(a)

\begin{align} y & = 3(x - 2)(x - 6) \phantom{0000000} [\text{Minimum curve: } \cup] \\ \\ \text{When } & x = 0, \\ y & = 3(0 - 2)(0 - 6) \\ y & = 36 \phantom{00000000} [y \text{-intercept}] \\ \\ \\ \text{When } & y = 0, \\ 0 & = 3(x - 2)(x - 6) \\ 0 & = (x - 2)(x - 6) \\ \\ x & = 2 \text{ or } 6 \phantom{000000} [x \text{-intercepts}] \\ \\ \\ x & = {2 + 6 \over 2} \\ x & = 4 \phantom{0000000000} [\text{Line of symmetry}] \\ \\ \\ \text{Substitute } & x = 4 \text{ into eqn of curve,} \\ y & = 3(4 - 2)(4 - 6) \\ y & = -12 \\ \\ \text{Minimum } & \text{point: } (4, -12) \end{align}

$$ \text{Minimum value} = -12 $$

(b)

\begin{align} y & = -2(x - 3)(x + 1) \phantom{0000000} [\text{Maximum curve: } \cap] \\ \\ \text{When } & x = 0, \\ y & = -2(0 - 3)(0 + 1) \\ y & = 6 \phantom{00000000} [y \text{-intercept}] \\ \\ \\ \text{When } & y = 0, \\ 0 & = -2(x - 3)(x + 1) \\ 0 & = (x - 3)(x + 1) \\ \\ x & = -1 \text{ or } 3 \phantom{000000} [x \text{-intercepts}] \\ \\ \\ x & = {-1 + 3 \over 2} \\ x & = 1 \phantom{0000000000} [\text{Line of symmetry}] \\ \\ \\ \text{Substitute } & x = 1 \text{ into eqn of curve,} \\ y & = -2(1 - 3)(1 + 1) \\ y & = 8 \\ \\ \text{Maximum } & \text{point: } (1, 8) \end{align}

$$ \text{Maximum value} = 8 $$

(c)

\begin{align} y & = (2x - 1)(x + 4) \\ y & = 2\left(x - {1 \over 2}\right)(x + 4) \phantom{0000000} [\text{Minimum curve: } \cup] \\ \\ \text{When } & x = 0, \\ y & = 2\left(0 - {1 \over 2}\right)(0 + 4) \\ y & = -4 \phantom{00000000} [y \text{-intercept}] \\ \\ \\ \text{When } & y = 0, \\ 0 & = 2\left(x - {1 \over 2}\right)(x + 4) \\ 0 & = \left(x - {1 \over 2}\right)(x + 4) \\ \\ x & = {1 \over 2} \text{ or } -4 \phantom{000000} [x \text{-intercepts}] \\ \\ \\ x & = {0.5 + (-4) \over 2} \\ x & = -1.75 \phantom{0000000000} [\text{Line of symmetry}] \\ \\ \\ \text{Substitute } & x = -1.75 \text{ into eqn of curve,} \\ y & = 2 (-1.75 - 0.5)(-1.75 + 4) \\ y & = -10.125 \\ \\ \text{Minimum } & \text{point: } (-1.75, -10.125) \end{align}

$$ \text{Minimum value} = -10.125 $$

(d)

\begin{align} y & = 30 + 13x - 3x^2 \\ y & = -3x^2 + 13x + 30 \\ y & = -(3x^2 - 13x - 30) \\ y & = -(3x + 5)(x - 6) \phantom{0000000} [\text{Maximum curve: } \cap] \\ \\ \text{When } & x = 0, \\ y & = -[3(0) + 5](0 - 6) \\ y & = 30 \phantom{00000000} [y \text{-intercept}] \\ \\ \\ \text{When } & y = 0, \\ 0 & = -(3x + 5)(x - 6) \\ 0 & = (3x + 5)(x - 6) \\ \\ x & = -{5 \over 3} \text{ or } 6 \phantom{000000} [x \text{-intercepts}] \\ \\ \\ x & = {-{5 \over 3} + 6 \over 2} \\ x & = 2{1 \over 6} \phantom{0000000000} [\text{Line of symmetry}] \\ \\ \\ \text{Substitute } & x = 2{1 \over 6} \text{ into eqn of curve,} \\ y & = -\left[ 3 \left(2{1 \over 6}\right) + 5 \right] \left(2{1 \over 6} -6 \right) \\ y & = 44{1 \over 12} \\ \\ \text{Maximum } & \text{point: } \left(2{1 \over 6}, 44{1 \over 12}\right) \end{align}

$$ \text{Maximum value} = 44{1 \over 12} $$

(a)

\begin{align} y & = (x + 1)(x + 5) \phantom{0000000} [\text{Minimum curve: } \cup] \\ \\ \text{When } & y = 0, \\ 0 & = (x + 1)(x + 5) \\ \\ x & = -1 \text{ or } -5 \phantom{000000} [x \text{-intercepts}] \\ \\ \\ x & = {-1 + (-5) \over 2} \\ x & = -3 \phantom{0000000000} [\text{Line of symmetry}] \\ \\ \\ \text{Substitute } & x = -3 \text{ into eqn of curve,} \\ y & = (-3 + 1)(-3 + 5) \\ y & = -4 \\ \\ \text{Minimum } & \text{point: } (-3, -4) \\ \\ \\ \therefore \text{Minimum } & \text{value} = -4 \end{align}

(b)

\begin{align} y & = -3(x - 5)(x + 7) \phantom{0000000} [\text{Maximum curve: } \cap] \\ \\ \text{When } & y = 0, \\ 0 & = -3(x - 5)(x + 7) \\ 0 & = (x - 5)(x + 7) \\ \\ x & = 5 \text{ or } -7 \phantom{000000} [x \text{-intercepts}] \\ \\ \\ x & = {5 + (-7) \over 2} \\ x & = -1 \phantom{0000000000} [\text{Line of symmetry}] \\ \\ \\ \text{Substitute } & x = -1 \text{ into eqn of curve,} \\ y & = -3(-1 - 5)(-1 + 7) \\ y & = 108 \\ \\ \text{Maximum } & \text{point: } (-1, 108) \\ \\ \\ \therefore \text{Maximum } & \text{value} = 108 \end{align}

(c)

\begin{align} y & = (3x + 1)(x - 4) \phantom{0000000} [\text{Minimum curve: } \cup] \\ \\ \text{When } & y = 0, \\ 0 & = (3x + 1)(x - 4) \\ 0 & = 3\left(x + {1 \over 3}\right)(x - 4) \\ \\ x & = -{1 \over 3} \text{ or } 4 \phantom{000000} [x \text{-intercepts}] \\ \\ \\ x & = {-{1 \over 3} + 4 \over 2} \\ x & = 1{5 \over 6} \phantom{0000000000} [\text{Line of symmetry}] \\ \\ \\ \text{Substitute } & x = 1{5 \over 6} \text{ into eqn of curve,} \\ y & = \left[ 3 \left(1{5 \over 6}\right) + 1 \right] \left(1{5 \over 6} - 4 \right) \\ y & = -14{1 \over 12} \\ \\ \text{Minimum } & \text{point: } \left(1{5 \over 6}, -14{1 \over 12}\right) \\ \\ \\ \therefore \text{Minimum } & \text{value} = -14{1 \over 12} \end{align}

(d)

\begin{align} y & = 12 - 13x - 4x^2 \\ y & = -4x^2 - 13x + 12 \\ y & = -(4x^2 + 13x - 12) \\ y & = -(4x - 3)(x + 4) \phantom{0000000} [\text{Maximum curve: } \cap] \\ \\ \text{When } & y = 0, \\ 0 & = -(4x - 3)(x + 4) \\ 0 & = -4 \left(x - {3 \over 4}\right)(x + 4) \\ 0 & = \left(x - {3 \over 4}\right)(x + 4) \\ \\ x & = {3 \over 4} \text{ or } -4 \phantom{000000} [x \text{-intercepts}] \\ \\ \\ x & = {{3 \over 4} + (-4) \over 2} \\ x & = -1{5 \over 8} \phantom{0000000000} [\text{Line of symmetry}] \\ \\ \\ \text{Substitute } & x = -1{5 \over 8} \text{ into eqn of curve,} \\ y & = 12 - 13 \left(-1{5 \over 8}\right) - 4 \left(-1{5 \over 8}\right)^2 \\ y & = 22{9 \over 16} \\ \\ \text{Maximum } & \text{point: } \left(-1{5 \over 8}, 22{9 \over 16} \right) \\ \\ \\ \therefore \text{Maximum } & \text{value} = 22{9 \over 16} \end{align}

(a)

\begin{align} x^2 + 14x & = x^2 + 14x + \left(14 \over 2\right)^2 - \left(14 \over 2\right)^2 \\ & = x^2 + 14x + 7^2 - 49 \\ & = (x + 7)^2 - 49 \end{align}

(b)

\begin{align} x^2 + 7x & = x^2 + 7x + \left(7 \over 2\right)^2 - \left(7 \over 2\right)^2 \\ & = x^2 + 7x + \left(7 \over 2\right)^2 - {49 \over 4} \\ & = \left(x + {7 \over 2}\right)^2 - {49 \over 4} \end{align}

(c)

\begin{align} x^2 - 6x & = x^2 - 6x + \left(6 \over 2\right)^2 - \left(6 \over 2\right)^2 \\ & = x^2 - 6x + 3^2 - 9 \\ & = (x - 3)^2 - 9 \end{align}

(d)

\begin{align} x^2 - 9x & = x^2 - 9x + \left(9 \over 2\right)^2 - \left(9 \over 2\right)^2 \\ & = x^2 - 9x + \left(9 \over 2\right)^2 - {81 \over 4} \\ & = \left(x - {9 \over 2}\right)^2 - {81 \over 4} \end{align}

(e)

\begin{align} x^2 + 4x + 3 & = x^2 + 4x + \left(4 \over 2 \right)^2 - \left(4 \over 2\right)^2 + 3 \\ & = x^2 + 4x + 2^2 - 4 + 3 \\ & = (x + 2)^2 - 1 \end{align}

(f)

\begin{align} x^2 - 9x - {5 \over 2} & = x^2 - 9x + \left(9 \over 2\right)^2 - \left(9 \over 2\right)^2 - {5 \over 2} \\ & = x^2 - 9x + \left(9 \over 2\right)^2 - {81 \over 4} - {5 \over 2} \\ & = \left(x - {9 \over 2}\right)^2 - {91 \over 4} \end{align}

(g)

\begin{align} 2x^2 - 3x - 1 & = 2 \left( x^2 - {3 \over 2}x \right) - 1 \\ & = 2 \left[ x^2 - {3 \over 2}x + \left(3 \over 4\right)^2 - \left(3 \over 4\right)^2 \right] - 1 \\ & = 2 \left[ x^2 - {3 \over 2}x + \left(3 \over 4\right)^2 - {9 \over 16} \right] - 1 \\ & = 2 \left[ \left(x - {3 \over 4}\right)^2 - {9 \over 16} \right] - 1 \\ & = 2 \left(x - {3 \over 4}\right)^2 - {9 \over 8} - 1 \\ & = 2 \left(x - {3 \over 4}\right)^2 - {17 \over 8} \end{align}

(h)

\begin{align} - 4x^2 + 2.8x + 0.5 & = -4 (x^2 - 0.7x) + 0.5 \\ & = -4 \left[ x^2 - 0.7x + \left(0.7 \over 2\right)^2 - \left(0.7 \over 2\right)^2 \right] + 0.5 \\ & = -4 \left[ x^2 - 0.7x + (0.35)^2 - 0.1225 \right] + 0.5 \\ & = -4 [ (x - 0.35)^2 - 0.1225 ] + 0.5 \\ & = -4 (x - 0.35)^2 + 0.49 + 0.5 \\ & = -4 (x - 0.35)^2 + 0.99 \end{align}

\begin{align} \text{For all real values of } x, \phantom{0} (x + 1)^2 & \ge 0 \\ 5(x + 1)^2 & \ge 5(0) \\ 5(x + 1)^2 & \ge 0 \\ 5(x + 1)^2 - 6 & \ge -6 \\ \\ \therefore 5(x + 1)^2 - 6 \text{ can never be } & \text{less than } -6 \end{align}

(i)

\begin{align} \text{For all real values of } x, \phantom{0} (x - 7)^2 & \ge 0 \\ -4(x - 7)^2 & \le -4(0) \\ -4(x - 7)^2 & \le 0 \\ -4(x - 7)^2 + {1 \over 2} & \le {1 \over 2} \\ \\ \therefore \text{Yes, the maximum value is } & {1 \over 2} \end{align}

(ii)

\begin{align} \text{When } & x = 7, \\ {1 \over 2} - 4(7 - 7)^2 & = {1 \over 2} - 4(0)^2 \\ & = {1 \over 2} - 0 \\ & = {1 \over 2} \\ \\ \\ \therefore x & = 7 \end{align}

(a)

\begin{align} y & = f(x) \\ y & = 6(x - 5)^2 - 24 \phantom{000000} [\text{Minimum curve: } \cup] \\ \\ \text{Minimum } & \text{point: } (5, -24) \\ \\ \text{Let } & x = 0, \\ y & = 6(0 - 5)^2 - 24 \\ y & = 126 \phantom{0000000000000000} [y \text{-intercept}] \end{align}

$$ \text{Minimum value} = -24 $$

(b)

\begin{align} y & = f(x) \\ y & = -4(x + 1)^2 + 8 \phantom{000000} [\text{Maximum curve: } \cap] \\ \\ \text{Maximum } & \text{point: } (-1, 8) \\ \\ \text{Let } & x = 0, \\ y & = -4(0 + 1)^2 + 8 \\ y & = 4 \phantom{00000000000000000.} [y \text{-intercept}] \end{align}

$$ \text{Maximum value} = 8 $$

(c)

\begin{align} y & = 12(3 - x)^2 - 6 \\ y & = 12[(-1)(-3 + x)]^2 - 6 \\ y & = 12(-1)^2 (x - 3)^2 - 6 \\ y & = 12(1) (x - 3)^2 - 6 \\ y & = 12(x - 3)^2 - 6 \phantom{000000} [\text{Minimum curve: } \cup] \\ \\ \text{Minimum } & \text{point: } (3, -6) \\ \\ \text{Let } & x = 0, \\ y & = 12(0 - 3)^2 - 6 \\ y & = 102 \phantom{0000000000000000} [y \text{-intercept}] \end{align}

$$ \text{Minimum value} = -6 $$

(d)

\begin{align} y & = 5x^2 + 20x - 4 \\ y & = 5(x^2 + 4x) - 4 \\ y & = 5 \left[ \left(x + {4 \over 2}\right)^2 - \left(4 \over 2\right)^2 \right] - 4 \phantom{000000} [\text{Complete the square}] \\ y & = 5 [ (x + 2)^2 - 4 ] - 4 \\ y & = 5(x + 2)^2 - 20 - 4 \\ y & = 5(x + 2)^2 - 24 \phantom{00000000000000000} [\text{Minimum curve: } \cup] \\ \\ \text{Minimum } & \text{point: } (-2, -24) \\ \\ \text{Let } & x = 0, \\ y & = 5(0 + 2)^2 - 24 \\ y & = -4 \phantom{000000000000000000000000000} [y \text{-intercept}] \end{align}

$$ \text{Minimum value} = -24 $$

(a)

\begin{align} f(x) & = x^2 + 5x + 5 \\ & = \left(x + {5 \over 2}\right)^2 - \left(5 \over 2\right)^2 + 5 \phantom{000000} [\text{Complete the square}] \\ & = \left(x + {5 \over 2}\right)^2 - {5 \over 4} \\ \\ \text{Mini} & \text{mum value} = - {5 \over 4} \end{align}

(b)

\begin{align} f(x) & = 3x^2 - 16x - 10 \\ & = 3 \left(x^2 - {16 \over 3}x\right) - 10 \\ & = 3 \left[ \left(x - {8 \over 3}\right)^2 - \left(8 \over 3\right)^2 \right] - 10 \phantom{000000} [\text{Complete the square}] \\ & = 3 \left[ \left(x - {8 \over 3}\right)^2 - {64 \over 9} \right] - 10 \\ & = 3 \left(x - {8 \over 3}\right)^2 - {64 \over 3} - 10 \\ & = 3 \left(x - {8 \over 3}\right)^2 - {94 \over 3} \\ \\ \text{Mini} & \text{mum value} = - {94 \over 3} \end{align}

(c)

\begin{align} y & = - 2x^2 - 4x + 44 \\ & = -2(x^2 + 2x) + 44 \\ & = -2 \left[ \left(x + {2 \over 2}\right)^2 - \left(2 \over 2\right)^2 \right] + 44 \phantom{000000} [\text{Complete the square}] \\ & = -2 [ (x + 1)^2 - 1 ]+ 44 \\ & = -2(x + 1)^2 + 2 + 44 \\ & = -2(x + 1)^2 + 46 \\ \\ \text{Maxi} & \text{mum value} = 46 \end{align}

(d)

\begin{align} y & = (6x + 1)(x - 5) - 1 \\ & = 6x^2 - 30x + x - 5 - 1 \\ & = 6x^2 - 29x - 6 \\ & = 6 \left(x^2 - {29 \over 6}x \right) - 6 \\ & = 6 \left[ \left(x - {29 \over 12}\right)^2 - \left(29 \over 12\right)^2 \right] - 6 \phantom{000000} [\text{Complete the square}] \\ & = 6 \left[ \left(x - {29 \over 12}\right)^2 - {841 \over 144} \right] - 6 \\ & = 6 \left(x - {29 \over 12}\right)^2 - {841 \over 24} - 6 \\ & = 6 \left(x - {29 \over 12}\right)^2 - {985 \over 24} \\ \\ \text{Mini} & \text{mum value} = -{985 \over 24} \end{align}

(a)

\begin{align} y & = -3(x + 4)^2 - 5 \phantom{000000} [\text{Maximum curve: } \cap] \\ \\ \text{Maximum } & \text{point: } (-4, -5) \\ \\ \text{Let } & x = 0, \\ y & = -3(0 + 4)^2 - 5 \\ y & = -53 \phantom{000000000000000} [y \text{-intercept}] \end{align}

$$ \text{Maximum value} = -5$$

(b)

\begin{align} y & = f(x) \\ y & = {1 \over 2} \left(x - {8 \over 3}\right)^2 + {8 \over 3} \phantom{000000} [\text{Minimum curve: } \cup] \\ \\ \text{Minimum } & \text{point: } \left({8 \over 3}, {8 \over 3} \right) \\ \\ \text{Let } & x = 0, \\ y & = {1 \over 2} \left(0 - {8 \over 3}\right)^2 + {8 \over 3} \\ & = 6{2 \over 9} \phantom{000000000000000000} [y \text{-intercept}] \end{align}

$$ \text{Minimum value} = 2{2 \over 3}$$

(a)

\begin{align} y & = 5x^2 - 20x + 23 \\ & = 5(x^2 - 4x) + 23 \\ & = 5 \left[ \left(x - {4 \over 2}\right)^2 - \left(4 \over 2\right)^2 \right] + 23 \phantom{000000} [\text{Complete the square}] \\ & = 5 [ (x - 2)^2 - 4 ] + 23 \\ & = 5 (x - 2)^2 - 20 + 23 \\ & = 5 (x - 2)^2 + 3 \\ \\ \\ \text{Mini} & \text{mum value} = 3, \text{ when } x = 2 \end{align}

(b)

\begin{align} f(x) & = - {1 \over 4} x^2 - {1 \over 4}x - {11 \over 16} \\ & = -{1 \over 4} (x^2 + x) - {11 \over 16} \\ & = -{1 \over 4} \left[ \left(x + {1 \over 2}\right)^2 - \left(1 \over 2\right)^2 \right] - {11 \over 16} \phantom{000000} [\text{Complete the square}] \\ & = -{1 \over 4} \left[ \left(x + {1 \over 2}\right)^2 - {1 \over 4} \right] - {11 \over 16} \\ & = -{1 \over 4} \left(x + {1 \over 2}\right)^2 + {1 \over 16} - {11 \over 16} \\ & = -{1 \over 4} \left(x + {1 \over 2}\right)^2 - {5 \over 8} \\ \\ \\ \text{Maxi} & \text{mum value} = -{5 \over 8}, \text{ when } x = -{1 \over 2} \end{align}

(a)

\begin{align} y & = -x^2 + 10x - 31 \phantom{0000000000000000} [\text{Maximum curve: } \cap] \\ y & = -(x^2 - 5x) - 31 \\ y & = - \left[ \left(x - {5 \over 2}\right)^2 - \left(5 \over 2\right)^2 \right] - 31 \phantom{00000} [\text{Complete the square}] \\ y & = - \left[ \left(x - {5 \over 2}\right)^2 - {25 \over 4} \right] - 31 \\ y & = - \left( x - {5 \over 2}\right)^2 + {25 \over 4} - 31 \\ y & = - \left(x - {5 \over 2}\right)^2 - {99 \over 4} \\ \\ \\ \text{Maxi} & \text{mum point: } \left({5 \over 2}, - {99 \over 4} \right) \\ \\ \text{Since } & \text{maximum point is below } x \text{-axis, curve is completely below } x \text{-axis} \end{align}

(b)

\begin{align} y & = {1 \over 4}x^2 + 2x + 7 \phantom{0000000000000000} [\text{Minimum curve: } \cup] \\ y & = {1 \over 4} (x^2 + 8x) + 7 \\ y & = {1 \over 4} \left[ \left(x + {8 \over 2}\right)^2 - \left(8 \over 2\right)^2 \right] + 7 \phantom{00000} [\text{Complete the square}] \\ y & = {1 \over 4} [ (x + 4)^2 - 16 ] + 7 \\ y & = {1 \over 4} (x + 4)^2 - 4 + 7 \\ y & = {1 \over 4} (x + 4)^2 + 3 \\ \\ \\ \text{Mini} & \text{mum point: } (-4, 3) \\ \\ \text{Since } & \text{minimum point is above } x \text{-axis, curve is completely above } x \text{-axis} \end{align}

(i)

\begin{align} y & = - 0.000 \phantom{.} 025x^2 + 0.003x + 0.06 \phantom{000000000} [\text{Surface resembles a maximum curve } \cap] \\ y & = - 0.000 \phantom{.} 025 (x^2 - 120x) + 0.06 \\ y & = - 0.000 \phantom{.} 025 \left[ \left(x - {120 \over 2}\right)^2 - \left(120 \over 2\right)^2 \right] + 0.06 \phantom{00000} [\text{Complete the square}] \\ y & = - 0.000 \phantom{.} 025 [ (x - 60)^2 - 3600 ] + 0.06 \\ y & = - 0.000 \phantom{.} 025 (x - 60)^2 + 0.09 + 0.06 \\ y & = - 0.000 \phantom{.} 025 (x - 60)^2 + 0.15 \\ \\ \text{Maximum} & \text{ point: } (60, 0.15) \\ \\ \text{Greatest} & \text{ height } (y) = 0.15 \text{ m} \\ \\ \text{Distance} & \text{ from one end } (x) = 60 \text{ m} \end{align}

(ii)

\begin{align} \text{From (i), distance from one end} & = 60 \text{ m} \\ \\ \therefore \text{Width of field} & = 60 \times 2 \\ & = 120 \text{ m} \end{align}

(i)

\begin{align} h & = ut + {1 \over 2} at^2 + 8 \\ \\ \text{When } & t = 0, \\ h & = u(0) + {1 \over 2} a(0)^2 + 8 \\ h & = 8 \\ \\ \therefore \text{Initial } & \text{height of the object is } 8 \text{ m} \end{align}

(ii)

\begin{align} h & = ut + {1 \over 2} at^2 + 8 \\ \\ \text{When } & u = 12 \text{ and } a = -10, \\ h & = 12t + {1 \over 2}(-10)t^2 + 8 \\ h & = 12t - 5t^2 + 8 \\ h & = -5t^2 + 12t + 8 \\ h & = -5(t^2 - 2.4t) + 8 \\ h & = -5 \left[ \left(t - {2.4 \over 2}\right)^2 - \left(2.4 \over 2\right)^2 \right] + 8 \phantom{000000} [\text{Complete the square}] \\ h & = -5 [ (t - 1.2)^2 - 1.44 ] + 8 \\ h & = -5(t - 1.2)^2 + 7.2 + 8 \\ h & = -5(t - 1.2)^2 + 15.2 \\ \\ \\ \text{Greatest } & \text{height} = 15.2 \text{ m}, \text{ when } t = 1.2 \text{ s} \end{align}

(iii)

\begin{align} h & = -5(t - 1.2)^2 + 15.2 \\ \\ \text{When } & h = 14, \\ 14 & = -5(t - 1.2)^2 + 15.2 \\ 5(t - 1.2)^2 & = 15.2 - 14 \\ 5(t - 1.2)^2 & = 1.2 \\ (t - 1.2)^2 & = {1.2 \over 5} \\ (t - 1.2)^2 & = 0.24 \\ t - 1.2 & = \pm \sqrt{0.24} \end{align} \begin{align} t & = \sqrt{0.24} + 1.2 && \text{ or } & t & = -\sqrt{0.24} + 1.2 \\ t & \approx 1.69 &&& t & \approx 0.710 \end{align}
\begin{align} & \text{At } t = 0.710 \text{ s, the object is moving upwards and is 14 m above the ground} \\ \\ & \text{From (ii), the object reaches maximum height at } t = 1.2 \text{ s and changes direction} \\ \\ & \text{Thus, at } t = 1.69 \text{ s, the object is moving downwards and is 14 m above the ground} \end{align}