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

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 + 2x – 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 – 36 + 8k = 0 k2 + 4k – 32 = 0 (k + 8)(k – 4) = 0  k = -8 and k = 4

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 – 9 3 f(x) = x2 + 2x – 9 f(x) = (x2 + 2x ) – 9 f(x) = (x + 1)2 – 10 1 = 1  1 is max x2 + 2x – 9 (x + 1)2 – 10 -10 + 1 – 1

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

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) 2 (a) f(x) = x2 + 6x + 11 f(x) = (x2 + 6x ) + 11 f(x) = (x + 3)2 + 2 + 9 – 9

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) 2 (b) f(x) = (x + 3)2 + 2   Min Turning Point (– 3, 2) y = f(x) f(x) = (x + 3)2 + 2 x = 0 : y = (0 + 3)2 + 2 y = 9 + 2 = 11  (0, 11) y (0, 11) (-3, 2) x

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) 2 (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 – 2/2 – 3/2 ] f(x) = 2[ (x + 1)2 – 5/2 ] f(x) = 2(x + 1)2 – 5 +1 – 1

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) 2 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)

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 ) + 5 f(x) = (x + 2)2 + 1 – 4 +4

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)2 + 1  Minimum Turning Point at (– 2, 1)

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)2 + 1] = 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

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) = 3 . (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 (3/(x + 1) ) + 1 = 3 3 + (x + 1) (x + 1) (x + 1) = 3 x + 4 (x + 1) Restriction is denominator ≠ 0 = 3 (x + 1) x + 4  {x: x є R, x ≠ -4}

Functions = 1 f(g(x)) = f (2x + 3) = 1 (2x + 3) – 4 2003 Paper 1 Q9. The function f(x) = 1 & 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) = 1 (2x + 3) – 4 Restriction is denominator ≠ 0 = 1 2x – 1 2x – 1 ≠ 0 2x ≠ 1 x ≠ ½  {x: x є R, x ≠ ½ }

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

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)) 2 Find g(g(x)) 2 (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

Graph Transformations 2003 Paper 2 Q2 y 5 Amplitude = 5-(-3) = 8 2 2 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

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

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

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

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

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

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)

Completing the Square, Functions & Graphs Total = 42