Unit 8 Review Exercises
Quadratic Functions
These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed.
1 Paper 1 A curve has equation \(y = -2x^2 + 8x + 3\).
(a) Show that the highest point on the curve is \((2, 11)\).
(b) Find the exact \(x\)-coordinates of the points where the curve meets the \(x\)-axis.
(a) \(-2(x^2 - 4x) + 3 = -2\big[(x - 2)^2 - 4\big] + 3 = -2(x - 2)^2 + 11\). Since \(-2(x - 2)^2 \leq 0\), \(y \leq 11\), with equality at \(x = 2\). So the highest point is \((2, 11)\).
(b) \(2(x - 2)^2 = 11 \Rightarrow x = 2 \pm \sqrt{\dfrac{11}{2}} = 2 \pm \dfrac{\sqrt{22}}{2}\). See 8.1 and 8.2.
2 Paper 1 The equation of a curve is \(y = x^2 + 4x + 10 - p\), where \(p\) is a constant.
(a) Find the range of values of \(p\) for which the curve lies completely above the \(x\)-axis.
(b) In the case \(p = 1\), find the coordinates of the turning point and the \(y\)-intercept.
3 Paper 1 The expression \(ax^2 + 6x + c\) is negative for all real values of \(x\).
(a) Find the conditions that \(a\) and \(c\) must satisfy.
(b) Give one pair of values of \(a\) and \(c\) that satisfy these conditions.
(a) We need \(a < 0\) and \(\Delta = 36 - 4ac < 0\), that is, \(a < 0\) and \(ac > 9\).
(b) For example \(a = -1\), \(c = -10\) (\(ac = 10 > 9\)). Then \(-x^2 + 6x - 10 = -(x - 3)^2 - 1 < 0\) ✓. See 8.3.
4 Paper 1 On part of a roller coaster, the height of a car above the ground is \(h = t^2 - 10t + 30\) metres, \(t\) seconds after it enters this part, for \(0 \leq t \leq 10\).
(a) Explain why the car never comes lower than \(5\) m above the ground on this part.
(b) Find the time when the car is at its lowest point.
5 Paper 1 Solve the inequality \(4x^2 - 4x - 3 \geq 0\).
\((2x - 3)(2x + 1) \geq 0\). Roots \(-\dfrac{1}{2}\) and \(\dfrac{3}{2}\); positive coefficient of \(x^2\), so outside the roots:
\(x \leq -\dfrac{1}{2}\) or \(x \geq \dfrac{3}{2}\). See 8.5.
6 Paper 1 The graph of \(y = f(x)\) is a parabola with \(x\)-intercepts \(-3\) and \(1\), and its maximum value is \(8\). Write \(f(x)\) in factorised form, in general form and in vertex form.
\(f(x) = a(x + 3)(x - 1)\). The axis is \(x = \dfrac{-3 + 1}{2} = -1\), and \(f(-1) = a(2)(-2) = -4a = 8\), so \(a = -2\).
Factorised: \(-2(x + 3)(x - 1)\). General: \(-2(x^2 + 2x - 3) = -2x^2 - 4x + 6\). Vertex: \(-2(x + 1)^2 + 8\). See 8.2.
7 Paper 1 Show that the equation \(kx^2 + (k + 3)x + 1 = 0\), \(k \neq 0\), has two distinct real roots for every value of \(k\).
\(\Delta = (k + 3)^2 - 4k = k^2 + 2k + 9 = (k + 1)^2 + 8\).
Since \((k + 1)^2 \geq 0\), \(\Delta \geq 8 > 0\). So there are always two distinct real roots. See 8.3.
8 Paper 1 A farmer has \(60\) m of fencing. She makes a rectangular pen against a long wall, so fencing is needed on three sides only. The two sides at right angles to the wall are each \(x\) m long.
(a) Show that the area of the pen is \(A = 60x - 2x^2\) square metres.
(b) By completing the square, find the maximum area and the value of \(x\) that gives it.
(c) Find the range of values of \(x\) for which the area is at least \(400\) m².
(a) The side parallel to the wall is \(60 - 2x\) m, so \(A = x(60 - 2x) = 60x - 2x^2\).
(b)
\[ \begin{aligned} A &= -2(x^2 - 30x) \\ &= -2\big[(x - 15)^2 - 225\big] \\ &= 450 - 2(x - 15)^2 . \end{aligned} \]
The maximum area is \(450\) m², when \(x = 15\).
(c) \(60x - 2x^2 \geq 400 \;\Rightarrow\; 2x^2 - 60x + 400 \leq 0 \;\Rightarrow\; x^2 - 30x + 200 \leq 0\).
\((x - 10)(x - 20) \leq 0\), so \(10 \leq x \leq 20\). (These values also satisfy \(0 < x < 30\), so all of them are possible.)
9 Paper 1 A set of experimental data is shown.
| \(x\) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| \(y\) | \(-2\) | \(1.5\) | \(6\) | \(11.5\) | \(18\) | \(25.5\) |
(a) Find a quadratic function \(y = ax^2 + bx + c\) that fits the data exactly. Explain your method.
(b) Using your model, find the range of values of \(x\) for which \(y < 0\).
(a) At \(x = 0\), \(y = -2\), so \(c = -2\). At \(x = 1\): \(a + b - 2 = 1.5\), so \(a + b = 3.5\). At \(x = 2\): \(4a + 2b - 2 = 6\), so \(2a + b = 4\). Subtract: \(a = 0.5\), \(b = 3\). So \(y = 0.5x^2 + 3x - 2\). (Check \(x = 5\): \(12.5 + 15 - 2 = 25.5\) ✓. Also, the second differences are all \(1 = 2a\).)
(b) \(0.5x^2 + 3x - 2 < 0 \Rightarrow x^2 + 6x - 4 < 0\). Roots: \(x = \dfrac{-6 \pm \sqrt{36 + 16}}{2} = -3 \pm \sqrt{13}\). So \(-3 - \sqrt{13} < x < -3 + \sqrt{13}\). See 8.2 and 8.5.
10 Paper 2 A ball is thrown upwards. Its height after \(t\) seconds is \(h = -4.9t^2 + 15t + 2\) metres.
(a) Find the greatest height of the ball.
(b) Find the time when the ball hits the ground.
(c) Find the length of time for which the ball is more than \(10\) m above the ground.
(a) Vertex at \(t = \dfrac{15}{9.8} = 1.53\) s; \(\;h = 13.5\) m (13.48 m).
(b) Solve \(-4.9t^2 + 15t + 2 = 0\) with the GDC: \(t = 3.19\) s (the other root is negative).
(c) \(-4.9t^2 + 15t + 2 > 10 \iff 4.9t^2 - 15t + 8 < 0\). The roots are \(t = 0.688\) and \(t = 2.37\). The ball is above \(10\) m for \(2.37 - 0.688 = 1.69\) s. See 8.4 and 8.5.