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This selection of past paper questions on completing the square, functions, and graphs will help you practice and improve your understanding. Topics include finding roots, expressing quadratics in different forms, sketching graphs, and applying transformations.
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Completing the Square, Functions & Graphs A Selection of Past Paper Questions Due 2nd December 2009
Completing the Square 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 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 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 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 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 0
Completing the Square 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 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 (0, – 3) (-1, -5)
Completing the Square 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 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 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 0 (x + 2) = ±3 x = -2 ±3 So roots are x = -5 and x = 1
Functions 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 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 2006 Paper 1 Q3. 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 As 16x2 ≥ 0 for all x 16x2 – 9 has a minimum value of -9 g (f(x)) = g(2x + 3) = 2(2x + 3) – 3 = 4x + 6 – 3 = 4x + 3
Functions 2007 Paper 1 Q3. 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 y 2003 Paper 2 Q2 5 Amplitude = 5-(-3) = 8 2 2 1 Amplitude, a = 4 π x 0 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) y 2003 Paper 2 Q5 4 (-4, 2) (4, 2) x -3 -1 0 1 3 -3 f(-x) change sign of x
Graph Transformations • Sketch y = f(-x) • On the same graph sketch y = 2f(-x) y 2003 Paper 2 Q5 4 (-4, 2) (4, 2) x -3 -1 0 1 3 -3 f(-x) change sign of x
Graph Transformations On the same graph sketch y = 2f(-x) y (4, 4) 2003 Paper 2 Q5 4 y = 2f(-x) change sign of x then double height (-4, 2) (4, 2) x -3 -1 0 1 3 -3 -6
Graph Transformations • Sketch y = -g(x) • On the same graph sketch y = 3 – g(x) y 2004 Paper 1 Q4 3 (b, 3) 1 x 0 (a, -2) -3 -g(x) change sign of y
Graph Transformations • Sketch y = -g(x) • On the same graph sketch y = 3 – g(x) y 2004 Paper 1 Q4 3 (b, 3) (a, 2) 1 x 0 -1 (a, -2) -3 (b, -3) -g(x) change sign of y
Graph Transformations (a, 5) • Sketch y = -g(x) • On the same graph sketch y = 3 – g(x) y 2004 Paper 1 Q4 3 (b, 3) (a, 2) 2 • Rearrange 3 – g(x) to – g(x) + 3 • change sign of y • push up 3 1 x 0 (b, 0) -1 (a, -2) -3 (b, -3)
Completing the Square, Functions & Graphs Total = 42
