Higher Maths 2.1.2 - Quadratic Functions

SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART

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

Sketching Quadratic Functions NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART 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

Perfect Squares NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART x 2 + 6 x + 2 x 2 + 6 x + 2 ( ) 2 + 9 – 9 ( ) x + 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 =  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 NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART 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 ) 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 SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART Solving Quadratic Equations NOTE Example f ( x ) = 0 Use the graph to solve x x = -2 - 2 5 x = 5 or Example 2 Solve 6 x 2 + x – 15 = 0 6 x 2 + x – 15 = 0 ( 2 x – 3 )( 3 x + 5 ) = 0 The trinomial can be factorised... or 2 x – 3 = 0 3 x + 5 = 0 2 x = 3 3 x = - 5 x = 3 2 x = - 5 3 ( f ) x

If a Quadratic Equation cannot be factorised, it is sometimes possible to solve by completing the square. SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART Solving Quadratic Equations by Completing the Square NOTE 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 – 2 ( ) 2 = 5 x – 2 x – 2 = 5 ± x = 2 5 ± Now solve for x ... x ≈ 4.24 - 0.24 or Not accurate! most accurate answer

Quadratic Inequations NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART 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 x - 4 3 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 12

The Quadratic Formula NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART x = - b b 2 – ( 4 ac ) ± 2 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 of are given by x root If roots cannot be found using the quadratic formula, they are impossible to find . x no roots LEARN THIS ( f ) x ( f ) x

Real and Imaginary Numbers NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART 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 . ? x 1 real root no real roots x x 2 real roots ( f ) x ( f ) x ( f ) x

• If • If The Discriminant NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART x = - b b 2 – ( 4 ac ) ± 2 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 NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART 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 > 2

Finding Unknown Coefficients using Quadratic Inequations NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART Example Find q given that x 2 + ( q – 3 ) x + q = 0 has non-real roots. 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: q 1 9 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 NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART 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 NOTE SLIDE Higher Maths 2 1 2 Quadratic Functions UNIT OUTCOME PART 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

Higher Maths 2.1.2 - Quadratic Functions

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