# 1Higher Maths 2 1 2 Quadratic Functions. Any function containing an term is called a Quadratic Function. The Graph of a Quadratic Function 2Higher Maths.

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1Higher Maths 2 1 2 Quadratic Functions

Any function containing an term is called a Quadratic Function. The Graph of a Quadratic Function 2Higher Maths 2 1 2 Quadratic Functions The graph of a Quadratic Function is a type of symmetrical curve called a parabola. x 2x 2 f ( x ) = ax 2 + bx + c General Equation of a Quadratic Function a > 0 a < 0 Minimum turning point Maximum turning point with a ≠ 0 turning point

Sketching Quadratic Functions 3Higher Maths 2 1 2 Quadratic Functions Before it is possible to sketch the graph of a quadratic function, the following information must be identified: the nature of the turning point i.e. the coordinates of the y -intercept the zeroes or ‘roots’ of the function i.e. the x -intercept(s), if any minimum or maximum the location of the axis of symmetry and coordinates of the turning point a > 0 or a < 0 substitute x = 0 solve f ( x ) = 0 : ax 2 + bx + c = 0 evaluate f ( x ) at axis of symmetry

The General Quadratic Equation Note (Back) Maths Intermediate 2 (Units 1, 2 and 3) Friday, 28 August 2015 Any quadratic equation can be written in the basic form Examples a The vertical size of the parabola. positive negative Writing the equation in this way makes it easy to sketch the parabola using, and. b a c b The -coordinate of the vertex. x ( x – b )²( x – b )² y x ( x + b )²( x + b )² y x positive negative c The -coordinate of the vertex. y y x ( 4, 3 )( 4, 3 ) y x ( -3, 7 ) y = ( x – 4 ) ² + 3 y = - ( x + 3 ) ² + 7 Note that the equation of the axis of symmetry is simply x =x = b. y = ( x – ) ² + b ac

Maths Intermediate 2 (Units 1, 2 and 3) Friday, 28 August 2015 If we know the equation of a parabola, it is also possible to calculate the -intercept. y Example The General Quadratic Equation Note (Back) Calculating the -intercept y Find the y -intercept of the graph with equation y x y = 2( x – 3 ) ² – 1 as shown opposite. At the y -intercept, x = 0x = 0 y = 2( 0 – 3 ) ² – 1 = 2 × ( - 3 ) ² – 1 = 2 × 9 – 1 = 17 17 a The vertical size of the parabola. positive negative b The -coordinate of the vertex. x ( x – b )²( x – b )² y x ( x + b )²( x + b )² y x positive negative c The -coordinate of the vertex. y

Perfect Squares 6Higher Maths 2 1 2 Quadratic Functions x 2 + 6 x + 2 ( ) 2 + 9– 9 () x + 3+ 3– 7 A perfect square is an expression that can be written in the form Example Complete the square for x 2 + 6 x + 2 = = (... ) 2 x 2 + 4 x + 4 ( ) 2 x + 2+ 2 =  x 2 + 5 x + 9 =  ( ) 2 x + ?+ ? Step One: Separate number term Step Two: Try to form a perfect square from remaining terms Step Three: Remember to balance the extra number, then write out result

Completing the Square 7Higher Maths 2 1 2 Quadratic Functions y = x 2 – 8 x + 19 = x 2 – 8 x + 19 = ( ) 2 + 16– 16 () x – 4+ 3 It is impossible for a square number to be negative. The minimum possible value of is zero. ( x – 4 ) 2 The minimum possible value of y is 3. x = 4. This happens when The minimum turning point is at ( 4, 3 )( 4, 3 ) A turning point can often be found by completing the square. Example y - coordinate x - coordinate Find the minimum turning point of

Quadratic Equations can be solved in several different ways: using a graph to identify roots factorising completing the square using the quadratic formula 8Higher Maths 2 1 2 Quadratic Functions Solving Quadratic Equations Example f ( x ) = 0f ( x ) = 0 Use the graph to solve x = -2 -2-2 5 x = 5x = 5 or Example 2 Solve 6 x 2 + x – 15 = 0 ( 2 x – 3 )( 3 x + 5 ) = 0 The trinomial can be factorised... or 2 x – 3 = 03 x + 5 = 0 2 x = 3 3 x = - 5 x = 3 2 x = - 5 3

If a Quadratic Equation cannot be factorised, it is sometimes possible to solve by completing the square. 9Higher Maths 2 1 2 Quadratic Functions Solving Quadratic Equations by Completing the Square Example Solve x 2 – 4 x – 1 = 0 The trinomial cannot be factorised so complete the square... x 2 – 4 x – 1 = 0 – 1 = 0 () ( ) 2 + 4– 4 x 2 – 4 x – 5 = 0 x – 2x – 2 ( ) 2 = 5 x – 2x – 2 x – 2x – 2 ± x = 2 5 ± Now solve for x... x ≈ 4.24 - 0.24 or Not accurate! most accurate answer

Quadratic Inequations 10Higher Maths 2 1 2 Quadratic Functions A Quadratic Inequation can be solved by using a sketch to identify where the function is positive or negative. Example Find the values of x for which 12 – 5 x – 2 x 2 > 0 Factorise 12 – 5 x – 2 x 2 = 0 ( 4 + x )( 3 – 2 x ) = 0 The graph has roots x = - 4 and and a y - intercept at 12 -4-4 3 2 1 2 Now sketch the graph: The graph is positive for - 4 < x < and negative for x < - 4 3 2 x > and 3 2 3 2

The Quadratic Formula 11Higher Maths 2 1 2 Quadratic Functions x = -b-b b 2 – ( 4 ac ) ± 2 a2 a with a ≠ 0 f ( x ) = ax 2 + bx + c If a quadratic function has roots, it is possible to find them using a formula. This is very useful if the roots cannot be found algebraically, i.e. by factorising or completing the square. The roots ofare given by root If roots cannot be found using the quadratic formula, they are impossible to find. no roots

Real and Imaginary Numbers 12Higher Maths 2 1 2 Quadratic Functions 36 = ±6 - 36 = It is impossible to find the square root of a negative number. In Mathematics, the square root of a negative number still exists and is called an imaginary number. It is possible for a quadratic equation to have roots which are not real. 1 real rootno real roots 2 real roots

If The Discriminant 13Higher Maths 2 1 2 Quadratic Functions x = -b-b b 2 – ( 4 ac ) ± 2 a2 a = b 2 – ( 4 ac ) The part of the quadratic formula inside the square root is known as the Discriminant and can be used to find the nature of the roots. The Discriminant If b 2 – ( 4 ac ) > 0 there are two real roots. b 2 – ( 4 ac ) = 0 there is only one real root. b 2 – ( 4 ac ) < 0 the roots cannot be calculated and are imaginary or non-real. (‘real and unequal’) (‘real and equal’)

Using the Discriminant to Find Unknown Coefficients 14Higher Maths 2 1 2 Quadratic Functions Example The quadratic equation 2 x 2 + 4 x + p = 0 Find all possible values of p. has real roots. b 2 – ( 4 ac ) 0 a = 2 b = 4 c = p For real roots, 16 – 8 p 0 16 8 p 8 p 16 p 2 The equation has real roots for. p 2 (for the roots are imaginary or non-real) p > 2p > 2

Finding Unknown Coefficients using Quadratic Inequations 15Higher Maths 2 1 2 Quadratic Functions Example Find q given that has non-real roots. x 2 + ( q – 3 ) x + q = 0 a = 1 b = ( q – 3 ) c = q b 2 – ( 4 ac ) < 0 For non-real roots, ( q – 3 ) 2 – 4 q < 0 q 2 – 6 q + 9 – 4 q < 0 q 2 – 10 q + 9 < 0 ( q – 9 ) ( q – 1 ) < 0 Sketch graph of the inequation: 19 q 2 – 10 q + 9 < 0 for 1 < q < 9 The roots of the original equation are non-real when 1 < < 9 q

Straight Lines and Quadratic Functions 16Higher Maths 2 1 2 Quadratic Functions When finding points of intersection between a line and a parabola: equate the functions rearrange into a quadratic equation equal to zero solve for substitute back to find x Example Find the points of intersection of y = x 2 + 3 x + 2 x 2 + 2 x = 0 y = x + 2 and x 2 + 3 x + 2 = x + 2 x ( x + 2 ) = 0 x = 0 x = - 2 or y = 2 y = 0 or y Points of intersection are ( - 2, 0 ) and ( 0, 2 ).

Tangents to Quadratic Functions 17Higher Maths 2 1 2 Quadratic Functions The discriminant can be used to find the number of points of intersection between a parabola and a straight line. equate the functions and rearrange into an equation equal to zero evaluate the discriminant of the new quadratic equation b 2 – ( 4 ac ) > 0 Two points of intersection b 2 – ( 4 ac ) = 0 One point of intersection b 2 – ( 4 ac ) < 0 No points of intersection the line is a tangent