1)(a - 3) = 0 a = 1, 3 Since a = 3x , 3x = 1 or 3x = 3 x = 0 or 1 LOGARITHMS Logarithms The logarithm is the inverse operation of the exponential. eg If ax = b then logab = x So, 53 = 125 is equivalent to log5 125 = 3 If log4 x = 3 then x = 64 (because 43 = x and x = 64) Logarithmic functions RETURN TO CONTENTS y What is the equation to this curve? No, think about it a bit more! 1 (5, 1) 0 1 5 What does the point (5, 1) tell you? When y = 1, x = 5, so… y = log5 x -1 x Domain and Range Domain means all of the x-values for which the relation exists. If you substitute an x-value into the equation and you get an answer for y, then that x-value is part of the domain. eg for the expression y = x , if x = 4 then y = 4 = 2 so x = 4 is part of the domain. However, if x = - 4 then y = 4 = undefined so x = - 4 is not part of the domain. The domain for y = x , is x 0 or [0, ) RETURN TO CONTENTS Domain and Range Range means all of the y-values for which the relation exists. If you substitute an x-value into the equation and you get an answer for y, then that y-value is part of the domain. eg for the expression y = x , if x = 4 then y = 4 = 2 so y = 2 is part of the range. Since it is impossible to get a real solution for y = x which is negative, y can only have non-negative values. The range for y = x , is y 0 or [0, ) Inverse Functions Gradient Differentiation is a technique for finding the gradient of any point on a curved line. Finding the gradient of a straight line is easy because the gradient is the same at every part of the line. y y2 Gradient of a straight line, y2 y1 rise m= = x2 x1 run You only need to know the coordinates of two points on the line, (x1, y1) and (x2, y2) (x1,y1) . x1 (x2,y2) . rise y1 run x2 x Average Gradient Average gradient of a curved line is just as easy because it is the same as the gradient of the straight line between the two points y2 y1 Average Gradient, m = x2 x1 y B(x2,y2) You only need to know the coordinates of two points on the curve, (x1, y1) and (x2, y2) A(x1,y1) x or (using f(x) instead of y) Average Gradient, m = f ( x h) f ( x ) ( x h) x f ( x h) f ( x ) Therefore, m = h You only need to know the coordinates of two points on the curve, (x, f(x)) and ((x+h), f(x+h)) f(x) B(x+h,f(x+h)) A(x,f(x)) x x+h x Differentiation is a technique for finding the gradient at one point on a curved line. Gradient of a straight line between two points is given by: m= y x = f ( x h) f ( x ) where h is the distance between h the x coordinates. However, to find the gradient at one point, the distance, h, between the x-coordinates has to become zero. The problem here is that we end up with zero in the denominator. y To get around this we must simplify the numerator so that we can cancel the h from the denominator Mathematically, this is written as: dy lim f ( x h) f ( x) m = dx = f’(x) = h 0 h A. x h h h x Differentiation is a technique for finding the gradient at one point on a curved line. The wonderful thing about differentiation is that there is a rule to make things easier: If y = axn then dy = naxn-1 dx If f(x) = axn and then f΄(x) =naxn-1 If y = x3 + 5x2 - 3x + 7 dy Then = 3x2 + 10x - 3 dx eg Gradients Differentiation is a technique for finding the gradient of a curve. The derivative is also called the Gradient Function - it gives the gradient at any point on the curve. eg. f(x) = x2 - 4x + 7 f΄(x) = 2x – 4 (gradient function) y The The gradient gradient at at xx = = 231 f΄(2) f΄(3) 2(3) f΄(1) = = 2(2) 2(1) –– 44 m==-022 m x Finding the coordinates of a point where the gradient is given eg What are the coordinates of the point on the curve y = x2 + 3x – 7 where the gradient is equal to 5? 1. Differentiate y΄ = 2x + 3 2. Equate to m 2x + 3 = 5 3. Solve for x 2x = 2 x =1 4. Substitute into the original equation y = (1)2 + 3(1) - 7 y=-3 The gradient of y = x2 + 3x – 7 is equal to 5 at (1, - 3) Finding the coordinates of a turning point (stationary point) Horizontal turning points have a gradient of zero eg What are the coordinates of the stationary points on the curve y = x3 + 3x2 – 9x ? 1. Differentiate y΄ = 3x2 + 6x – 9 2. Equate to zero 3x2 + 6x – 9 = 0 (turning point) 3. Solve for x x2 + 2x - 3 = 0 (x + 3)(x – 1) = 0 x = 1, -3 4. Substitute into the original equation At x = 1, y = (1)3 + 3(1)2 – 9(1) y=-5 There are turning points at (1, -5) and at (-3, 27) At x = -3, y = (-3)3 + 3(-3)2 – 9(-3) y = 27 Antidifferentiation – the indefinite integral Antidifferentiation “undoes” differentiation. It turns the derivative back into the original function. axn dx = ax n 1 n 1 +c The indefinite integral reverses the differentiation but c must always be added to make up for the constant which disappears during differentiation. eg (5x3 – 3x2) dx 5x4 = 4 = 5x4 4 - 3x 3 3 +c - x3 + c The Indefinite Integral – yields yields the a function original with an function unknown when constant, you have been c. given the derivative. eg.The Findvalue the equation thefound function whose the of c canto2be by knowing gradient is given by 6x 4x – 3, given that the coordinates of a point the original curve. graph of the function passeson through the point (2, 3) Gradient means dy dx so, the original function, y = dy dx .dx y = (6x2 - 4x – 3) dx and hence, y = 2x3 - 2x2 – 3x + c Now, since this function passes through the point (2, 3), we can say: 3 = 2(2)3 - 2(2)2 – 3(2) + c (Substitute into x and y) Now, solving for c, c = 3 – (16 – 8 – 6) = 1 And so, y = 2x3 - 2x2 – 3x + 1 The Indefinite Integral – yields the original function when you have been given the derivative. Try this: If f(x) = 3x2 + 8x and f(-3) = 2, find f(x). f(x) = f (x) dx = f(3x2 + 8x) dx = x3 + 4x2 + c If f(-3) = 2, then (-3)3 + 4(-3)2 + c = 2 c = 2 +27 –36 = - 7 So, f(x) = x3 + 4x2 - 7 Definite Integral The Definite Integral gives the integral between two limits. Area RETURN TO CONTENTS