This document provides an overview of quadratic functions including:
- The graph of a quadratic function is a parabola.
- The general form of a quadratic function is f(x) = ax2 + bx + c.
- Key aspects that must be identified before sketching the graph are the nature of the turning point, y-intercept, x-intercepts, and axis of symmetry.
- Quadratic equations can be solved by factorizing, completing the square, or using the quadratic formula. The discriminant determines the nature of the roots.
2. 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
3. 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
4. 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
5. 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
6. 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
7. 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
8. 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
9. 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
10. 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
11. • 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’)
12. 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
13. 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
14. 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 ) .
15. 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