(i)

\begin{align*} \text{When } & x = 0, \\ y & = {1 \over 1000} (0 - 300)^2 + 10 \\ & = 100 \\ \\ \text{Height of each tower} & = 100 \text{ m} \end{align*}

(ii)

\begin{align*} \text{When } & y = 100, \\ 100 & = {1 \over 1000} (x - 300)^2 + 10 \\ 100 - 10 & = {1 \over 1000} (x - 300)^2 \\ 90 & = {1 \over 1000} (x - 300)^2 \\ 1000(90) & = (x - 300)^2 \\ 90 000 & = (x - 300)^2 \\ \pm \sqrt{90000} & = x - 300 \\ \pm 300 & = x - 300 \\ \\ x & = 300 + 300 \text{ or } -300 + 300 \\ & = 600 \text{ or } 0 \\ \\ \therefore \text{Distance between two towers} & = 600 \text{ m} \end{align*}

(iii)

\begin{align*} \text{When } & y = 20, \\ 20 & = {1 \over 1000} (x - 300)^2 + 10 \\ 20 - 10 & = {1 \over 1000} (x - 300)^2 \\ 10 & = {1 \over 1000} (x - 300)^2 \\ 1000(10) & = (x - 300)^2 \\ 10 000 & = (x - 300)^2 \\ \pm \sqrt{10000} & = x - 300 \\ \pm 100 & = x - 300 \\ \\ x & = 100 + 300 \text{ or } -100 + 300 \\ & = 400 \text{ or } 200 \\ \\ \therefore \text{Distance of car from tower } P & = 200 \text{ m} \text{ or } 400 \text{ m} \end{align*}

(i)

\begin{align*} y & = {1 \over 3}x^2 - 2x + 8 \\ y & = {1 \over 3}(x^2 - 6x) + 8 \\ y & = {1 \over 3} \left[ \left( x - {6 \over 2} \right)^2 - \left(6 \over 2\right)^2 \right] + 8 \\ y & = {1 \over 3} [ (x - 3)^2 - 9 ] + 8 \\ y & = {1 \over 3} (x - 3)^2 - 3 + 8 \\ y & = {1 \over 3} (x - 3)^2 + 5 \end{align*}

(ii)

\begin{align*} y & = {1 \over 3} (x - 3)^2 + 5 \\ \\ \text{Coordinates of mini} & \text{mum point is } (3, 5) \\ \\ \therefore \text{Rider's minimum height} & = 5 \text{ m} \end{align*}

(iii)

\begin{align*} y & = {1 \over 3} (x - 3)^2 + 5 \\ \\ \text{When } & y = 8, \\ 8 & = {1 \over 3} (x - 3)^2 + 5 \\ 8 - 5 & = {1 \over 3} (x - 3)^2 \\ 3 & = {1 \over 3} (x - 3)^2 \\ 3(3) & = (x - 3)^2 \\ 9 & = (x - 3)^2 \\ \pm \sqrt{9} & = x - 3 \\ \pm 3 & = x - 3 \\ \\ x & = 3 + 3 \text{ or } - 3 + 3 \\ x & = 6 \text{ or } 0 \\ \\ \therefore \text{Rider's horizontal distance from start} & = 6 \text{ m} \end{align*}

(i)

\begin{align*} y & = -{1 \over 200} x^2 + 5x \\ \\ \text{When } & y = 0, \\ 0 & = -{1 \over 200}x^2 + 5x \\ 0 & = x \left(- {1 \over 200}x + 5 \right) \\ \\ x & = 0 \phantom{0} \text{ or } -{1 \over 200}x + 5 = 0 \\ & \phantom{0000000000(} -{1 \over 200}x = - 5 \\ & \phantom{00000000000-.} {1 \over 200}x = 5 \\ & \phantom{0000000000000000(} x = 200(5) \\ & \phantom{0000000000000000(} x = 1000 \\ \\ \text{Distance of enemy} & \text{ frigate from battleship} = 1000 \text{ m} \end{align*}

(ii)

\begin{align*} \text{Line of symmetry, } x & = {0 + 1000 \over 2} \\ x & =500 \\ \\ \text{When } & x = 500, \\ y & = -{1 \over 200}(500)^2 + 5(500) \\ y & = 1250 \\ \implies \text{Coordinate} & \text{ of maximum point is } (500, 1250) \\ \\ \therefore \text{Maximum height of first shell} & = 1250 \text{ m} \end{align*}

(iii)

\begin{align*} y & = -{1 \over 5000} (x - 1500)^2 + 450 \\ \\ \text{When } & y = 0, \\ 0 & = -{1 \over 5000} (x - 1500)^2 + 450 \\ {1 \over 5000}(x - 1500)^2 & = 450 \\ (x - 1500)^2 & = 5000(450) \\ (x - 1500)^2 & = 2,250,000 \\ x - 1500 & = \pm \sqrt{2,250,000} \\ x - 1500 & = \pm 1500 \\ x & = 1500 + 1500 \text{ or } -1500 + 1500 \\ x & = 3000 \text{ or } 0 \\ \\ \text{Horizontal distance travelled by second shell} & = 3000 \text{ m} \\ & = 3 \text{ km} \\ \\ \therefore \text{Second shell will} & \text{ hit the second enemy frigate} \end{align*}

(i)

\begin{align*} y & = a(x - h)^2 + k \\ \\ \text{Since mini} & \text{mum point is } (60, 0), \\ \\ y & = a(x - 60)^2 + 0 \\ y & = a(x - 60)^2 \\ \\ \text{Using } & \text{the point } (0, 20), \\ 20 & = a(0 - 60)^2 \\ 20 & = a(3600) \\ {20 \over 3600} & = a \\ {1 \over 180} & = a \\ \\ \therefore y & = {1 \over 180}(x - 60)^2 \end{align*}

(ii)

\begin{align*} y & = {1 \over 180}(x - 60)^2 \\ \\ \text{When } & x = 20, \\ y & = {1 \over 180}(20 - 60)^2 \\ y & = 8{8 \over 9} \\ \\ \therefore \text{Height of cable above the roadway} & = 8{8 \over 9} \text{ m} \end{align*}

(i)

\begin{align*} y & = a(x - h)^2 + k \\ \\ \text{Since mini} & \text{mum point is } (100, 10), \\ \\ y & = a(x - 100)^2 + 10 \\ \\ \text{Using } & \text{the point } (0, 40), \\ 40 & = a(0 - 100)^2 + 10 \\ 40 & = a(10000) + 10 \\ 40 - 10 & = 10000a \\ 30 & = 10000a \\ {30 \over 10000} & = a \\ {3 \over 1000} & = a \\ \\ \therefore y & = {3 \over 1000}(x - 100)^2 + 10 \end{align*}

(ii)

\begin{align*} y & = {3 \over 1000}(x - 100)^2 + 10 \\ \\ \text{When } & x = 30, \\ y & = {3 \over 1000} (30 - 100)^2 + 10 \\ y & = 24.7 \\ \\ \therefore \text{Length of vertical supporting wire} & = 24.7 \text{ m} \end{align*}

(i)

\begin{align*} y & = 0.25x^2 - x + 2 \\ \\ \text{When } & x =0, \\ y & = 0.25(0)^2 - (0) + 2 \\ y & = 2 \\ \implies & y \text{-intercept is } (0, 2) \\ \\ y & = 0.25(x^2 - 4x) + 2 \\ y & = 0.25 \left[ \left( x - {4 \over 2} \right)^2 - \left(4 \over 2\right)^2 \right] + 2 \\ y & = 0.25 [ (x - 2)^2 - 4 ] + 2 \\ y & = 0.25(x - 2)^2 - 1 + 2 \\ y & = 0.25(x - 2)^2 + 1 \\ \\ \therefore \text{Coordinates} & \text{ of turning point is } (2, 1) \end{align*}

(ii) From (i), the coordinates of the minimum point is (2, 1). Recall that y denotes the height of the ramp above the ground (in meters).

\begin{align*} \text{Maximum depth} & = 1 \text{ m} \end{align*}

(iii)

\begin{align*} y & = 0.25(x - 2)^2 + 1 \\ \\ \text{When } & y = 2, \\ 2 & = 0.25(x- 2)^2 + 1 \\ 2- 1 & = 0.25(x - 2)^2 \\ 1 & = 0.25(x - 2)^2 \\ {1 \over 0.25} & = (x - 2)^2 \\ 4 & = (x - 2)^2 \\ \pm \sqrt{4} & = x - 2 \\ \pm 2 & = x - 2\\ \\ x & = 2 + 2 \text{ or } -2 + 2 \\ x & = 4 \text{ or } 0 \\ \\ \therefore \text{Maximum possible width} & = 4 \text{ m} \end{align*}

(i)

\begin{align*} y & = -{1 \over 16}x^2 + {1 \over 8}x + {7 \over 4} \\ y & = -{1 \over 16} (x^2 - 2x) + {7 \over 4} \\ y & = -{1 \over 16} \left[ \left(x - {2 \over 2} \right)^2 - \left(2 \over 2\right)^2 \right] + {7 \over 4} \\ y & = -{1 \over 16} [ (x - 1)^2 - 1 ] + {7 \over 4} \\ y & = -{1 \over 16} (x - 1)^2 + {1 \over 16} + {7 \over 4} \\ y & = -{1 \over 16} (x - 1)^2 + {29 \over 16} \end{align*}

(ii)

\begin{align*} y & = -{1 \over 16} (x - 1)^2 + {29 \over 16} \\ \\ \text{Coordinates} & \text{ of turning point is } \left(1, {29 \over 16} \right) \\ \\ \text{Let } & x =0, \\ y & = -{1 \over 16} (0 - 1)^2 + {29 \over 16} \\ y & = {7 \over 4} \\ \implies & y\text{-intercept is } \left(0, {7 \over 4} \right) \\ \\ \text{Let } & y = 0, \\ 0 & = -{1 \over 16} (x - 1)^2 + {29 \over 16} \\ {1 \over 16}(x - 1)^2 & = {29 \over 16} \\ (x - 1)^2 & = 29 \\ x - 1 & = \pm \sqrt{29} \\ x & = \sqrt{29} + 1 \text{ or } -\sqrt{29} + 1 \\ x & = 6.3851 \text{ or } -4.38516 \\ x & \approx 6.39 \text{ or } -4.39 \\ \implies & x \text{-intercepts are } (6.39, 0) \text{ and } (-4.39, 0) \end{align*}

(iii) From (ii), the coordinates of the maximum point is $\left(1, {29 \over 16}\right)$. Recall that y denotes the height of the shot above the ground

\begin{align*} \text{Maximum height} & = {29 \over 16} \\ & = 1{13 \over 16} \text{ m} \end{align*}

(iv) From (ii), the coordinates of the second x-intercept is (6.39, 0). Recall that x denotes the horizontal distance travelled by the shot.

\begin{align*} \text{Horizontal distance} & = 6.39 \text{ m} \end{align*}

(i)

\begin{align*} h & = 8t - 4t^2 \\ h & = -4t^2 + 8t \\ h & = -4t(t - 2) \\ h & = -4(t - 0)(t - 2) \end{align*}

(ii)

\begin{align*} h & = -4(t - 0)(t - 2) \\ \\ \text{Let } & t = 0, \\ h & = -4(0 - 0)(0 - 2) \\ h & = 0 \\ \implies & h \text{-intercept is } (0, 0) \\ \\ \text{Let } & h = 0, \\ 0 & = -4(t - 0)(t - 2) \\ 0 & = -4(t)(t - 2) \\ 0 & = t(t - 2) \\ \\ t & = 0 \text{ or } 2 \\ \implies & t \text{-intercepts are } (0, 0) \text{ and } (2, 0) \\ \\ \text{Line of symmetry, } t & = {0 + 2 \over 2} \\ t & = 1 \\ \\ \text{When } & t = 1, \\ h & = -4(1 - 0)(1 - 2) \\ h & = 4 \\ \implies & \text{Coordinates of turning point is } (1, 4) \end{align*}

(iii) From (ii), the coordinates of the maximum point is (1, 4). Recall that h denotes the height of the ball above the ground.

\begin{align*} \text{Maximum height} & = 4 \text{ m} \end{align*}

(iv) From (ii), the coordinates of the second x-intercept is (2, 0). Recall that t denotes time

\begin{align*} \text{Time taken} & = 2 \text{ seconds} \end{align*}

(v)

\begin{align*} h & = 8t - 4t^2 \\ \\ \text{When } & h = 1.8, \\ 1.8 & = 8t - 4t^2 \\ 0 & = -4t^2 + 8t - 1.8 \\ 0 & = -20t^2 + 40t - 9 \\ 0 & = 20t^2 - 40t + 9 \\ \\ t & = {-b \pm \sqrt{ b^2 - 4ac } \over 2a} \\ t & = {-(-40) \pm \sqrt{ (-40)^2 - 4(20)(9)} \over 2(20) } \\ t & = {40 \pm \sqrt{880} \over 40} \\ t & = 1.7416 \text{ or } 0.25838 \\ t & \approx 1.74 \text{ or } 0.258 \\ \\ \therefore \text{Earliest time} & = 1.74 \text{ s} \end{align*}
The path of the ball resembles a maximum curve (∩). Thus, the ball is rising when t = 0.258s and falling when t = 1.74 s.

(i)

\begin{align*} \text{Length of rectangle} & = {40 - x - x \over 2} \\ & = {40 - 2x \over 2} \\ & = {40 \over 2} - {2x \over 2} \\ & = (20 - x) \text{ cm} \\ \\ A & = \text{Length} \times \text{Width} \\ & = (20 - x) \times x \\ & = x(20 - x) \end{align*}

(ii)

\begin{align*} A & = x(20 - x) \\ \\ \text{When } & A = 0, \\ 0 & = x(20 - x) \\ \\ x & = 0 \phantom{0} \text{ or } \phantom{0} 20 - x = 0 \\ & \phantom{00000000000.} 20 = x \\ \implies & x \text{-intercepts are } (0, 0) \text{ and } (20, 0) \end{align*}

(iii)

\begin{align*} \text{Line of symmetry, } x & = {0 + 20 \over 2} \\ x & = 10 \\ \\ \text{When } & x = 10, \\ A & = (10)(20 - 10) \\ A & = 100 \\ \implies \text{Coordinates} & \text{ of maximum point is } (10, 100) \\ \\ \therefore \text{Maximum area} & = 100 \text{ cm}^2 \end{align*}

(iv)

\begin{align*} \text{Maximum area occurs when } & x = 10 \\ \\ \text{Width of rectangle} & = 10 \text{ cm} \\ \\ \text{Length of rectangle} & = 20 - 10 \\ & = 10 \text{ cm} \\ \\ \therefore \text{Since all sides have the same length},& \text{ shape formed is a square} \end{align*}

(i)

\begin{align*} y & = -5x^2 + 20x + c \\ \\ \text{When } & y = 50, \\ 50 & = -5x^2 + 20x + c \\ 0 & = -5x^2 + 20x + (c - 50) \\ \\ b^2 - 4ac & = (20)^2 - 4(-5)(c - 50) \\ & = 400 + 20(c - 50) \\ & = 400 + 20c - 1000 \\ & = 20c - 600 \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots, since baseball did not reach the height of 50m}] \\ 20c - 600 & < 0 \\ 20c & < 600 \\ c & < {600 \over 20} \\ c & < 30 \end{align*}

(ii)

\begin{align*} y & = -5x^2 + 20x + 20 \\ y & = -5(x^2 - 4x) + 20 \\ y & = -5 \left[ \left( x - {4 \over 2} \right)^2 - \left(4 \over 2\right)^2 \right] + 20 \\ y & = -5 [ (x - 2)^2 - 4] + 20 \\ y & = -5(x - 2)^2 + 20 + 20 \\ y & = -5(x - 2)^2 + 40 \\ \\ \therefore \text{Coordinates} & \text{ of maximum point is } (2, 40) \end{align*}

(iii)

From (i), when c < 30, the baseball does not reach a height of 50 m.

In (ii), when c = 20, the maximum height reached by the baseball is 40m. This validates the conclusion reached in (i).

(i)

\begin{align*} A: y & = -0.3x^2 + 3 \\ \\ \text{Let } & y = 0, \\ 0 & = -0.3x^2 + 3 \\ 0.3x^2 & = 3 \\ x^2 & = {3 \over 0.3} \\ x^2 & = 10 \\ x & = \pm \sqrt{10} \\ \\ \\ B: y & = -0.2x^2 + 1.8 \\ \\ \text{Let } & y = 0, \\ 0 & = -0.2x^2 + 1.8 \\ 0.2x^2 & = 1.8 \\ x^2 & = {1.8 \over 0.2} \\ x^2 & = 9 \\ x & = \pm \sqrt{9} \\ x & = \pm 3 \\ \\ \\ C: y & = -0.08x^2 + 2.4 \\ \\ \text{Let } & y = 0, \\ 0 & = -0.08x^2 + 2.4 \\ 0.08x^2 & = 2.4 \\ x^2 & = {2.4 \over 0.08} \\ x^2 & = 30 \\ x & = \pm \sqrt{30} \\ \\ \\ \therefore & \text{Water jet } C \text{ will send water the farthest} \end{align*}

(ii)

\begin{align*} A: y & = -0.3(x - 0)^2 + 3 \implies \text{Maximum point } (0, 3) \\ B: y & = -0.2(x - 0)^2 + 1.8 \implies \text{Maximum point } (0, 1.8) \\ C: y & = -0.08(x - 0)^2 + 2.4 \implies \text{Maximum point } (0, 2.4) \\ \\ \therefore & \text{Water jet } A \text{ will send water the highest} \end{align*}

(iii)

Water jet $B$, since it sends water to the lowest height (1.8 m) and to the shortest horizontal distance (3 m).

(iv)

\begin{align*} y & = -0.5x^2 + 2x + k \\ \\ \text{When } & y = 10, \\ 10 & = -0.5x^2 + 2x + k \\ 0 & = -0.5x^2 + 2x + (k - 10) \\ \\ b^2 - 4ac & = (2)^2 - 4(-0.5)(k - 10) \\ & = 4 + 2(k - 10) \\ & = 4 + 2k - 20 \\ & = 2k - 16 \\ \\ b^2 - 4ac & \ge 0 \\ 2k - 16 & \ge 0 \\ 2k & \ge 16 \\ k & \ge {16 \over 2} \\ k & \ge 8 \end{align*}