(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 - 8.25 & \ge - 8.25 \\ (x + 1.5)^2 - 8.25 & \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*}

(a)

\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*}

(b)

\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 \\ \implies & y\text{-intercept is } 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} & = x \text{ (repeated root)} \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{Since the curve } y = -2x^2 + 2x + 4 \text{ has two } x \text{-intercepts } (-1 \text{ and } 2), \\ & -2x^2 + 2x + 4 = 0 \text{ has two distinct real roots.} \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) The arch is symmetrical about x = 1.5, so a width of 1 m runs from x = 1 to x = 2.

\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*}

(i)

\begin{align*} f(3) & = 2(3)^2 + 1 \\ & = 19 \end{align*}

(ii)

\begin{align*} f(-2) & = 2(-2)^2 + 1 \\ & = 9 \end{align*}

(iii)

\begin{align*} f(x) & = 2x^2 + 1 \\ \\ \text{Since } x^2 \ge 0, & \phantom{0} 2x^2 + 1 \ge 1 \\ \\ \therefore f(x) & \ge 1 \end{align*}

(a)

\begin{align*} x^2 + 12x & = \left(x + {12 \over 2}\right)^2 - \left(12 \over 2\right)^2 \\ & = (x + 6)^2 - 36 \end{align*}

(b)

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

(c)

\begin{align*} 3x^2 + 3x - 2 & = 3(x^2 + x) - 2 \\ & = 3 \left[ \left(x + {1 \over 2}\right)^2 - \left( {1 \over 2}\right)^2 \right] - 2 \\ & = 3 \left[ \left(x + {1 \over 2}\right)^2 - {1 \over 4} \right] - 2 \\ & = 3 \left( x + {1 \over 2} \right)^2 - {3 \over 4} - 2 \\ & = 3 \left(x + {1 \over 2} \right)^2 - {11 \over 4} \end{align*}

(d)

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

(e)

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

(f)

\begin{align*} -x^2 - 4x + 1 & = -(x^2 + 4x) + 1 \\ & = - \left[ \left( x + {4 \over 2} \right)^2 - \left(4 \over 2\right)^2 \right] + 1 \\ & = - [ (x + 2)^2 - 4] + 1 \\ & = -(x + 2)^2 + 4 + 1 \\ & = -(x + 2)^2 + 5 \end{align*}

(g)

\begin{align*} -{1 \over 2}x^2 - 3x + 5 & = -{1 \over 2} \left( x^2 + 6x \right) + 5 \\ & = -{1 \over 2} \left[ \left(x + {6 \over 2} \right)^2 - \left(6 \over 2\right)^2 \right] + 5 \\ & = -{1 \over 2} [ (x + 3)^2 - 9] + 5 \\ & = -{1 \over 2} (x + 3)^2 + {9 \over 2} + 5 \\ & = -{1 \over 2} (x + 3)^2 + {19 \over 2} \end{align*}

(h)

\begin{align*} -2x^2 + 5x + 3 & = -2 \left( x^2 - {5 \over 2}x \right) + 3 \\ & = -2 \left[ \left(x - {5 \over 4}\right)^2 - \left(5 \over 4\right)^2 \right] + 3 \\ & = -2 \left[ \left(x - {5 \over 4}\right)^2 - {25 \over 16} \right] + 3 \\ & = -2 \left( x - {5 \over 4} \right)^2 + {25 \over 8} + 3 \\ & = -2 \left( x - {5 \over 4} \right)^2 + {49 \over 8} \end{align*}

For a quadratic expression in the form a(x - h)² + k,

• If a > 0, the expression has a minimum value of k, when x = h
• If a < 0, the expression has a maximum value of k, when x = h

(a)

$$ \text{Minimum value} = -7, \text{ when } x = -1$$

(b)

$$ \text{Maximum value} = 4, \text{ when } x = 1$$

(c)

$$ \text{Maximum value} = {4 \over 5}, \text{ when } x = {1 \over 2}$$

(d)

$$ \text{Minimum value} = 4.5, \text{ when } x = -3$$

\begin{align*} \text{For all real } & \text{values of } x, \\ (x + 3)^2 & \ge 0 \\ (x + 3)^2 + 1 & \ge 1 \\ \\ \therefore (x + 3)^2 + 1 \text{ is } & \text{positive for all real values of } x \end{align*}

(a)

\begin{align*} f(x) & = 2x^2 - 4x - 5 \\ & = 2(x^2 - 2x) - 5 \\ & = 2 \left[ \left(x - {2 \over 2} \right)^2 - \left( 2 \over 2 \right)^2 \right] - 5 \\ & = 2 [ (x - 1)^2 - 1 ] - 5 \\ & = 2(x - 1)^2 - 2 - 5 \\ & = 2(x - 1)^2 - 7 \\ \\ \text{Minimum value} & = -7, \text{ when } x = 1 \end{align*}

(b)

\begin{align*} f(x) & = -x^2 - 2x + 3 \\ & = -(x^2 + 2x) + 3 \\ & = - \left[ \left( x + {2 \over 2} \right)^2 - \left(2 \over 2\right)^2 \right] + 3 \\ & = - [ (x + 1)^2 - 1 ] + 3 \\ & = -(x + 1)^2 + 1 + 3 \\ & = -(x + 1)^2 + 4 \\ \\ \text{Maximum value} & = 4, \text{ when } x = -1 \end{align*}

(c)

\begin{align*} f(x) & = 2 - 3x + 3x^2 \\ & = 3x^2 - 3x + 2 \\ & = 3(x^2 - x) + 2 \\ & = 3 \left[ \left(x - {1 \over 2} \right)^2 - \left(1 \over 2\right)^2 \right] + 2 \\ & = 3 \left[ \left(x - {1 \over 2} \right)^2 - {1 \over 4} \right] + 2 \\ & = 3 \left(x - {1 \over 2}\right)^2 - {3 \over 4} + 2 \\ & = 3\left(x - {1 \over 2} \right)^2 + {5 \over 4} \\ \\ \text{Minimum value} & = {5 \over 4}, \text{ when } x = {1 \over 2} \end{align*}

(d)

\begin{align*} f(x) & = 3x + 4 - 2x^2 \\ & = -2x^2 + 3x + 4 \\ & = -2 \left( x^2 - {3 \over 2}x \right) + 4 \\ & = -2 \left[ \left(x - {3 \over 4}\right)^2 - \left(3 \over 4\right)^2 \right] + 4 \\ & = -2 \left[ \left(x - {3 \over 4}\right)^2 - {9 \over 16} \right] + 4 \\ & = -2 \left(x - {3 \over 4} \right)^2 + {9 \over 8} + 4 \\ & = -2 \left(x - {3 \over 4} \right)^2 +{41 \over 8} \\ \\ \text{Maximum value} & = {41 \over 8}, \text{ when } x = {3 \over 4} \end{align*}

\begin{align*} f(x) & = x^2 -2x + 3 \\ & = \left(x - {2 \over 2} \right)^2 - \left(2 \over 2\right)^2 + 3 \\ & = (x - 1)^2 - 1 + 3 \\ & = (x - 1)^2 + 2 \\ \\ \text{For all real } & \text{values of } x, \\ (x - 1)^2 & \ge 0 \\ (x - 1)^2 + 2 & \ge 2 \\ \\ \therefore f(x) \ge 2 \text{ for} & \text{ all real values of } x \end{align*}

\begin{align*} 4x - x^2 - 7 & = -x^2 + 4x - 7 \\ & = -(x^2 - 4x) - 7 \\ & = -\left[ \left(x - {4 \over 2}\right)^2 - \left(4 \over 2\right)^2 \right] - 7 \\ & = -[ (x - 2)^2 - 4] - 7 \\ & = -(x - 2)^2 + 4 - 7 \\ & = -(x - 2)^2 - 3 \\ \\ \text{For all real val} & \text{ues of } x, \\ (x - 2)^2 & \ge 0 \\ -(x - 2)^2 & \le 0 \\ -(x - 2)^2 - 3 & \le -3 \\ \\ \therefore 4x - x^2 - 7 & \text{ is always negative} \end{align*}

\begin{align*} 4x^2 - 12x + 10 & = 4(x^2 - 3x) + 10 \\ & = 4 \left[ \left( x - {3 \over 2} \right)^2 - \left(3 \over 2\right)^2 \right] + 10 \\ & = 4 \left[ \left( x - {3 \over 2} \right)^2 - {9 \over 4} \right] + 10 \\ & = 4 \left(x - {3 \over 2} \right)^2 - 9 + 10 \\ & = 4 \left(x - {3 \over 2} \right)^2 + 1 \\ \\ \text{For all real val} & \text{ues of } x, \\ \left(x - {3 \over 2}\right)^2 & \ge 0 \\ 4\left(x - {3 \over 2}\right)^2 & \ge 0 \\ 4\left(x - {3 \over 2}\right)^2 + 1 & \ge 1 \\ \\ \therefore 4x^2 - 12x + 10 & \text{ is always positive for all real values of } x \end{align*}

By factorisation:
\begin{align*} 3x^2 - 10x - 8 & = 0 \\ (x - 4)(3x + 2) & = 0 \\ \\ x = 4 \phantom{3} \text{ or } \phantom{3} & 3x = -2 \\ & \phantom{3} x = -{2 \over 3} \end{align*}

By completing the square:
\begin{align*} 3x^2 - 10x - 8 & = 0 \\ 3 \left( x^2 - {10 \over 3}x \right) - 8 & = 0 \\ 3 \left[ \left(x - {10 \over 6} \right)^2 - \left(10 \over 6\right)^2 \right] - 8 & = 0 \\ 3 \left[ \left(x - {5 \over 3} \right)^2 - {25 \over 9} \right] - 8 & = 0 \\ 3 \left(x - {5 \over 3} \right)^2 - {25 \over 3} - 8 & = 0 \\ 3 \left(x - {5 \over 3} \right)^2 - {49 \over 3} & = 0 \\ 3 \left(x - {5 \over 3} \right)^2 & = {49 \over 3} \\ \left(x - {5 \over 3} \right)^2 & = {49 \over 9} \\ x - {5 \over 3} & = \pm \sqrt{49 \over 9} \\ x - {5 \over 3} & = \pm {7 \over 3} \\ x & = \pm {7 \over 3} + {5 \over 3} \\ x & = 4 \text{ or } -{2 \over 3} \end{align*}

By quadratic formula:
\begin{align*} 3x^2 - & 10x - 8 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-10) \pm \sqrt{ (-10)^2 - 4(3)(-8)} \over 2(3)} \\ & = {10 \pm \sqrt{196} \over 6} \\ & = {10 \pm 14 \over 6} \\ & = 4 \text{ or } - {2 \over 3} \end{align*}

Factorisation is the fastest here, but only works when the quadratic factorises easily. Completing the square and the formula always work, but take more steps. Preferred method is factorisation, since it is the fastest method.

$$ y = a(x - h)^2 + k$$

In this form, the turning point of the graph is \((h, k)\).

(a)

\begin{align*} \text{Turning point: } & (-2, -8) \\ \\ \text{Let } & x = 0, \\ y & = 2(0 + 2)^2 - 8 \\ & = 0 \\ \implies & y \text{-intercept is } (0, 0) \\ \\ \text{Let } & y = 0, \\ 0 & = 2(x + 2)^2 - 8 \\ 8 & = 2(x + 2)^2 \\ {8 \over 2} & = (x + 2)^2 \\ 4 & = (x + 2)^2 \\ \pm \sqrt{4} & = x + 2 \\ \pm 2 & = x + 2 \\ \pm 2 - 2 & = x \\ -4 \text{ or } 0 & = x \\ \implies & x \text{-intercepts are } (0, 0) \text{ and } (-4, 0) \end{align*}

(b)

\begin{align*} \text{Turning point: } & (2, 3) \\ \\ \text{Let } & x = 0, \\ y & = -3(0 - 2)^2 + 3 \\ & = -9 \\ \implies & y \text{-intercept is } (0, -9) \\ \\ \text{Let } & y = 0, \\ 0 & = -3(x - 2)^2 + 3 \\ -3 & = -3(x - 2)^2 \\ {-3 \over -3} & = (x - 2)^2 \\ 1 & = (x - 2)^2 \\ \pm \sqrt{1} & = x - 2 \\ \pm 1 & = x - 2 \\ \pm 1 + 2 & = x \\ 3 \text{ or } 1 & = x \\ \implies & x \text{-intercepts are } (1, 0) \text{ and } (3, 0) \end{align*}

(c)

\begin{align*} \text{Turning point: } & (-3, -2) \\ \\ \text{No } x \text{-intercepts since} & \text{ maximum point is below } x \text{-axis} \\ \\ \text{Let } & x = 0, \\ y & = -{1 \over 2}(0 + 3)^2 - 2 \\ & = -6.5 \\ \implies & y \text{-intercept is } (0, -6.5) \end{align*}

(d)

\begin{align*} \text{Turning point: } & (1, -12) \\ \\ \text{Let } & x = 0, \\ y & = 3(0 - 1)^2 - 12 \\ & = -9 \\ \implies & y \text{-intercept is } (0, -9) \\ \\ \text{Let } & y = 0, \\ 0 & = 3(x - 1)^2 - 12 \\ 12 & = 3(x - 1)^2 \\ {12 \over 3} & = (x - 1)^2 \\ 4 & = (x - 1)^2 \\ \pm \sqrt{4} & = x - 1 \\ \pm 2 & = x - 1 \\ \pm 2 + 1 & = x \\ 3 \text{ or } -1 & = x \\ \implies & x \text{-intercepts are } (-1, 0) \text{ and } (3, 0) \end{align*}

(e)

\begin{align*} \text{Turning point: } & (-1, -3) \\ \\ \text{Let } & x = 0, \\ y & = {1 \over 3}(0 + 1)^2 - 3 \\ & = - {8 \over 3} \\ \implies & y \text{-intercept is } \left( 0, -{8 \over 3} \right) \\ \\ \text{Let } & y = 0, \\ 0 & = {1 \over 3}(x + 1)^2 - 3 \\ 3 & = {1 \over 3}(x + 1)^2 \\ {3 \over {1 \over 3}} & = (x + 1)^2 \\ 9 & = (x + 1)^2 \\ \pm \sqrt{9} & = x + 1 \\ \pm 3 & = x + 1 \\ \pm 3 - 1 & = x \\ 2 \text{ or } -4 & = x \\ \implies & x \text{-intercepts are } (-4, 0) \text{ and } (2, 0) \end{align*}

(f)

\begin{align*} \text{Turning point: } & (-2, 9) \\ \\ \text{Let } & x = 0, \\ y & = -4(0 + 2)^2 + 9 \\ & = -7 \\ \implies & y \text{-intercept is } (0, -7) \\ \\ \text{Let } & y = 0, \\ 0 & = -4(x + 2)^2 + 9 \\ -9 & = -4(x + 2)^2 \\ {-9 \over -4} & = (x + 2)^2 \\ {9 \over 4} & = (x + 2)^2 \\ \pm \sqrt{9 \over 4} & = x + 2 \\ \pm 1.5 & = x + 2 \\ -0.5 \text{ or } -3.5 & = x \\ \implies & x \text{-intercepts are } (-3.5, 0) \text{ and } (-0.5, 0) \end{align*}

(a)

\begin{align*} \text{Let } y = 0, \phantom{0} 0 & = 2(x - 1)(x + 2) \\ \\ x & = 1 \text{ or } - 2 \\ \implies & x \text{-intercepts are } (-2, 0) \text{ and } (1, 0) \\ \\ \text{Line of symmetry, } x & = {-2 + 1 \over 2} \\ & = -0.5 \\ \\ \text{Let } x = -0.5, \phantom{0} y & = 2(-0.5 - 1)(-0.5 + 2) \\ & = -4.5 \\ \implies & \text{Turning point is } (-0.5, -4.5) \end{align*}

(b)

\begin{align*} \text{Let } y = 0, \phantom{0} 0 & = -3(2x - 1)(3x - 2) \\ \\ 0 & = 2x -1 \phantom{0} \text{ or } \phantom{0} 0 = 3x - 2 \\ 1 & = 2x \phantom{10 or 000(.} 2 = 3x \\ \\ x & = {1 \over 2} \text{ or } {2 \over 3} \\ \implies & x \text{-intercepts are } \left( {1 \over 2}, 0 \right) \text{ and } \left( {2 \over 3}, 0 \right) \\ \\ \text{Line of symmetry, } x & = {{1 \over 2} + {2 \over 3} \over 2} \\ & = {7 \over 12} \\ \\ \text{Let } x = {7 \over 12}, \phantom{0} y & = -3 \left[ 2 \left(7 \over 12\right) -1 \right] \left[ 3 \left(7 \over 12\right) - 2 \right] \\ & = {1 \over 8} \\ \implies & \text{Turning point is } \left( {7 \over 12}, {1 \over 8} \right) \end{align*}

(c)

\begin{align*} \text{Let } y = 0, \phantom{0} 0 & = -(x + 2)(2x - 3) \\ \\ 0 & = x + 2 \phantom{0} \text{ or } \phantom{0} 0 = 2x - 3 \\ -2 & = x \phantom{000000-.} 3 = 2x \\ \\ x & = -2 \text{ or } 1.5 \\ \implies & x \text{-intercepts are } (-2, 0) \text{ and } (1.5, 0) \\ \\ \text{Line of symmetry, } x & = {-2 + 1.5 \over 2} \\ & = -0.25 \\ \\ \text{Let } x = -0.25, \phantom{0} y & = -(-0.25 + 2)[2(-0.25) -3] \\ & = 6.125 \\ \implies & \text{Turning point is } (-0.25, 6.125) \end{align*}

(d)

\begin{align*} \text{Let } y = 0, \phantom{0} 0 & = 3(x - 4)(2x - 5) \\ \\ 0 & = x - 4 \phantom{0} \text{ or } \phantom{0} 0 = 2x - 5 \\ 4 & = x \phantom{000000-.} 5 = 2x \\ \\ x & = 4 \text{ or } 2.5 \\ \implies & x \text{-intercepts are } (2.5, 0) \text{ and } (4, 0) \\ \\ \text{Line of symmetry, } x & = {2.5 + 4 \over 2} \\ & = 3.25 \\ \\ \text{Let } x = 3.25, \phantom{0} y & = 3(3.25 - 4)[2(3.25) - 5] \\ & = -3.375 \\ \implies & \text{Turning point is } (3.25, -3.375) \end{align*}

(e)

\begin{align*} \text{Let } y = 0, \phantom{0} 0 & = -2(2x + 5)(3 - x) \\ \\ 0 & = 2x + 5 \phantom{0} \text{ or } \phantom{0} 0 = 3 - x \\ -5 & = 2x \phantom{00000000} x = 3 \\ \\ x & = -2.5 \text{ or } 3 \\ \implies & x \text{-intercepts are } (-2.5, 0) \text{ and } (3, 0) \\ \\ \text{Line of symmetry, } x & = {-2.5 + 3 \over 2} \\ & = 0.25 \\ \\ \text{Let } x = 0.25, \phantom{0} y & = -2[2(0.25) + 5](3 - 0.25) \\ & = -30.25 \\ \implies & \text{Turning point is } (0.25, -30.25) \end{align*}

(f)

\begin{align*} \text{Let } y = 0, \phantom{0} 0 & = 2(3x - 4)(5 - x) \\ \\ 0 & = 3x - 4 \phantom{0} \text{ or } \phantom{0} 0 = 5 - x \\ 4 & = 3x \phantom{10 or 000(.} x = 5 \\ \\ x & = {4 \over 3} \text{ or } 5 \\ \implies & x \text{-intercepts are } \left( {4 \over 3}, 0 \right) \text{ and } (5, 0) \\ \\ \text{Line of symmetry, } x & = {{4 \over 3} + 5 \over 2} \\ & = {19 \over 6} \\ \\ \text{Let } x = {19 \over 6}, \phantom{0} y & = 2 \left[ 3 \left({19 \over 6}\right) - 4 \right] \left( 5 - {19 \over 6} \right) \\ & = 20{1 \over 6} \\ \implies & \text{Turning point is } \left( 3{1 \over 6}, 20{1 \over 6} \right) \end{align*}

(i)

\begin{align*} y & = -3x^2 - 2x + k \\ \\ [a = -3 &, \phantom{.} b = -2, c = k] \\ \\ b^2 - 4ac & = (-2)^2 - 4(-3)(k) \\ & = 4 + 12k \end{align*}
\begin{align*} b^2 - 4ac & < 0 \\ 4 + 12k & < 0 \\ 12k & < - 4 \\ k & < {-4 \over 12} \\ k & < -{1 \over 3} \end{align*}

(ii)

\begin{align*} b^2 - 4ac & = 0 \\ 4 + 12k & = 0 \\ 12k & = -4 \\ k & = {-4 \over 12} \\ & = -{1 \over 3} \end{align*}

(iii)

\begin{align*} b^2 - 4ac & > 0 \\ 4 + 12k & > 0 \\ 12k & > -4 \\ k & > {-4 \over 12} \\ k & > -{1 \over 3} \end{align*}

\begin{align*} \text{For } y = 2x^2 + bx + c, \phantom{0} \text{Discriminant} & = (b)^2 - 4(2)(c) \\ & = b^2 - 8c \\ \\ \text{For } y = 2x^2 - bx + c, \phantom{0} \text{Discriminant} & = (-b)^2 - 4(2)(c) \\ & = b^2 - 8c \\ \\ \text{Yes. Since both functions have the same} & \text{ discriminant, they have the same number of } x \text{-intercepts} \end{align*}

\begin{align*} y & = ax^2 + bx + c \\ \\ \text{Discriminant} & = b^2 - 4ac \\ \\ b^2 - 4ac & > 0 \phantom{00000} [\text{2 } x \text{-intercepts}] \\ -4ac & > -b^2 \\ 4ac & < b^2 \\ ac & < {b^2 \over 4} \text{ (Shown)} \end{align*}

(i)

\begin{align*} y & = 2x^2 + 2x - 1 \\ & = 2(x^2 + x) - 1 \\ & = 2 \left[ \left( x + {1 \over 2} \right)^2 - \left(1 \over 2\right)^2 \right] - 1 \\ & = 2 [ (x + 0.5)^2 - 0.25 ] - 1 \\ & = 2(x + 0.5)^2 - 0.5 - 1 \\ & = 2(x + 0.5)^2 - 1.5 \\ \\ \implies & \text{Turning point is } (-0.5, -1.5) \\ \\ \implies & \text{Line of symmetry is } x = -0.5 \end{align*}

(ii)

\begin{align*} 2x^2 + 2x - 1 & = 0 \\ 2(x + 0.5)^2 - 1.5 & = 0 \\ 2(x + 0.5)^2 & = 1.5 \\ (x + 0.5)^2 & = {1.5 \over 2} \\ (x + 0.5)^2 & = {3 \over 4} \\ x + 0.5 & = \pm \sqrt{3 \over 4} \\ x + {1 \over 2} & = \pm {\sqrt{3} \over 2} \\ x & = \pm {\sqrt{3} \over 2} - {1 \over 2} \\ x & = {\sqrt{3} - 1 \over 2} \text{ or } {-\sqrt{3} - 1 \over 2} \end{align*}

(i)

\begin{align*} y & = -2x^2 + 4x + 3 \\ & = -2(x^2 - 2x) + 3 \\ & = -2 \left[ \left( x - {2 \over 2} \right)^2 - \left(2 \over 2\right)^2 \right] + 3 \\ & = -2 [ (x - 1)^2 - 1] + 3 \\ & = -2(x - 1)^2 + 2 + 3 \\ & = -2(x - 1)^2 + 5 \\ \\ \implies & \text{Turning point is } (1, 5) \\ \\ \text{Let } x = 0, \phantom{.} y & = -2(0 - 1)^2 + 5 \\ & = 3 \\ \implies & y \text{-intercept is } (0, 3) \\ \\ \text{Let } y = 0, \phantom{.} 0 & = -2(x - 1)^2 + 5 \\ 2(x - 1)^2 & = 5 \\ (x - 1)^2 & = {5 \over 2} \\ x - 1 & = \pm \sqrt{5 \over 2} \\ x & = \pm \sqrt{5 \over 2} + 1 \\ x & = 2.5811 \text{ or } -0.58114 \\ x & \approx 2.58 \text{ or } -0.581 \\ \implies & x \text{-intercepts are } (2.58, 0) \text{ and } (-0.581, 0) \end{align*}

(ii)

\begin{align*} -2x^2 + 4x & = -3 \\ -2x^2 + 4x + 3 & = 0 \\ \\ \text{Since the graph of } y = -2x^2 + 4x + 3 & \text{ has two }x \text{-intercepts, the equation has two distinct real roots} \end{align*}

(a)

\begin{align*} px^2 - 6x & + p = 0 \\ \\ [a = p, b & = - 6, c = p] \\ \\ b^2 - 4ac & = 0 \phantom{00000} [\text{Equal real roots}] \\ (-6)^2 - 4(p)(p) & = 0 \\ 36 - 4p^2 & = 0 \\ -4p^2 & = -36 \\ p^2 & = {-36 \over -4} \\ & = 9 \\ p & = \pm \sqrt{9} \\ & = \pm 3 \end{align*}

(b)

\begin{align*} 3x^2 + 2x & - p = 0 \\ \\ [a = 3, b & = 2, c = -p] \\ \\ b^2 - 4ac & > 0 \phantom{00000} [\text{Two distinct roots}] \\ (2)^2 - 4(3)(-p) & > 0 \\ 4 + 12p & > 0 \\ 12p & > -4 \\ p & > {-4 \over 12} \\ p & > -{1 \over 3} \end{align*}

(c)

\begin{align*} 2x^2 + 3x & + 2p = 0 \\ \\ [a = 2, b & = 3, c = 2p] \\ \\ b^2 - 4ac & \ge 0 \phantom{00000} [\text{Two distinct or equal roots}] \\ (3)^2 - 4(2)(2p) & \ge 0 \\ 9 - 16p & \ge 0 \\ -16p & \ge -9 \\ p & \le {-9 \over -16} \\ p & \le {9 \over 16} \end{align*}

(d)

\begin{align*} px^2 - x - 4 & = 0 \\ \\ [a = p, b & = -1, c = - 4] \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ (-1)^2 - 4(p)(-4) & < 0 \\ 1 + 16p & < 0 \\ 16p & < -1 \\ p & < -{1 \over 16} \end{align*}

(a) Always positive implies that the entire graph is above the x-axis and does not intersect the x-axis

\begin{align*} y & = x^2 + 2x + (k + 1) \\ \\ [a & = 1, b = 2, c = k + 1] \\ \\ \text{Since } a > 0, & \text{ the graph is a minimum curve } (\cup) \\ \\ b^2 - 4ac & = (2)^2 - 4(1)(k + 1) \\ & = 4 - 4(k + 1) \\ & = 4 - 4k - 4 \\ & = -4k \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ -4k & < 0 \\ k & > {0 \over -4} \\ k & > 0 \end{align*}

(b) Always negative implies that the entire graph is below the x-axis and does not intersect the x-axis

\begin{align*} & -x^2 + 4x + k \\ \\ [a & = -1, b = 4, c = k] \\ \\ \text{Since } a < 0, & \text{ the graph is a maximum curve } (\cap) \\ \\ b^2 - 4ac & = (4)^2 - 4(-1)(k) \\ & = 16 - 4(-k) \\ & = 16 + 4k \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ 16 + 4k & < 0 \\ 4k & < -16 \\ k & < {-16 \over 4} \\ k & < -4 \end{align*}

(a) Always positive implies that the entire graph is above the x-axis and does not intersect the x-axis

\begin{align*} & 2x^2 + 2x + k \\ \\ [a & = 2, b = 2, c = k] \\ \\ \text{Since } a > 0, & \text{ the graph is a minimum curve } (\cup) \\ \\ b^2 - 4ac & = (2)^2 - 4(2)(k) \\ & = 4 - 8k \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ 4 - 8k & < 0 \\ -8k & < - 4 \\ k & > {-4 \over -8} \\ k & > {1 \over 2} \end{align*}

(b) Always negative implies that the entire graph is below the x-axis and does not intersect the x-axis

\begin{align*} & -3x^2 + 6x + (k - 1) \\ \\ [a & = -3, b = 6, c = k - 1] \\ \\ \text{Since } a < 0, & \text{ the graph is a maximum curve } (\cap) \\ \\ b^2 - 4ac & = (6)^2 - 4(-3)(k - 1) \\ & = 36 - 4(-3k + 3) \\ & = 36 + 12k - 12 \\ & = 24 + 12k \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ 24 + 12k & < 0 \\ 12k & < - 24 \\ k & < {-24 \over 12} \\ k & < -2 \end{align*}

(c) If the graph lies entirely above the x-axis, the graph does not intersect the x-axis

\begin{align*} y & = 2x^2 + x - 2k \\ \\ [a & = 2, b = 1, c = -2k] \\ \\ \text{Since } a > 0, & \text{ the graph is a minimum curve } (\cup) \\ \\ b^2 - 4ac & = (1)^2 - 4(2)(-2k) \\ & = 1 -4(-4k) \\ & = 1 + 16k \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ 1 + 16k & < 0 \\ 16k & < -1 \\ k & < -{1 \over 16} \end{align*}

The graph can intersect the x-axis once or twice, thus b² - 4ac ≥ 0

\begin{align*} y & = 3x^2 - 2x + c - 1 \\ \\ [a & = 3, b = -2, 'c' = c - 1] \\ \\ \text{Discriminant} & = (-2)^2 - 4(3)(c - 1) \\ & = 4 - 4(3c - 3) \\ & = 4 - 12c + 12 \\ & = 16 - 12c \\ \\ b^2 - 4ac & \ge 0 \phantom{00000} [\text{Real roots}] \\ 16 - 12c & \ge 0 \\ -12c & \ge -16 \\ c & \le {-16 \over -12} \\ c & \le {4 \over 3} \end{align*}

\begin{align*} \text{When } k = -5, \phantom{.} y & = 2x^2 + 4x - (-5) + 1 \\ y & = 2x^2 +4x + 6 \\ \\ b^2 - 4ac & = (4)^2 - 4(2)(6) \\ & = -32 \\ \\ \text{Since } b^2 - 4ac < 0, & \text{ the graph of } y = 2x^2 + 4x + 6 \text{ has no } x \text{-intercepts} \end{align*}

(i)

\begin{align*} 2x^2 + 2x & = 2(x^2 + x) \\ & = 2 \left[ \left( x + {1 \over 2} \right)^2 - \left( 1 \over 2 \right)^2 \right] \\ & = 2 [ (x + 0.5)^2 - 0.25 ] \\ & = 2(x + 0.5)^2 - 0.5 \\ \\ \text{For all real} & \text{ values of } x, \\ (x + 0.5)^2 & \ge 0 \\ 2(x + 0.5)^2 & \ge 0 \\ 2(x + 0.5)^2 - 0.5 & \ge -0.5 \\ \\ \therefore 2(x + 0.5)^2 - 0.5 & > -4 \\ \therefore 2x^2 + 2x & > -4 \text{ for all real values of } x \end{align*}

(ii)

\begin{align*} \text{Since } 2x^2 + 2x & > -4 \text{ for all real values of } x, \\ \\ 2x^2 + 2x + 4 & > -4 + 4 \\ 2x^2 + 2x + 4 & > 0 \\ \\ \text{Since the graph of } y & = 2x^2 + 2x + 4 \text{ is always positive, the graph has no } x\text{-intercepts} \end{align*}

(a)

\begin{align*} y & = (x + 1)(2x - 1) - (p - 2) \\ & = 2x^2 - x + 2x - 1 - p + 2 \\ & = 2x^2 + x + 1 - p \\ \\ b^2 - 4ac & = (1)^2 - 4(2)(1 - p) \\ & = 1 - 4(2 - 2p) \\ & = 1 - 8 + 8p \\ & = 8p - 7 \\ \\ 8p - 7 & > 0 \\ 8p & > 7 \\ p & > {7 \over 8} \end{align*}

(b)

\begin{align*} y & = p(x + 1)(x - 3) - x + 4p + 2 \\ & = p(x^2 - 3x + x - 3) - x + 4p + 2 \\ & = p(x^2 - 2x - 3) - x + 4p + 2 \\ & = px^2 - 2px - 3p - x + 4p + 2 \\ & = px^2 - 2px - x + p + 2 \\ & = px^2 + (-2p - 1)x + (p + 2) \\ \\ b^2 - 4ac & = (-2p - 1)^2 - 4(p)(p + 2) \\ & = (-2p)^2 - 2(-2p)(1) + (1)^2 - 4p(p + 2) \\ & = 4p^2 + 4p + 1 - 4p^2 - 8p \\ & = 1 - 4p \\ \\ b^2 - 4ac & < 0 \\ 1 - 4p & < 0 \\ -4p & < - 1 \\ p & > {-1 \over -4} \\ p & > {1 \over 4} \end{align*}

The graph can intersect the x-axis once or twice, thus b² - 4ac ≥ 0

\begin{align*} y & = x^2 - 2kx + k^2 - (3 + x) \\ & = x^2 - 2kx + k^2 - 3 - x \\ & = x^2 - 2kx - x + k^2 - 3 \\ & = x^2 + (-2k - 1)x + (k^2 - 3) \\ \\ b^2 - 4ac & = (-2k - 1)^2 - 4(1)(k^2 - 3) \\ & = (-2k)^2 - 2(-2k)(1) + (1)^2 - 4(k^2 - 3) \\ & = 4k^2 + 4k + 1 - 4k^2 + 12 \\ & = 4k + 13 \\ \\ b^2 - 4ac & \ge 0 \phantom{00000} [\text{Real roots}] \\ 4k + 13 & \ge 0 \\ 4k & \ge -13 \\ k & \ge -{13 \over 4} \\ k & \ge -3{1 \over 4} \\ \\ \text{Least value of } k & = -3{1 \over 4} \end{align*}

(a) > 0 means the expression is always positive

\begin{align*} & x^2 - 10x + (4 - m) \\ \\ \text{Since } a > 0, & \text{ the graph is a minimum curve } (\cup) \\ \\ b^2 - 4ac & = (-10)^2 - 4(1)(4 - m) \\ & = 100 - 4(4 -m ) \\ & = 100 - 16 + 4m \\ & = 84 + 4m \\ \\ b^2 - 4ac & < 0 \\ 84 + 4m & < 0 \\ 4m & < -84 \\ m & < {-84 \over 4} \\ m & < -21 \end{align*}

(b) < 0 means the expression is always negative

\begin{align*} - x^2 + 4kx - 4(k - 1)(k + 2) & = -x^2 + 4kx - 4(k^2 + 2k - k - 2) \\ & = -x^2 + 4kx - 4(k^2 + k - 2) \\ & = -x^2 + 4kx - 4k^2 - 4k + 8 \\ & = -x^2 + 4kx + (-4k^2 - 4k + 8) \\ \\ \text{Since } a < 0, & \text{ the graph is a maximum curve } (\cap) \\ \\ b^2 - 4ac & = (4k)^2 - 4(-1)(-4k^2 - 4k + 8) \\ & = 16k^2 + 4(-4k^2 -4k + 8) \\ & = 16k^2 - 16k^2 - 16k + 32 \\ & = -16k + 32 \\ \\ b^2 - 4ac & < 0 \\ -16k + 32 & < 0 \\ -16k & < -32 \\ k & > {-32 \over -16} \\ k & > 2 \end{align*}

(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 are } (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 \phantom{00000} [\text{Reject } x = 0 \text{, since the ride has started}] \\ \\ \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{Coordinates} & \text{ of maximum point are } (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 are } (2, 1) \end{align*}

(ii) From (i), the y-intercept is (0, 2) and the coordinates of the minimum point are (2, 1). Recall that y denotes the height of the ramp above the ground (in metres). So the top of the ramp is 2 m above the ground and its lowest point is 1 m above the ground.

\begin{align*} \text{Maximum depth} & = 2 - 1 \\ & = 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 - 0 \\ & = 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 are } \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 & = 6.38516 \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 are \(\left(1, {29 \over 16}\right)\). 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 x-intercepts are (6.39, 0) and (-4.39, 0). Recall that x denotes the horizontal distance travelled by the shot, so we reject x = -4.39 since distance cannot be negative.

\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 are } (1, 4) \end{align*}

(iii) From (ii), the coordinates of the maximum point are (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 t-intercept are (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 \end{align*}
The graph of h against t resembles a maximum curve (∩). Thus, the ball is rising when t = 0.258 s and falling when t = 1.74 s. \begin{align*} \therefore \text{Earliest time} & = 1.74 \text{ s} \end{align*}

(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 are } (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 are } (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 its path has the smallest width at ground level. Using the x-intercepts from (i): \begin{align*} \text{Width of path of } A & = 2\sqrt{10} \approx 6.32 \text{ m} \\ \text{Width of path of } B & = 2(3) = 6 \text{ m} \\ \text{Width of path of } C & = 2\sqrt{30} \approx 11.0 \text{ m} \end{align*}

(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 \phantom{00000} [\text{Real roots, since the water reaches a height of 10 m}] \\ 2k - 16 & \ge 0 \\ 2k & \ge 16 \\ k & \ge {16 \over 2} \\ k & \ge 8 \end{align*}

(i)

\begin{align*} d & = 0.002m^2 + 0.02m + 0.005 \\ \\ \text{When } & m = 200, \\ d & = 0.002(200)^2 + 0.02(200) + 0.005 \\ d & = 84.005 \\ \\ \text{Deflection of plastic rod} & = 84.005 \text{ mm} \end{align*}

(ii)

\begin{align*} d & = 0.002m^2 + 0.02m + 0.005 \\ \\ 100 \text{ kg} & = (100 \times 1000) \text{ g} \\ & = 100\,000 \text{ g} \\ \\ \text{When } & m = 100\,000, \\ d & = 0.002(100\,000)^2 + 0.02(100\,000) + 0.005 \\ d & = 20 \phantom{.} 002 \phantom{.} 000.005 \text{ mm} \\ & \approx 20 \text{ km} \\ \\ \text{Since deflection is} & \text{ too large, the model is not suitable for a load of 100 kg} \end{align*}

(i) I think there's a typo in the question. It should be 'the highest point on the curve' (instead of lowest point) since this curve is a maximum curve (∩)

\begin{align*} y & = -2x^2 +4x + 3 \\ y & = -2(x^2 - 2x) + 3 \\ y & = -2 \left[ \left( x - {2 \over 2} \right)^2 - \left( 2 \over 2 \right)^2 \right] + 3 \\ y & = -2 [ (x - 1)^2 - 1 ] + 3 \\ y & = -2(x - 1)^2 + 2 + 3 \\ y & = -2(x - 1)^2 + 5 \\ \\ \implies & \text{Coordinates of maximum point are } (1, 5) \text{ (Shown)} \end{align*}

(ii)

\begin{align*} y & = -2(x - 1)^2 + 5 \\ \\ \text{Let } & y = 0, \\ 0 & = -2(x - 1)^2 + 5 \\ 2(x - 1)^2 & = 5 \\ (x - 1)^2 & = {5 \over 2} \\ (x - 1)^2 & = 2.5 \\ x - 1 & = \pm \sqrt{2.5} \\ x & = \sqrt{2.5} + 1 \text{ or } -\sqrt{2.5} + 1 \\ x & = 2.5811 \text{ or } -0.58114 \\ x & \approx 2.58 \text{ or } -0.581 \end{align*}

(i)

\begin{align*} y & = x^2 + 2x + (8 - p) \\ \\ [a & = 1, b = 2, c = 8 - p] \\ \\ \text{Since } & a > 0, \text{ curve is a minimum curve } (\cup) \\ \\ b^2 - 4ac & = (2)^2 - 4(1)(8 - p) \\ & = 4 - 4(8 - p) \\ & = 4 - 32 + 4p \\ & = 4p - 28 \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ 4p - 28 & < 0 \\ 4p & < 28 \\ p & < {28 \over 4} \\ p & < 7 \end{align*}

(ii)

\begin{align*} y & = x^2 + (2)x + 8 - (2) \\ y & = x^2 + 2x + 6 \\ y & = \left(x + {2 \over 2} \right)^2 - \left(2 \over 2\right)^2 + 6 \\ y & = (x + 1)^2 - 1 + 6 \\ y & = (x + 1)^2 + 5 \\ \\ \implies & \text{Coordinates of turning point are } (-1, 5) \\ \\ \text{Let } & x = 0, \\ y & = (0 + 1)^2 + 5 \\ y & = 6 \\ \implies & y\text{-intercept is } (0, 6) \end{align*}

(i)

\begin{align*} \text{Condition 1: } \phantom{0} a & > 0 \phantom{00000} [\text{Minimum curve } (\cup)] \\ \\ \text{Discriminant} & = (-4)^2 - 4(a)(-c) \\ & = 16 + 4ac \\ \\ \text{Discriminant} & < 0 \phantom{00000} [\text{No real roots}] \\ 16 + 4ac & < 0 \\ 4ac & < -16 \\ ac & < {-16 \over 4} \\ \text{Condition 2: } \phantom{0} ac & < - 4 \end{align*}

(ii) Note that there are many possible answers - ensure that a > 0 and the value of ac is smaller than -4

\begin{align*} a & = 2, c = -3 \\ \\ a & = 10, c = -1 \\ \\ & \text{(and so on...)} \end{align*}

(i)

\begin{align*} h & = t^2 - 12t + 40 \\ h & = \left( t - {12 \over 2} \right)^2 - \left(12 \over 2\right)^2 + 40 \\ h & = (t - 6)^2 - 36 + 40 \\ h & = (t - 6)^2 + 4 \\ \\ \implies \text{Coordinates }& \text{of minimum point are } (6, 4) \\ \\ \text{Minimum height} & = 4 \text{ m} \\ \\ \therefore \text{Capsule } & \text{cannot reach height of 2m} \end{align*}

(ii) The coordinates of the minimum point are (6, 4). In the context of the question, the capsule reaches its lowest height of 4 m when t = 6 seconds.

\begin{align*} \text{Time taken} & = 6 \text{ seconds} \end{align*}

(i) Let the function be \(y = ax^2 + bx + c\). Substitute three points from the table to find \(a\), \(b\) and \(c\), then check the function using the remaining three points.

\begin{align*} y & = ax^2 + bx + c \\ \\ \text{Using } & (0, 1), \\ 1 & = a(0)^2 + b(0) + c \\ 1 & = c \\ \\ y & = ax^2 + bx + 1 \\ \\ \text{Using } & (2, 7), \\ 7 & = a(2)^2 + b(2) + 1 \\ 7 & = 4a + 2b + 1 \\ 6 & = 4a + 2b \\ 3 & = 2a + b \\ \\ b & = 3 - 2a \phantom{000} \text{ --- (1)} \\ \\ \text{Using } & (4, 17), \\ 17 & = a(4)^2 + b(4) + 1 \\ 17 & = 16a + 4b + 1 \\ 16 & = 16a + 4b \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & (1) \text{ into } (2), \\ 16 & = 16a + 4(3 - 2a) \\ 16 & = 16a + 12 - 8a \\ 4 & = 8a \\ {4 \over 8} & = a \\ {1 \over 2} & = a \\ \\ \text{Substitute } & a = {1 \over 2} \text{ into } (1), \\ b & = 3 - 2\left(1 \over 2\right) \\ b & = 2 \\ \\ \therefore y & = {1 \over 2}x^2 + 2x + 1 \\ \\ \text{Check with } & \text{the remaining three points:} \\ \text{When } x = 1, \phantom{.} y & = {1 \over 2}(1)^2 + 2(1) + 1 = 3.5 \\ \text{When } x = 3, \phantom{.} y & = {1 \over 2}(3)^2 + 2(3) + 1 = 11.5 \\ \text{When } x = 5, \phantom{.} y & = {1 \over 2}(5)^2 + 2(5) + 1 = 23.5 \\ \\ \text{These values } & \text{agree with the table} \end{align*}

(ii)

\begin{align*} y & = {1 \over 2}x^2 + 2x + 1 \\ \\ \text{Let } & y = 0, \\ 0 & = {1 \over 2}x^2 + 2x + 1 \\ 0 & = x^2 + 4x + 2 \\ 0 & = \left(x + {4 \over 2} \right)^2 - \left(4 \over 2\right)^2 + 2 \\ 0 & = (x + 2)^2 - 4 + 2 \\ 0 & = (x + 2)^2 - 2 \\ 2 & = (x + 2)^2 \\ \pm \sqrt{2} & = x + 2 \\ \\ x & = \sqrt{2} - 2 \text{ or } -\sqrt{2} - 2 \\ \\ \text{Since the graph} & \text{ of the function is a minimum curve } (\cup), \\ \\ \text{for } y < 0, & \phantom{.} -\sqrt{2} - 2 < x < \sqrt{2} - 2 \end{align*}

(i)

\begin{align*} y & = 2x^2 - 10x + m \\ y & = 2(x^2 - 5x) + m \\ y & = 2 \left[ \left( x - {5 \over 2} \right)^2 - \left(5 \over 2\right)^2 \right] + m \\ y & = 2 [ (x - 2.5)^2 - 6.25] + m \\ y & = 2(x - 2.5)^2 - 12.5 + m \\ y & = 2(x - 2.5)^2 + (m - 12.5) \end{align*}

(ii)

\begin{align*} y & = 2(x - 2.5)^2 + (m - 12.5) \\ \\ \implies & \text{Coordinates of minimum point are } (2.5, m - 12.5) \\ \\ m - 12.5 & = -{19 \over 2} \\ m & = -{19 \over 2} + 12.5 \\ m & = 3 \end{align*}

(iii)

\begin{align*} y & = 2x^2 - 10x + m \\ \\ b^2 - 4ac & = (-10)^2 - 4(2)(m) \\ & = 100 - 8m \\ \\ b^2 - 4ac & > 0 \phantom{00000} [\text{2 real and distinct roots}] \\ 100 - 8m & > 0 \\ -8m & > -100 \\ 8m & < 100 \\ m & < {100 \over 8} \\ m & < 12.5 \end{align*}

(i)

\begin{align*} y & = ax^2 - x + c \\ \\ \text{Using } & (-2, -3), \\ -3 & = a(-2)^2 - (-2) + c \\ -3 & = a(4) + 2 + c \\ -3 & = 4a + 2 + c \\ -5 & = 4a + c \\ \\ c & = -5 - 4a \phantom{000} \text{ --- (1)} \\ \\ \text{Using } & (1, 3), \\ 3 & = a(1)^2 - (1) + c \\ 3 & = a(1) - 1 + c \\ 3 & = a - 1 + c \\ 4 & = a + c \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute} & \text{ (1) into (2),} \\ \\ 4 & = a + (-5 - 4a) \\ 4 & = a - 5 - 4a \\ 4 + 5 & = a - 4a \\ 9 & = -3a \\ {9 \over -3} & = a \\ -3 & = a \\ \\ \text{Substitute } & a = -3 \text{ into (1),} \\ \\ c & = -5 - 4(-3) \\ c & = 7 \end{align*}

(ii)

\begin{align*} y & = -3x^2 - x + 7 \\ y & = -3 \left( x^2 + {1 \over 3}x \right) + 7 \\ y & = -3 \left[ \left(x + {1 \over 6} \right)^2 - \left(1 \over 6\right)^2 \right] + 7 \\ y & = -3 \left[ \left(x + {1 \over 6} \right)^2 - {1 \over 36} \right] + 7 \\ y & = -3 \left( x + {1 \over 6} \right)^2 + {1 \over 12} + 7 \\ y & = -3 \left( x + {1 \over 6} \right)^2 + 7{1 \over 12} \\ \\ \implies & \text{Coordinates of maximum point are } \left(-{1 \over 6}, 7{1 \over 12} \right) \\ \\ \text{Since maximum} & \text{ value of } y = 7{1 \over 12}, \text{ the value of } y \text{ can never exceed } 8 \end{align*}

(i)

\begin{align*} h & = 4t - t^2 + 12 \\ \\ \text{When } & t =3, \\ h & = 4(3) - (3)^2 + 12 \\ h & = 15 \\ \\ \text{Height of ball} & = 15 \text{ m} \end{align*}

(ii)

\begin{align*} h & = 4t - t^2 + 12 \\ h & = -t^2 + 4t + 12 \\ h & = -(t^2 - 4t) + 12 \\ h & = - \left[ \left( t - {4 \over 2} \right)^2 - \left(4 \over 2\right)^2 \right] + 12 \\ h & = - [ (t - 2)^2 - 4] + 12 \\ h & = - (t - 2)^2 + 4 + 12 \\ h & = - (t - 2)^2 + 16 \\ \\ \implies & \text{Coordinates of maximum point are } (2, 16) \\ \\ \therefore \text{Maximum height} & = 16 \text{ m} \end{align*}

(iii)

\begin{align*} h & = - (t - 2)^2 + 16 \\ \\ \text{When } & h = 0, \\ 0 & = -(t - 2)^2 + 16 \\ (t - 2)^2 & = 16 \\ t - 2 & = \pm \sqrt{16} \\ t - 2 & = \pm 4 \\ t & = 4 + 2 \text{ or } - 4 + 2 \\ t & = 6 \text{ or } -2 \end{align*}

(iv)

We reject t = -2, since time cannot be negative. Thus, t = 6 means the ball reaches the ground 6 seconds after it is thrown.

(i)

\begin{align*} 2x + 2y & = 16 \\ 2y & = 16 - 2x \\ y & = 8 - x \end{align*}

(ii)

\begin{align*} A & = xy \\ \\ \text{Since } & y = 8 - x, \\ A & = x(8 - x) \end{align*}

(iii)

\begin{align*} A & = x(8 - x) \\ A & = 8x - x^2 \\ A & = -x^2 + 8x \\ A & = -(x^2 - 8x) \\ A & = - \left[ \left( x - {8 \over 2} \right)^2 - \left(8 \over 2\right)^2 \right] \\ A & = - [ (x - 4)^2 - 16] \\ A & = - (x - 4)^2 + 16 \\ \\ \implies & \text{Coordinates of (maximum) turning point are } (4, 16) \\ \\ \therefore \text{Greatest area} & = 16 \text{ cm}^2 \text{ when } x = 4 \\ \\ \text{When } x = 4, \phantom{.} y & = 8 - (4) \\ y & = 4 \\ \\ \therefore \text{Greatest area} & \text{ occurs when } x = y \end{align*}

\begin{align*} \text{Line of symmetry, } x & = {0 + 180 \over 2} \\ x & = 90 \\ \\ \implies \text{Coordinates of mini} & \text{mum point are } (90, 30) \\ \\ y & = a(x - h)^2 + k \\ y & = a(x - 90)^2 + 30 \\ \\ \text{Using } & (0, 75), \\ 75 & = a(0 - 90)^2 + 30 \\ 75 & = a(8100) + 30 \\ 75 & = 8100a + 30 \\ 75 - 30 & = 8100a \\ 45 & = 8100a \\ {45 \over 8100} & = a \\ {1 \over 180} & = a \\ \\ \therefore y & = {1 \over 180}(x - 90)^2 + 30 \end{align*}

(a)

\begin{align*} \text{Condition 1: } \phantom{0} a & > 0 \phantom{00000} [\text{Minimum curve } (\cup)] \\ \\ b^2 - 4ac & = (4)^2 - 4(a)(6) \\ & = 16 - 24a \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ 16 - 24a & < 0 \\ -24a & < -16 \\ 24a & > 16 \\ a & > {16 \over 24} \\ \text{Condition 2: } \phantom{0} a & > {2 \over 3} \\ \\ \\ \therefore \text{Smallest integer value of } a & = 1 \end{align*}

(b)

\begin{align*} \text{Condition 1 met: } \phantom{0} a & = -6 \phantom{00000} [\text{Maximum curve } (\cap)] \\ \\ b^2 - 4ac & = (b)^2 - 4(-6)(-3) \\ & = b^2 - 72 \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No real roots}] \\ b^2 - 72 & < 0 \\ b^2 & < 72 \text{ (Shown)} \end{align*}