1. Completing
the
Square

2. 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
x a2
2. Multiply the two terms then double your answer
x2 + 2ax + a2

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

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

5. → (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
x2 + 6x is part of (x + 3)2
Still equal to x2 + 6x
= (x + 3) ] [
– 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)

6. → (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) – 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)

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

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

9. Completing the Square
and Fractions

10. 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

11. [ ] ] [ Express x2 + 6x + 13 in the form (x + p)2 + r.
Hence state the maximum value of
1 of 2 [ ]
x2 + 6x
x2 + 6x is part of (x + 3)2
= (x + 3) ] [
– 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)

12. 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

13. [ ] 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]
x2 + 2x is part of (x + 1)2
= –4[(x + 1) ] + 1
– 1
x2 + 2x + 1
(–1, 5)
= –4(x + 1)
= –4(x + 1)2 + 5
 MAX value is 5
So MIN value of f(x) is 10/5 = 2

14. Past Paper
Examples

15. 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) ] – 10
– 1
x2 + 2x + 1
= 3(x + 1)2 – 3 – 10
= 3(x + 1)2 – 13
So q is – 13

16. 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) ] – 8
– 9
x2 + 6x + 9
= (x + 3)2 – 9 – 8
= (x + 1)2 – 17
So turning point is (–1, –17)
(–1, –17)

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

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

19. 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)