(i)

\begin{align*} f(-1) & = -2(-1)^2 + 4 \\ & = 2 \end{align*}

(ii)

\begin{align*} \text{For all real} & \text{ values of } x, \\ x^2 & \ge 0 \\ -2x^2 & \le 0 \\ -2x^2 + 4 & \le 0 + 4 \\ -2x^2 + 4 & \le 4 \\ \therefore f(x) & \le 4 \end{align*}

(i)

\begin{align*} x^2 + 3x - 6 & = x^2 + 3x + \left(3 \over 2\right)^2 - \left(3 \over 2\right)^2 - 6 \\ & = \underbrace{x^2 + 3x + 1.5^2}_{a^2 + 2ab + b^2 = (a + b)^2} - 1.5^2 - 6 \\ & = (x + 1.5)^2 - 8.25 \\ \\ \text{For all real} & \text{ values of } x, \\ (x + 1.5)^2 & \ge 0 \\ (x + 1.5)^2 - 8.25 & \ge 0 - 8.25 \\ (x + 1.5)^2 & \ge - 8.25 \\ (x + 1.5)^2 & \ge -{33 \over 4} \\ \therefore f(x) & \ge - {33 \over 4} \end{align*}

(ii)

\begin{align*} \text{Since } f(x) & \ge - {33 \over 4}, \text{ minimum value} = -{33 \over 4} \end{align*}

(iii)

\begin{align*} f(x) & = (\underbrace{x + 1.5}_{=0})^2 - {33 \over 4} \\ \\ x & = -1.5 \end{align*}

(i)

\begin{align*} f(x) & = 2x^2 - 3x - 1 \\ & = 2(x^2 - 1.5x) - 1 \\ & = 2 \left[ x^2 - 1.5x + \left(1.5 \over 2\right)^2 - \left(1.5 \over 2\right)^2 \right] - 1 \\ & = 2 \left( \underbrace{x^2 - 1.5x + 0.75^2}_{a^2 - 2ab + b^2 = (a - b)^2} - 0.75^2 \right) - 1 \\ & = 2 \left[ (x - 0.75)^2 - 0.5625 \right] - 1 \\ & = 2(x - 0.75)^2 - 1.125 - 1 \\ & = 2(x - 0.75)^2 - 2.125 \\ \\ \text{For all real} & \text{ values of } x, \\ (x - 0.75)^2 & \ge 0 \\ 2(x - 0.75)^2 & \ge 0 \\ 2(x - 0.75)^2 -2.125 & \ge -2.125 \\ \\ \therefore \text{Minimum value of } f(x) & = -2.125 = -{17 \over 8} \end{align*}

(ii)

\begin{align*} f(x) & = -3x^2 + 4x - 2 \\ & = -3 \left( x^2 - {4 \over 3}x \right) - 2 \\ & = -3 \left[ x^2 - {4 \over 3}x + \left({4 \over 3} \div 2\right)^2 - \left({4 \over 3} \div 2\right)^2 \right] - 2 \\ & = -3 \left[ \underbrace{x^2 - {4 \over 3}x + \left(2 \over 3\right)^2}_{a^2 - 2ab + b^2 = (a - b)^2} - \left(2 \over 3\right)^2 \right] - 2 \\ & = -3 \left[ \left(x - {2 \over 3}\right)^2 - {4 \over 9}\right] - 2 \\ & = -3 \left(x - {2 \over 3}\right)^2 + {4 \over 3} - 2 \\ & = -3 \left(x - {2 \over 3}\right)^2 - {2 \over 3} \\ \\ \text{For all real} & \text{ values of } x, \\ \left(x - {2 \over 3}\right)^2 & \ge 0 \\ -3 \left(x - {2 \over 3}\right)^2 & \le 0 \\ -3 \left(x - {2 \over 3}\right)^2 - {2 \over 3} & \le - {2 \over 3} \\ \\ \therefore \text{Maximum value of } f(x) & = -{2 \over 3} \end{align*}

\begin{align*} -x^2 - 4x - 10 & = - (x^2 + 4x) - 10 \\ & = - \left[ x^2 + 4x + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 \right] - 10 \\ & = - ( \underbrace{ x^2 + 4x + 2^2 }_{a^2 + 2ab + b^2 = (a + b)^2} - 2^2 ) - 10 \\ & = - [ (x + 2)^2 - 4 ] - 10 \\ & = - (x + 2)^2 + 4 - 10 \\ & = - (x + 2)^2 - 6 \\ \\ \text{For all real} & \text{ values of } x, \\ (x + 2)^2 & \ge 0 \\ -(x + 2)^2 & \le 0 \\ -(x + 2)^2 - 6 & \le -6 \\ \\ \therefore -x^2 - 4x - 10 & \text{ is always negative} \end{align*}

(i)

\begin{align*} y & = 4x^2 - 12x + 9 \\ y & = 4(x^2 - 3x) + 9 \\ y & = 4 \left[ \underbrace{x^2 - 3x + \left(3 \over 2\right)^2}_{a^2 - 2ab + b^2 = (a - b)^2} - \left(3 \over 2\right)^2 \right] +9 \\ y & = 4 \left[ \left(x - {3 \over 2}\right)^2 - {9 \over 4} \right] + 9 \\ y & = 4 \left(x - {3 \over 2}\right)^2 - 9 + 9 \\ y & = 4 \left(x - {3 \over 2}\right)^2 + 0 \\ \\ \text{Turning } & \text{point: } \left({3 \over 2}, 0\right) \\ \\ \text{Let } & x = 0, \\ y & = 4 \left(0 - {3 \over 2}\right)^2 + 0 \\ y & = 9 \end{align*}

(ii)

\begin{align*} y & = 4 \left(x - {3 \over 2}\right)^2 \\ \\ \text{Let } & y = 0, \\ 0 & = 4 \left(x - {3 \over 2}\right)^2 \\ {0 \over 4} & = \left(x - {3 \over 2}\right)^2 \\ 0 & = \left(x - {3 \over 2}\right)^2 \\ \pm \sqrt{0} & = x - {3 \over 2} \\ 0 & = x - {3 \over 2} \\ {3 \over 2} & = 0 \end{align*}

(i)

\begin{align*} y & = - 2x^2 + 2x + 4 \\ \\ \text{Let } & y = 0, \\ 0 & = - 2x^2 + 2x + 4 \\ 0 & = x^2 - x - 2 \\ 0 & = (x + 1)(x - 2) \end{align*} \begin{align*} x + 1 & = 0 && \text{ or } & x - 2 & = 0 \\ x & = -1 &&& x & = 2 \end{align*}
\begin{align*} \text{Line of symmetry, } x & = {-1 + 2 \over 2} \\ x & = 0.5 \\ \\ \text{Substitute } & x = 0.5 \text{ into eqn of curve,} \\ y & = -2(0.5)^2 + 2(0.5) + 4 \\ y & = 4.5 \\ \\ \text{Turning } & \text{point: } (0.5, 4.5) \end{align*}

(ii)

\begin{align*} \text{The curve } y = -2x^2 + 2x + 4 \text{ has two } x \text{-intercepts.} \end{align*}

(a)

\begin{align*} y & = x^2 + 2x - 6 \\ \\ b^2 - 4ac & = (2)^2 - 4(1)(-6) \\ & = 28 > 0 \\ \\ \text{Graph of } y = x^2 & + 2x - 6 \text{ has two } x \text{-intercepts.} \end{align*}

(b)

\begin{align*} y & = (1 - x)^2 + 2 \\ y & = \underbrace{ (1)^2 - 2(1)(x) + (x)^2 }_{ (a - b)^2 = a^2 - 2ab + b^2 } + 2 \\ y & = 1 - 2x + x^2 + 2 \\ y & = x^2 - 2x + 3 \\ \\ b^2 - 4ac & = (-2)^2 - 4(1)(3) \\ & = -8 < 0 \\ \\ \text{Graph of } y & = (1 - x)^2 + 2 \text{ has no } x \text{-intercepts.} \end{align*}

\begin{align*} y & = 3x^2 - px + 12 \\ \\ b^2 - 4ac & = (-p)^2 - 4(3)(12) \\ & = p^2 - 144 \\ \\ b^2 - 4ac & = 0 \phantom{000000} [\text{Only 1 } x\text{-intercept}] \\ p^2 - 144 & = 0 \\ p^2 & = 144 \\ p & = \pm \sqrt{144} \\ p & = \pm 12 \end{align*}

\begin{align*} y & = x^2 - 2x + 2 - p \\ \\ b^2 - 4ac & = (-2)^2 - 4(1)(2 - p) \\ & = 4 - 4(2 - p) \\ & = 4 - 8 + 4p \\ & = 4p - 4 \\ & = 4(p - 1) \\ \\ p & > 1 \\ p - 1 & > 0 \\ 4(p - 1) & > 4(0) \\ 4(p - 1) & > 0 \\ \implies b^2 - 4ac & > 0 \\ \\ \therefore \text{For } p > 1, \text{ the curve } & \text{intersects the } x \text{-axis at two points} \end{align*}

(a)

\begin{align*} y & = - x^2 + 2kx - (k^2 + 1) \\ y & = \underbrace {- x^2}_{ \text{Maximum curve } \cap} + 2kx - k^2 - 1 \\ \\ b^2 - 4ac & = (2k)^2 - 4(-1)(-k^2 - 1) \\ & = 4k^2 + 4(-k^2 - 1) \\ & = 4k^2 - 4k^2 - 4 \\ & = -4 < 0 \\ \\ \implies \text{The graph is a } & \text{maximum curve } \cap \text{ that does not meet the } x \text{-axis.} \\ \therefore \text{The graph lies } & \text{completely below the } x \text{-axis}. \end{align*}

(b)

\begin{align*} \text{Consider } y & = \underbrace {{1 \over 4} x^2}_{ \text{Minimum curve } \cup} + kx + k^2 + 1 \\ \\ b^2 - 4ac & = (k)^2 - 4\left(1 \over 4\right)(k^2 + 1) \\ & = k^2 - (k^2 + 1) \\ & = k^2 - k^2 - 1 \\ & = - 1 < 0 \\ \\ \implies \text{Graph of } y = {1 \over 4}x^2 + kx + k^2 + 1 & \text{ is a minimum curve } \cup \text{ that does not meet the } x \text{-axis.} \\ \therefore \text{Graph of } y = {1 \over 4}x^2 + kx + k^2 + 1 & \text{ lies completely above the } x \text{-axis} \\ \\ \therefore {1 \over 4}x^2 + kx + k^2 + 1 & \text{ is always positive} \end{align*}

(i)

\begin{align*} h & = - {1 \over 2500} x^2 + {2 \over 25}x + 3 \\ \\ \text{Let } & x = 0, \\ h & = -{1 \over 2500}(0)^2 + {2 \over 25}(0) + 3 \\ h & = 3 \text{ m} \end{align*}

(ii)

\begin{align*} h & = - {1 \over 2500} x^2 + {2 \over 25}x + 3 \\ h & = - {1 \over 2500} (x^2 - 200x) + 3 \\ h & = - {1 \over 2500} \left[ x^2 - 200x + \left(200 \over 2\right)^2 - \left(200 \over 2\right)^2 \right] + 3 \\ h & = - {1 \over 2500} ( \underbrace{x^2 - 200x + 100^2}_{a^2 - 2ab + b^2 = (a - b)^2} - 10 \phantom{.} 000 ) + 3 \\ h & = - {1 \over 2500} [ (x - 100)^2 - 10 \phantom{.} 000 ] + 3 \\ h & = - {1 \over 2500} (x - 100)^2 + 4 + 3 \\ h & = - {1 \over 2500} (x - 100)^2 + 7 \\ \\ \text{Maximum } & \text{point: } (100, 7) \\ \\ \text{Greatest height} & = 7 \text{ m} \end{align*}

(iii)

\begin{align*} h & = - {1 \over 2500} x^2 + {2 \over 25}x + 3 \\ \\ \text{Let } & x = 150, \\ h & = - {1 \over 2500}(150)^2 + {2 \over 25}(150) + 3 \\ h & = 6 > 5 \\ \\ \text{Projectile } & \text{will not smash the structure} \end{align*}

(i)

\begin{align*} \text{Maximum } & \text{point: } (5, 5) \\ \\ x \text{-intercepts: } & 0 \text{ and } 10 \\ \\ y & = a(x - p)(x - q) \\ y & = a(x - 0)(x - 10) \\ y & = ax(x - 10) \\ \\ \text{Using } & (5, 5), \\ 5 & = a(5)(5 - 10) \\ 5 & = 5a(-5) \\ 5 & = -25a \\ {5 \over -25} & = a \\ -{1 \over 5} & = a \\ \\ \therefore y & = -{1 \over 5}x(x - 10) \end{align*}

(ii)

\begin{align*} y & = -{1 \over 5}x(x - 10) \\ \\ \text{When } x & = 2, \\ y & = -{1 \over 5}(2)(2 - 10) \\ y & = 3{1 \over 5} \text{ m} \end{align*}

(iii)

\begin{align*} y & = -{1 \over 5}x(x - 10) \\ \\ \text{When } y & = 4.2, \\ 4.2 & = -{1 \over 5}x(x - 10) \\ 4.2 & = -{1 \over 5}x^2 + 2 x \\ {1 \over 5}x^2 - 2x + 4.2 & = 0 \\ x^2 - 10x + 21 & = 0 \\ (x - 3)(x - 7) & = 0 \end{align*} \begin{align*} x - 3 & = 0 && \text{ or } & x - 7 & = 0 \\ x & = 3 &&& x & = 7 \end{align*} \begin{align*} \text{Width of tunnel} & = 7 - 3 \\ & = 4 \text{ m} \end{align*}

(i)

\begin{align*} y & = a(x - h)^2 + k \\ \\ \text{Maximum } & \text{point: } (1.5, 2) \\ \\ \implies y & = a(x - 1.5)^2 + 2 \\ \\ x & \text{-intercepts: } 0, 3 \\ \\ \text{Using } & (0, 0), \\ 0 & = a(0 - 1.5)^2 + 2 \\ 0 & = a(2.25) + 2 \\ -2.25a & = 2 \\ a & = {2 \over -2.25} \\ a & = -{8 \over 9} \\ \\ \therefore y & = -{8 \over 9} \left(x - {3 \over 2}\right)^2 + 2 \end{align*}

(ii)

\begin{align*} y & = -{8 \over 9} \left(x - {3 \over 2}\right)^2 + 2 \\ \\ \text{When } & x = 1, \\ y & = -{8 \over 9} \left(1 - {3 \over 2}\right)^2 + 2 \\ y & = 1 {7 \over 9} \text{ m} \end{align*}

(iii)

\begin{align*} \text{When } & x = 1, y = 1 {7 \over 9} \approx 1.78 \text{ m} < 1.8 \text{ m} \\ \\ \therefore \text{The boat} & \text{ is unable to navigate through the underpass.} \end{align*}