Completing the Square

We need to revise some algebra before going any further There is a quick way to square a bracket without using FOIL (x + a)2 = (x + a)(x + a) = x2 + ax + ax + a2 1. Square each term = x2 + 2ax + a2 x2 + a2 2. Multiply the two terms then double your answer x2 + 2ax + a2

There is a quick way to square a bracket (x + 5)2 1. Square each term x2 + a2 x2 + 25 2. Multiply the two terms then double your answer 2 × 5 × x x2 + 2ax + a2 x2 + 10x + 25

There is a quick way to square a bracket (3x – 2)2 1. Square each term x2 + a2 9x2 + 4 2. Multiply the two terms then double your answer 2 × (–2) × 3x x2 + 2ax + a2 9x2 – 12x + 4

→ (x + a)2 [ ] ] [ Isolate the x and x2 terms in square brackets Completing the square reverses this process x2 + 2ax + a2 → (x + a)2 Isolate the x and x2 terms in square brackets [ ] x2 + 6x + 1 x2 + 6x is part of (x + 3)2 Still equal to x2 + 6x = (x + 3)2 + 1 ] [ – 9 x2 + 6x + 9 = (x + 3)2 – 8 y = (x + 3)2 – 8 A ‘HAPPY’ parabola with minimum turning point x = –3 , y = –8 (–3, –8)

→ (x + a)2 [ ] ] [ Isolate the x and x2 terms in square brackets x2 + 2ax + a2 Isolate the x and x2 terms in square brackets [ ] x2 – 2x – 7 x2 – 2x is part of (x – 1)2 Still equal to x2 – 2x ] = (x – 1)2 – 7 [ – 1 x2 – 2x + 1 = (x – 1)2 – 8 y = (x – 1)2 – 8 A ‘HAPPY’ parabola with minimum turning point x = 1 , y = –8 (1, –8)

A ‘HAPPY’ parabola with minimum turning point x = 2 , y = –7 [ ] 3x2 – 12x + 5 Take out common factor 3 = 3[x2 – 4x] + 5 x2 – 4x is part of (x – 2)2 = 3[(x – 2)2 ] + 5 – 4 x2 – 4x + 4 = 3(x – 2)2 – 12 + 5 = 3(x – 2)2 – 7 y = 3(x – 2)2 – 7 A ‘HAPPY’ parabola with minimum turning point x = 2 , y = –7 (2, –7)

A ‘SAD’ parabola with maximum turning point x = 5 , y = 26 [ ] 1 + 10x – x2 Take out common factor –1 = –1[x2 – 10x] + 1 x2 – 10x is part of (x – 5)2 = –1[(x – 5)2 ] + 1 – 25 x2 – 10x + 25 = –(x – 5)2 + 25 + 1 = –(x – 5)2 + 26 (5, 26) y = –(x – 5)2 + 26 A ‘SAD’ parabola with maximum turning point x = 5 , y = 26

Completing the Square and Fractions

The maximum value of the fraction will occur when f(x) is a minimum The minimum value of the fraction will occur when f(x) is a maximum

[ ] ] [ Express x2 + 6x + 13 in the form (x + p)2 + r. Hence state the maximum value of 1 of 2 [ ] x2 + 6x + 13 x2 + 6x is part of (x + 3)2 = (x + 3)2 + 13 ] [ – 9 x2 + 6x + 9 = (x + 3)2 + 4 y = (x + 3)2 + 4 A ‘HAPPY’ parabola with minimum turning point x = –3 , y = 4 (–3, 4)

Express x2 + 6x + 13 in the form (x + p)2 + r. Hence state the maximum value of 2 of 2 y = (x + 3)2 + 4 The maximum value of a fraction occurs when the denominator is as small as possible, ie a minimum. (–3, 4) The minimum value of x2 + 6x + 13 is 4 So MAX f(x) is 1/4

[ ] Express 3 – 8x – 4x2 in the form p(x + q)2 + r. Hence state the minimum value of [ ] 3 – 8x – 4x2 Take out common factor –4 = –4[x2 + 2x] + 3 x2 + 2x is part of (x + 1)2 = –4[(x + 1)2 ] + 1 – 1 x2 + 2x + 1 (–1, 5) = –4(x + 1)2 + 4 + 1 = –4(x + 1)2 + 5  MAX value is 5 So MIN value of f(x) is 10/5 = 2

Past Paper Examples

If 3x2 + 6x – 10 is expressed in the form 3(x + p)2 + q then the value of ‘q’ is ? [ ] 3x2 + 6x – 10 Take out common factor 3 = 3[x2 + 2x] – 10 x2 + 2x is part of (x + 1)2 = 3[(x + 1)2 ] – 10 – 1 x2 + 2x + 1 = 3(x + 1)2 – 3 – 10 = 3(x + 1)2 – 13 So q is – 13

So turning point is (–1, –17) A parabola has equation y = x2 + 6x – 8 State the coordinates of the minimum turning point? [ ] x2 + 6x – 8 = [x2 + 6x] – 8 x2 + 6x is part of (x + 3)2 = [(x + 3)2 ] – 8 – 9 x2 + 6x + 9 = (x + 3)2 – 9 – 8 = (x + 1)2 – 17 So turning point is (–1, –17) (–1, –17)

[ ] When 2x2 – 12x + 13 is written in the form 2(x + q)2 + r, the values of q and r are? [ ] 2x2 – 12x + 13 Take out common factor 2 = 2[x2 – 6x] + 13 x2 – 6x is part of (x – 3)2 = 2[(x – 3)2 ] + 13 – 9 x2 – 6x + 9 = 2(x – 3)2 – 18 + 13 = 2(x – 3)2 – 5 So q = –3 and r = –5

If x2 – 8x + 7 is written in the form (x – p)2 + q, what is the value of q? [ ] x2 – 8x + 7 x2 – 8x is part of (x – 4)2 = [(x – 4)2 ] + 7 – 16 x2 – 8x + 16 = (x – 4)2 – 16 + 7 = (x – 4)2 – 9 So q = –9

A ‘SAD’ parabola so p < 0 The diagram shows the graph of the function f where f(x) = p(x – q)2 + r . The line x = 0 is an axis of symmetry of the curve. What can you say about p, q and r ? x y A ‘SAD’ parabola so p < 0 x = 0 is the y-axis so q = 0 Maximum value is positive so r > 0 Remember turning point is (–q, r)