(a)

\begin{align*} x^2 + 16x & = \left(x + {16 \over 2} \right)^2 - \left(16 \over 2\right)^2 \\ & = (x + 8)^2 - 64 \end{align*}

(b)

\begin{align*} -x^2 + 5x + 10 & = -(x^2 - 5x) + 10 \\ & = - \left[ \left(x - {5 \over 2}\right)^2 - \left(5 \over 2\right)^2 \right] + 10 \\ & = - \left[ \left(x - {5 \over 2}\right)^2 - {25 \over 4} \right] + 10 \\ & = - \left(x - {5 \over 2}\right)^2 + {25 \over 4} + 10 \\ & = - \left(x - {5 \over 2}\right)^2 + {65 \over 4} \end{align*}

(c)

\begin{align*} 8x^2 - x + {1 \over 4} & = 8 \left(x^2 - {1 \over 8}x \right) + {1 \over 4} \\ & = 8 \left[ \left( x - {1 \over 16} \right)^2 - \left(1 \over 16\right)^2 \right] + {1 \over 4} \\ & = 8 \left[ \left(x - {1 \over 16} \right)^2 - {1 \over 256} \right] + {1 \over 4} \\ & = 8 \left( x - {1 \over 16} \right)^2 - {1 \over 32} + {1 \over 4} \\ & = 8 \left( x - {1 \over 16} \right)^2 + {7 \over 32} \end{align*}

(d)

\begin{align*} 7 - 9x - 3x^2 & = -3x^2 - 9x + 7 \\ & = -3 (x^2 + 3x) + 7 \\ & = -3 \left[ \left( x + {3 \over 2} \right)^2 - \left(3 \over 2\right)^2 \right] + 7 \\ & = -3 \left[ \left( x + {3 \over 2} \right)^2 - {9 \over 4} \right] + 7 \\ & = -3 \left(x + {3 \over 2}\right)^2 + {27 \over 4} + 7 \\ & = -3 \left(x + {3 \over 2}\right)^2 + {55 \over 4} \end{align*}

(a)

\begin{align*} y & = -2(x + 7)^2 + 8 \phantom{000} [\text{Maximum curve } (\cap)] \\ \\ \text{Coordinates} & \text{ of maximum point is } (-7, 8) \\ \\ \therefore \text{Maximum value} & = 8, \text{ when } x = -7 \end{align*}

(b)

\begin{align*} y & = {1 \over 4}(x - 1)^2 + {\pi \over 6} \phantom{00000} [\text{Minimum curve } (\cup)] \\ \\ \text{Coordinates} & \text{ of minimum point is } \left(1 , {\pi \over 6}\right) \\ \\ \therefore \text{Minimum value} & = {\pi \over 6}, \text{ when } x = 1 \end{align*}

(c)

\begin{align*} y & = 5x - 9 - 1.5x^2 \\ y & = -1.5x^2 + 5x - 9 \\ y & = -{3 \over 2} \left(x^2 - {10 \over 3}x \right) - 9 \\ y & = -{3 \over 2} \left[ \left( x - {5 \over 3} \right)^2 - \left(5 \over 3\right)^2 \right] - 9 \\ y & = -{3 \over 2} \left[ \left( x - {5 \over 3} \right)^2 - {25 \over 9} \right] - 9 \\ y & = -{3 \over 2} \left(x - {5 \over 3}\right)^2 + {25 \over 6} - 9 \\ y & = -{3 \over 2} \left(x - {5 \over 3}\right)^2 - {29 \over 6} \phantom{00000} [\text{Maximum curve } (\cap)] \\ \\ \text{Coordinates} & \text{ of maximum point is } \left( {5 \over 3}, -{29 \over 6}\right) \\ \\ \therefore \text{Maximum value} & = -{29 \over 6}, \text{ when } x = {5 \over 3} \end{align*}

(d)

\begin{align*} y & = (3x - 2)(x - 7) + 18 \\ y & = 3x^2 - 21x - 2x + 14 + 18 \\ y & = 3x^2 - 23x + 32 \\ y & = 3 \left( x^2 - {23 \over 3}x \right) + 32 \\ y & = 3 \left[ \left( x - {23 \over 6} \right)^2 - \left(23 \over 6\right)^2 \right] + 32 \\ y & = 3 \left[ \left( x - {23 \over 6} \right)^2 - {529 \over 36} \right] + 32 \\ y & = 3 \left(x - {23 \over 6} \right)^2 - {529 \over 12} + 32 \\ y & = 3 \left(x - {23 \over 6} \right)^2 - {145 \over 12} \phantom{00000} [\text{Minimum curve } (\cup)] \\ \\ \text{Coordinates} & \text{ of minimum point is } \left( {23 \over 6}, -{145 \over 12}\right) \\ \\ \therefore \text{Minimum value} & = -{145 \over 12}, \text{ when } x = {23 \over 6} \end{align*}

\begin{align*} 4x^2 + 9x - 2 & = 4 \left(x^2 + {9 \over 4}x\right) - 2 \\ & = 4 \left[ \left(x + {9 \over 8}\right)^2 - \left(9 \over 8\right)^2 \right] - 2 \\ & = 4 \left[ \left(x + {9 \over 8}\right)^2 - {81 \over 64} \right] - 2 \\ & = 4 \left(x + {9 \over 8}\right)^2 - {81 \over 16} - 2 \\ & = 4 \left(x + {9 \over 8}\right)^2 - {113 \over 16} \\ \\ \therefore \text{Minimum value} & = -{113 \over 16}, \text{ when } x = -{9 \over 8} \end{align*}

\begin{align*} \text{Condition 1: } a & < 0, \text{ since the expression has a minimum value} \\ \\ ax^2 + 10x + c & = a \left(x^2 + {10 \over a}x \right) + c \\ & = a \left[ \left(x + {5 \over a}\right)^2 - \left(5 \over a\right)^2 \right] + c \\ & = a \left[ \left(x + {5 \over a}\right)^2 - {25 \over a^2} \right] + c \\ & = a \left(x + {5 \over a}\right)^2 - {25 \over a} + c \\ \\ \text{Maximum value} & = c - {25 \over a} = 7 \\ \\ \text{Condition 2} : c & - {25 \over a} = 7 \\ \\ \\ \text{If } & a = -1, \\ c - {25 \over (-1)} & = 7 \\ c + 25 & = 7 \\ c & = 7 - 25 \\ c & = -18 \\ \\ \therefore a & = -1, c = -18 \end{align*}

(i)

\begin{align*} y & = a(x - h)^2 + k \\ \\ \text{Maximum height} & = 16 \text{ m} \text{ when } x = 2 \\ \implies \text{Coordinates } & \text{of maximum point is } (2, 16) \\ \\ y & = a(x - 2)^2 + 16 \\ \\ \text{Since ball lands after} & \text{ 4 seconds, when } x = 4 \text{ and } y = 0, \\ 0 & = a(4 - 2)^2 + 16 \\ 0 & = a(4) + 16 \\ 0 & = 4a + 16 \\ -4a & = 16 \\ a & = {16 \over -4} \\ a & = -4 \\ \\ y & = -4(x - 2)^2 + 16 \\ \\ \therefore a & = -4, h = 2, k = 16 \end{align*}

(ii)

(a)

As x increases from 30 to 70, y decreases from 7.6 to 6.0. As x increases from 76 to 90, y increases from 6.0 to 6.4.

This implies that the a minimum curve (∪) is suitable for this set of data.

(b)(i)

\begin{align*} 0.001x^2 - 0.14x + 10.9 & = 0.001(x^2 - 140x) + 10.9 \\ & = 0.001 \left[ \left( x - {140 \over 2} \right)^2 - \left(140 \over 2\right)^2 \right] + 10.9 \\ & = 0.001 [ (x - 70)^2 - 4900 ] + 10.9 \\ & = 0.001(x - 70)^2 - 4.9 + 10.9 \\ & = 0.001(x - 70)^2 + 6 \\ \\ \text{Least amount of petrol} & = 6 \text{ litres, when speed} = 70 \text{ km/h} \\ \\ \therefore \text{Disagree } & \text{with Vani} \end{align*}

(b)(ii)

\begin{align*} \text{Optimal speed} & = 70 \text{ km/h} \end{align*}

(i)

$$ c = 0 $$
c is the braking distance when the speed of the car is 0 m/s. Since the car is not moving, the braking distance is equals to 0.

(ii)

\begin{align*} d & = {v^2 \over k} + c \\ d & = {v^2 \over k} \\ \\ \text{When } & v = 12.5 \text{ and } d = 7.8, \\ 7.8 & = {(12.5)^2 \over k} \\ 7.8k & = 156.25 \\ k & = {156.25 \over 7.8} \\ k & = {3125 \over 156} \\ \\ d & = {v^2 \over {3125 \over 156}} \\ d & = {156 v^2 \over 3125} \\ \\ 58 \text{ km/h} & = {58 \text{ km} \over 1 \text{ hour}} \\ & = {58000 \text{ m} \over 3600 \text{ s}} \\ & = 16{1 \over 9} \text{ s} \\ \\ \text{When } & v = 16{1 \over 9}, \\ d & = {156 \left(16 {1 \over 9}\right)^2 \over 3125} \\ d & = 12.957 < 15 \\ \\ \therefore \text{Cheryl} & \text{ is able to brake in time to avoid the debris} \end{align*}