1. Completing the Square, Functions & Graphs
A Selection of Past Paper Questions
Due 2nd December 2009

2. Completing the Square x2 – kx + 2x – 2k = – 9
Pg 225 Q23
The roots of the equation (x + 2)(x – k) = – 9 are equal.
Find the values of k.
(x + 2)(x – k) = – 9
x2 – kx x – 2k = – 9
1x2 + (2 – k)x + (9 – 2k) = 0
If equal roots  b2 – 4ac = 0
(2 – k)2 – 4(1)(9 – 2k) = 0
4 – 4k + k2 – 4(9 – 2k) = 0
4 – 4k + k2 – k = 0
k2 + 4k – 32 = 0
(k + 8)(k – 4) = 0
 k = -8 and k = 4

3. Completing the Square + 1 – 1 f(x) = (x2 + 2x ) – 9
Pg 225 Q26
Express x2 + 2x – 9, in the form (x + a)2 + b and hence state the
maximum value of 1
x2 + 2x –
f(x) = x2 + 2x – 9
f(x) = (x2 + 2x ) – 9
f(x) = (x + 1)2 – 10
1 =  is max
x2 + 2x – (x + 1)2 –
+ 1
– 1

4. Completing the Square f(x) = x2 + 2x – 8 f(x) = (x2 + 2x ) – 8
2001 Paper 1 Q4.
Given f(x) = x2 + 2x – 8, express f(x) in the form (x + a)2 – b 2
f(x) = x2 + 2x – 8
f(x) = (x2 + 2x ) – 8
f(x) = (x + 1)2 – 9
+ 1
– 1

5. Completing the Square (a) f(x) = x2 + 6x + 11 f(x) = (x2 + 6x ) + 11
2003 Paper 1 Q2.
Write f(x) = x2 + 6x + 11 in the form f(x) = (x + a)2 + b 2
Hence or otherwise sketch y = f(x)
(a) f(x) = x2 + 6x + 11
f(x) = (x2 + 6x )
f(x) = (x + 3)2 + 2
+ 9
– 9

6. Completing the Square (b) f(x) = (x + 3)2 + 2 
2003 Paper 1 Q2.
Write f(x) = x2 + 6x + 11 in the form f(x) = (x + a)2 + b 2
Hence or otherwise sketch y = f(x)
(b) f(x) = (x + 3) 
 Min Turning Point (– 3, 2)
y = f(x)
f(x) = (x + 3)2 + 2
x = 0 : y = (0 + 3)2 + 2
y = = 11  (0, 11) y
(0, 11)
(-3, 2) x

7. Completing the Square – 1 +1 (a) f(x) = 2x2 + 4x – 3
2006 Paper 1 Q8.
Express f(x) = 2x2 + 4x – 3 in the form f(x) = a(x + b)2 + c 2
Hence or otherwise sketch y = f(x)
(a) f(x) = 2x2 + 4x – 3
f(x) = 2[x2 + 2x – 3/2 ]
f(x) = 2[(x2 + 2x ) – 3/2 ]
f(x) = 2[ (x + 1) – 2/2 – 3/2 ]
f(x) = 2[ (x + 1)2 – 5/2 ]
f(x) = 2(x + 1)2 – 5 +1
– 1

8. Completing the Square f(x) = 2x2 + 4x – 3
2006 Paper 1 Q8.
Express f(x) = 2x2 + 4x – 3 in the form f(x) = a(x + b)2 + c 2
Hence or otherwise sketch y = f(x)
f(x) = 2x2 + 4x – 3
 Min Turning Point (– 1, – 5)
f(x) = 2(x + 1)2 – 5
x = 0 : y = 2(0 + 1)2 – 5
y = 2 – 5 = – 3  (0, – 3)
= 2(x + 1)2 – 5  y x
(0, – 3)
(-1, -5)

9. Completing the Square – 4 (a) f(x) = x2 + 4x + 5 f(x) = (x2 + 4x ) + 5
2002 Paper 1 Q7.
Express f(x) = x2 + 4x + 5 in the form f(x) = (x + a)2 + b 2
Write down the coordinates of the turning point 1
Find the range of values for which 10 – f(x) is positive 1
(a) f(x) = x2 + 4x + 5
f(x) = (x2 + 4x )
f(x) = (x + 2)
– 4 +4

10. Completing the Square (b) f(x) = x2 + 4x + 5 f(x) = (x + 2)2 + 1 
2002 Paper 1 Q7.
Express f(x) = x2 + 4x + 5 in the form f(x) = (x + a)2 + b 2
Write down the coordinates of the turning point 1
Find the range of values for which 10 – f(x) is positive 1
(b) f(x) = x2 + 4x + 5
f(x) = (x + 2) 
Minimum
Turning Point at (– 2, 1)

11. Completing the Square (c) 2002 Paper 1 Q7.
Express f(x) = x2 + 4x + 5 in the form f(x) = (x + a)2 + b 2
Write down the coordinates of the turning point 1
Find the range of values for which 10 – f(x) is positive 1
(c)
Find roots of quadratic
10 – f(x) = 0 y
10 – [(x + 2) ] = 0
10 – (x + 2)2 – 1 = 0
Positive when
10 – f(x) ≥ 0
range is when -5 ≤ x ≤ 1
9 – (x + 2)2 = 0 x
(x + 2)2 = 9
– 5 1
(x + 2) = ±3
x = -2 ±3
So roots are x = -5 and x = 1

12. Functions = 3 = 3 f(f(x)) = f ( 3/(x + 1)) = 3 (3/(x + 1) ) + 1
2002 WD Paper 1 Q9.
The function f, defined on a suitable domain, is given by f(x) =
(x + 1)
Find an expression for h(x), h(x) = f(f(x)) 3
Describe any restrictions on the domain of h. 1
f(f(x)) = f ( 3/(x + 1)) =
(3/(x + 1) ) + 1
= 3
(x + 1)
(x + 1) (x + 1)
= 3
x (x + 1)
Restriction is
denominator ≠ 0
= 3 (x + 1)
x + 4
 {x: x є R, x ≠ -4}

13. Functions = 1 f(g(x)) = f (2x + 3) = 1 (2x + 3) – 4
2003 Paper 1 Q9.
The function f(x) = & g(x) = 2x + 3
(x – 4)
Find an expression for h(x), h(x) = f(g(x)) 2
Write down any restrictions on the domain of h. 1
f(g(x)) = f (2x + 3) =
(2x + 3) – 4
Restriction is
denominator ≠ 0
= 1
2x – 1
2x – 1 ≠ 0
2x ≠ 1
x ≠ ½
 {x: x є R, x ≠ ½ }

14. Functions -9 2006 Paper 1 Q3. f(g(x)) = f (2x – 3) = 2(2x – 3) + 3
Two functions f and g are defined by f(x) = 2x + 3 & g(x) = 2x – 3 , where x is a
real number.
Find an expression for (i) f(g(x)) and (ii) g(f(x)) 3
Determine the least possible value of the product of f(g(x)) x g(f(x)) 1
f(g(x)) = f (2x – 3) = 2(2x – 3) + 3
(b) f(g(x)) x g(f(x))
= 4x – 6 + 3
= (4x – 3) x (4x – 3)
= 4x – 3
= 16x2 – 9
g (f(x)) = g(2x + 3) = 2(2x + 3) – 3
As 16x2 ≥ 0 for all x
 16x2 – 9 has a minimum value of -9
= 4x + 6 – 3
= 4x + 3

15. Functions 2007 Paper 1 Q3. (b) (a) g (g(x)) = g(1 – 2x)
Functions f and g are defined on suitable domains, f(x) = x2 + 1 & g(x) = 1 – 2x
Find g(f(x))
Find g(g(x))
(b)
(a)
g (g(x)) = g(1 – 2x)
g(f(x)) = g(x2 + 1)
= 1 – 2(x2 + 1)
= 1 – 2(1 – 2x)
= 1 – 2x2 – 2
= 1 – 2 + 4x
= – 2x2 – 1
= 4x – 1

16. Graph Transformations
2003 Paper 2 Q2 y 5
Amplitude = 5-(-3) = 8 1
 Amplitude, a = 4 π x
Sine usually has a period of 2π, but here we have a period of π -3
 will have 2 repetitions of the wave , b = 2
Y = a Sin(bx) + c
Sine usually starts at zero
 y = 4 Sin(2x) + 1
pushed up 1 position
c = 1

17. Graph Transformations
Sketch y = f(-x)
On the same graph sketch y = 2f(-x)
2003 Paper 2 Q5 y 4
(-4, 2)
(4, 2) x -3 -1 1 3
Original coord
f(-x)
 (-x, y)
(-4, 2)
(4, 2)
(-3, 0)
(3, 0)
(0, -3)
(1, 0)
(-1, 0) -3
f(-x)  change sign of x

18. Graph Transformations
Sketch y = f(-x)
On the same graph sketch y = 2f(-x)
2003 Paper 2 Q5 y 4
(-4, 2)
(4, 2) x -3 -1 1 3
Original coord
f(-x)
 (-x, y)
(-4, 2)
(4, 2)
(-3, 0)
(3, 0)
(0, -3)
(1, 0)
(-1, 0) -3
f(-x)  change sign of x

19. Graph Transformations
On the same graph
sketch y = 2f(-x)
2003 Paper 2 Q5 y
(4, 4) 4
y = 2f(-x)
 change sign of x
then double height
(-4, 2)
(4, 2) x -3 -1 1 3
Orig
coord
f(-x)
 (-x, y)
2f(-x)
 (-x, 2y)
(-4, 2)
(4, 2)
(4, 4)
(-3, 0)
(3, 0)
(0, -3)
(0, -6)
(1, 0)
(-1, 0) -3 -6

20. Graph Transformations
Sketch y = -g(x)
On the same graph sketch y = 3 – g(x)
2004 Paper 1 Q4 y 3
(b, 3) 1 x
Original coord
-g(x)
 (x, -y)
(a, -2)
(a, 2)
(b, 3)
(b, -3)
(0, 1)
(0, -1)
(a, -2) -3
-g(x)  change sign of y

21. Graph Transformations
Sketch y = -g(x)
On the same graph sketch y = 3 – g(x)
2004 Paper 1 Q4 y 3
(b, 3)
(a, 2) 1 x -1
Original coord
-g(x)
 (x, -y)
(a, -2)
(a, 2)
(b, 3)
(b, -3)
(0, 1)
(0, -1)
(a, -2) -3
(b, -3)
-g(x)  change sign of y

22. Graph Transformations
2004 Paper 1 Q4
Sketch y = -g(x)
On the same graph sketch y = 3 – g(x) y 3
(b, 3)
(a, 2) 2
Rearrange 3 – g(x) to – g(x) + 3
change sign of y
push up 3 1 x
(b, 0) -1
Orig coord
-g(x)
 (x, -y)
3 – g(x) -g(x)+3
(a, -2)
(a, 2)
(a, 5)
(b, 3)
(b, -3)
(b, 0)
(0, 1)
(0, -1)
(0, 2)
(a, -2) -3
(b, -3)

23. Completing the Square, Functions & Graphs
Total = 42