(a)

\begin{align} 3x^2 + 12x - 7 & = 3(x^2 + 4x) - 7 \\ & = 3\left[ x^2 + 4x + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 \right] - 7 \\ & = 3 (x^2 + 4x + 2^2 - 2^2 ) - 7 \\ & = 3 [ (x + 2)^2 - 4 ] - 7 \\ & = 3 (x + 2)^2 - 12 - 7 \\ & = 3 (x + 2)^2 - 19 \\ \\ \text{Min. value} & = -19, \text{ when } x = -2 \end{align}

(b)

\begin{align} -2x^2 - x - 11 & = -2 \left(x^2 + {1 \over 2}x \right) - 11 \\ & = -2 \left[ x^2 + {1 \over 2}x + \left(1 \over 4\right)^2 - \left(1 \over 4\right)^2 \right] - 11 \\ & = -2 \left[ \left(x + {1 \over 4}\right)^2 - {1 \over 16} \right] - 11 \\ & = -2 \left(x + {1 \over 4} \right)^2 + {1 \over 8} - 11 \\ & = -2 \left(x + {1 \over 4} \right)^2 - {87 \over 8} \\ \\ \text{Max. value} & = -{87 \over 8}, \text{ when } x = -{1 \over 4} \end{align}

(a)

\begin{align} y & = {1 \over 4}(x + 12)(x + 8) \phantom{000000} [ \text{Minimum curve } \cup] \\ \\ \text{Let } & x = 0, \\ y & = {1 \over 4}(0 + 12)(0 + 8) \\ & = 24 \phantom{000000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = {1 \over 4}(x + 12)(x + 8) \\ 0 & = (x + 12)(x + 8) \\ \\ x & = -12 \text{ or } - 8 \phantom{00000000000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } & x = {-12 + (-8) \over 2} \\ & = -10 \\ \\ \text{When } & x = -10, \\ y & = {1 \over 4} (-10 + 12)(-10 + 8) \\ y & = - 1 \\ \\ \text{Turning } & \text{point: } (-10, -1) \end{align}

(b)

\begin{align} y & = (6 - 5x)(4x - 3) \\ & = 24x - 18 - 20x^2 + 15x \\ & = - 20x^2 + 39x - 18 \phantom{000000} [ \text{Maximum curve } \cap] \\ \\ \text{Let } & x = 0, \\ y & = [6 - 5(0)][4(0) - 3] \\ y & = -18 \phantom{000000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = (6 - 5x)(4x - 3) \end{align} \begin{align} 6 - 5x & = 0 && \text{ or } & 4x - 3 & = 0 \\ -5x & = -6 &&& 4x & = 3 \\ x & = {-6 \over -5} &&& x & = {3 \over 4} \\ x & = 1.2 &&& x & = 0.75 \phantom{0000000} [x \text{-intercepts}] \end{align}
\begin{align} \text{Line of symmetry, } x & = {1.2 + 0.75 \over 2} \\ & = 0.975 \\ \\ \text{When } & x = 0.975, \\ y & = [6 - 5(0.975)][4(0.975) - 3] \\ & = 1.0125 \\ \\ \text{Turning point: } & (0.975, 1.0125) \end{align}

(c)

\begin{align} y & = 0.5(x + 6)^2 + 9 \phantom{000000} [ \text{Minimum curve } \cup] \\ \\ \text{Turning} & \text{ point: } (-6, 9) \\ \\ \text{Let } & x = 0, \\ y & = 0.5(0 + 6)^2 + 9 \\ & = 27 \phantom{000000000000000000} [y \text{-intercept}] \end{align}

(d)

\begin{align} y & = 2x^2 - 10x + 17 \phantom{000000} [ \text{Minimum curve } \cup] \\ & = 2(x^2 - 5x) + 17 \\ & = 2 \left[ x^2 - 5x + \left(5 \over 2\right)^2 - \left(5 \over 2\right)^2 \right] + 17 \\ & = 2 \left[ \left( x - {5 \over 2} \right)^2 - {25 \over 4} \right] + 17 \\ & = 2 \left(x - {5 \over 2} \right)^2 - {25 \over 2} + 17 \\ & = 2 \left(x - {5 \over 2} \right)^2 + {9 \over 2} \\ \\ \text{Turning} & \text{ point: } \left({5 \over 2}, {9 \over 2}\right) \\ \\ \text{Let } & x = 0, \\ y & = 2 \left(0 - {5 \over 2}\right)^2 + {9 \over 2} \\ & = 17 \phantom{000000000000000000} [y \text{-intercept}] \end{align}

(a)

\begin{align} y & = 4x^2 + 5x + 1 \\ & = 4 \left(x^2 + {5 \over 4}x \right) + 1 \\ & = 4 \left[x^2 + {5 \over 4}x + \left(5 \over 8\right)^2 - \left(5 \over 8\right)^2 \right] + 1 \\ & = 4 \left[ \left(x + {5 \over 8} \right)^2 - {25 \over 64} \right] + 1 \\ & = 4 \left(x + {5 \over 8} \right)^2 - {25 \over 16} + 1 \\ & = 4 \left(x + {5 \over 8} \right)^2 - {9 \over 16} \\ \\ \\ \text{Shape: } & \text{Minimum curve } \cup \\ \\ \text{Minimum point: } & \left(-{5 \over 8}, -{9 \over 16}\right) \end{align}

(b)

\begin{align} y & = 4x^2 + 5x + 1 \\ & = (4x + 1)(x + 1) \\ \\ \\ \text{Shape: } & \text{Minimum curve } \cup \\ \\ \text{Let } & y = 0, \\ 0 & = (4x + 1)(x + 1) \\ \\ x \text{-intercepts: } & -{1 \over 4}, -1 \end{align}

(i)

\begin{align} \text{Condition #1: } \phantom{0} & h < 0 \\ \\ \\ hx^2 + 6x + k & = h \left(x^2 + {6 \over h} x \right) + k \\ & = h \left[ x^2 + {6 \over h} + \left(3 \over h\right)^2 - \left(3 \over h\right)^2 \right] + k \\ & = h \left[ \left(x + {3 \over h}\right)^2 - {9 \over h^2} \right] + k \\ & = h \left(x + {3 \over h}\right) - {9h \over h^2} + k \\ & = h \left(x + {3 \over h}\right) - {9 \over h} + k \\ \\ \text{Max. value} & = -{9 \over h} + k, \text{ when } x = -{3 \over h} \\ \\ -{9 \over h} + k & < 0 \phantom{000000} [\text{Max. value is negative}] \\ \text{Condition #2: } \phantom{0} k & < {9 \over h} \end{align}

(ii)

\begin{align} \text{Condition #1: } \phantom{0} & h < 0 \\ \\ \text{Condition #2: } \phantom{0} k & < {9 \over h} \\ \\ \\ \text{Let } & h = -1, \\ k & < {9 \over -1} \\ k & < -9 \\ \\ \\ \text{Possible set: } & h = -1, k = -10 \end{align}

\begin{align} \text{Condition #1: } \phantom{0} & a > 0 \phantom{000000} [\text{Minimum curve } \cup] \\ \\ \\ y & = ax^2 + bx + c \\ & = a \left(x^2 + {b \over a}x\right) + c \\ & = a \left[ x^2 + {b \over a}x + \left(b \over 2a\right)^2 - \left(b \over 2a\right)^2 \right] + c \\ & = a \left[ \left(x + {b \over 2a} \right)^2 - {b^2 \over 4a^2} \right] + c \\ & = a \left(x + {b \over 2a} \right)^2 - {ab^2 \over 4a^2} + c \\ & = a \left(x + {b \over 2a} \right)^2 - {b^2 \over 4a} + c \\ \\ \text{Turning point: } & \left(-{b \over 2a}, -{b^2 \over 4a} + c \right) \\ \\ \text{Condition #2: } \phantom{0} & -{b^2 \over 4a} + c > 0 \\ \\ \\ \text{Condition #3: } \phantom{0} & abc < 0 \\ \\ \\ \text{Let } & a = 2 \phantom{000000} [\text{to meet condition #1}] \\ \\ \text{Let } & b = -4 \\ \\ -{(-4)^2 \over 4(2)} + c & > 0 \phantom{0000000} [\text{condition #2}] \\ \\ -2 + c & > 0 \\ c & > 2 \\ \\ \\ \text{Possible values: } & a = 2, b = -4, c = 3 \phantom{0000000} [\text{satisfies condition #3}] \end{align}

(i)

\begin{align} 3x^2 + 12x - 5 & = 3(x^2 + 4x) - 5 \\ & = 3 \left[ x^2 + 4x + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 \right] - 5 \\ & = 3 (x^2 + 4x + 2^2 - 2^2 ) - 5 \\ & = 3 [(x + 2)^2 - 4] - 5 \\ & = 3(x + 2)^2 - 12 - 5 \\ & = 3(x + 2)^2 - 17 \\ \\ y & = 3x^2 + 12x - 5 \phantom{0000000} [\text{Minimum curve } \cup] \\ y & = 3(x + 2)^2 - 17 \\ \\ \text{Turning} & \text{ point: } (-2, -17) \\ \\ \text{Let } & x = 0, \\ y & = 3(0 + 2)^2 - 17 \\ & = -5 \phantom{00000000000000000} [y \text{-intercept}] \end{align}

$$ \text{Minimum value} = -17, \text{ when } x = -2 $$

(ii)

\begin{align} {1 \over -17} & = -{1 \over 17} \\ \\ \text{Maximum point: } & \left( -2, - {1 \over 17} \right) \end{align}

(a)

\begin{align} & \text{Insufficient information to determine that } a = 4 \end{align}

(b)

\begin{align} \text{Turning} & \text{ point: } (-2, 5) \\ \\ y = f(x) & \text{ is a minimum curve } \cup \text{ given Statement B} \\ \\ \text{Since tu} & \text{rning point lies above } x \text{-axis, statement is true} \end{align}

(c)

\begin{align} \text{Turning} & \text{ point: } (-2, 5) \\ \\ y = f(x) & \text{ is a minimum curve } \cup \text{ given Statement B} \\ \\ \therefore \text{Statem} & \text{ent is true} \end{align}

(i)

\begin{align} y & = -{1 \over 8} x^2 + x + c \\ & = -{1 \over 8} (x^2 - 8x) + c \\ & = -{1 \over 8} \left[ x^2 - 8x + \left(8 \over 2\right)^2 - \left(8 \over 2\right)^2 \right] + c \\ & = -{1 \over 8} ( x^2 - 8x + 4^2 - 4^2 ) + c \\ & = -{1 \over 8} [ (x - 4)^2 - 16] + c \\ & = -{1 \over 8}(x - 4)^2 + 2 + c \\ \\ \text{Turning} & \text{ point: } (4, 2 + c) \\ \\ 2 + c & = 2.4 \\ c & = 2.4 - 2 \\ c & = 0.4 \\ \\ \text{Horizontal } & \text{distance} = 4 \text{ m} \end{align}

(ii)

\begin{align} y & = -{1 \over 8}(x - 4)^2 + 2 + c \\ & = -{1 \over 8}(x - 4)^2 + 2 + 0.4 \\ & = -{1 \over 8}(x - 4)^2 + 2.4 \phantom{00000000} [\text{Maximum curve } \cap] \\ \\ \text{Turning} & \text{ point: } (4, 2.4) \\ \\ \text{Let } & x = 0, \\ y & = -{1 \over 8}(0 - 4)^2 + 2.4 \\ & = 0.4 \phantom{00000000000000000000} [y \text{-intercept}] \end{align}

(iii)

\begin{align} y & = -{1 \over 8}(x - 4)^2 + 2.4 \\ \\ \text{Let } & y = 0, \\ 0 & = -{1 \over 8}(x - 4)^2 + 2.4 \\ {1 \over 8}(x - 4)^2 & = 2.4 \\ (x - 4)^2 & = {2.4 \over {1 \over 8}} \\ (x - 4)^2 & = 19.2 \\ x - 4 & = \pm \sqrt{19.2} \\ x - 4 & = \pm 4.3817 \\ x & = 4.3817 + 4 \phantom{0} \text{ or } \phantom{0} -4.3817 + 4 \\ x & = 8.3817 \phantom{0} \text{ or } \phantom{0} -0.3817 \text{ (Reject)} \\ x & \approx 8.38 \\ \\ \text{The ball lands } & \text{on the ground } (y = 0) \text{ after travelling 8.38 m horizontally} \end{align}

(i)

\begin{align} N & = 25t^2 - 200t + 500 \\ \\ \text{Let } & t = 24, \\ N & = 25(24)^2 - 200(24) + 500 \\ & = 10 \phantom{.} 100 \end{align}

(ii)

\begin{align} N & = 25t^2 - 200t + 500 \phantom{000000} [\text{Minimum curve } \cup] \\ & = 25(t^2 - 8t) + 500 \\ & = 25 \left[ t^2 - 8t + \left(8 \over 2\right)^2 - \left(8 \over 2\right)^2 \right] + 500 \\ & = 25 \left( t^2 - 8t + 4^2 - 4^2 \right) + 500 \\ & = 25 [ (t - 4)^2 - 16 ] + 500 \\ & = 25 (t - 4)^2 - 400 + 500 \\ & = 25 (t - 4)^2 + 100 \\ \\ \text{Turning } & \text{point: } (4, 100) \\ \\ \text{Let } & t = 0, \\ N & = 25(0 - 4)^2 + 100 \\ & = 500 \end{align}

\begin{align} & \text{Not accurate to conclude since the number of bacteria decreases for the first 4 hours} \end{align}

(i)

\begin{align} & \text{When } x = 0, y = -1.4 \\ \\ & \implies \text{Fixed cost of operation is } \$14 \phantom{.} 000 \end{align}

(ii)

\begin{align} \text{General equation: } & y = a(x - p)(x - q) \\ \\ \text{Since one } & \text{x-intercept is } 15, \\ y & = a(x - 15)(x - q) \\ \\ \text{Using } & (0, -1.4), \\ -1.4 & = a(-15)(-q) \\ -1.4 & = 15aq \phantom{000} \text{--- (1)} \\ \\ \text{Using } & (4, 3.4), \\ 3.4 & = a(4 - 15)(4 - q) \\ 3.4 & = a(-11)(4- q) \\ 3.4 & = -11a(4 - q) \\ 3.4 & = -44a + 11aq \phantom{000} \text{--- (2)} \\ \\ \text{From (1), } -{1.4 \over 15} & = aq \\ -{7 \over 75} & = aq \\ \\ \text{Substitute} & \text{ into (2),} \\ 3.4 & = -44a + 11 \left(-7 \over 75\right) \\ 3.4 & = -44a - {77 \over 75} \\ 44a & = -{77 \over 75} - 3.4 \\ 44a & = -{332 \over 75} \\ a & = -{83 \over 825} \\ \\ aq & = -{7 \over 75} \\ -{83 \over 825} q & = -{7 \over 75} \\ q & = {- {7 \over 75} \over -{83 \over 825}} \\ & = {77 \over 83} \\ \\ \\ y & = -{83 \over 825} (x - 15)\left(x - {77 \over 83}\right) \end{align}

(iii)

\begin{align} y & = -{83 \over 825} (x - 15)\left(x - {77 \over 83}\right) \\ \\ \text{Line of symmetry, } x & = {15 + {77 \over 83} \over 2} \\ & = 7{80 \over 83} \\ \\ \text{No. of workers} & \approx 8 \\ \\ \text{When } & x = 8, \\ y & = -{83 \over 825} (8 - 15)\left(8 - {77 \over 83}\right) \\ & = 4.9806 \\ & \approx 4.98 \\ \\ \text{Expected revenue} & = 4.98 \times 10 \phantom{.} 000 \\ & = \$ 49 \phantom{.} 800 \end{align}