(a)

\begin{align*} y & = 6(x + 1)(x - 3) \phantom{000} [\text{Minimum curve } (\cup)] \\ \\ \text{Let } & y = 0, \\ 0 & = 6(x + 1)(x - 3) \\ \\ x + 1 = 0 \phantom{0.} & \text{ or } \phantom{0} x - 3 = 0 \\ x = -1 \phantom{} & \phantom{0000000} x = 3 \\ \\ \text{Line of symmetry, } x & = {-1 + 3 \over 2} \\ x & = 1 \\ \\ \text{When } & x = 1, \\ y & = 6(1 + 1)(1 - 3) \\ y & = -24 \\ \implies & \text{Coordinates of minimum point is } (1, -24) \\ \\ \therefore \text{Minimum value} & = -24 \end{align*}

(b)

\begin{align*} y & = -{1 \over 2}(x + 5)(x + 2) \phantom{000} [\text{Maximum curve } (\cap)] \\ \\ \text{Let } & y = 0, \\ 0 & = -{1 \over 2}(x + 5)(x + 2) \\ \\ x + 5 = 0 \phantom{0.} & \text{ or } \phantom{0} x +2 = 0 \\ x = -5 \phantom{} & \phantom{0000000} x = -2 \\ \\ \text{Line of symmetry, } x & = {-5 + (-2) \over 2} \\ x & = -3.5 \\ \\ \text{When } & x = -3.5, \\ y & = -{1 \over 2}[(-3.5) + 5][(-3.5) + 2] \\ y & = 1.125 \\ \implies & \text{Coordinates of maximum point is } (-3.5, 1.125) \\ \\ \therefore \text{Maximum value} & = 1.125 \end{align*}

(c)

\begin{align*} y & = (8x - 9)(x - 4) \phantom{000} [\text{Minimum curve } (\cup)] \\ \\ \text{Let } & y = 0, \\ 0 & = (8x - 9)(x - 4) \\ \\ 8x - 9 = 0 \phantom{0.} & \text{ or } \phantom{0} x - 4= 0 \\ 8x = 9 \phantom{0.} & \phantom{0000000} x = 4 \\ x = {9 \over 8} \phantom{0} & \\ \\ \text{Line of symmetry, } x & = {{9 \over 8} + 4 \over 2} \\ x & = {41 \over 16} \\ \\ \text{When } & x = {41 \over 16}, \\ y & = \left[ 8 \left(41 \over 16\right) - 9 \right] \left( {41 \over 16} - 4 \right) \\ y & = -{529 \over 32} \\ \implies & \text{Coordinates of minimum point is } \left({41 \over 16}, -{529 \over 32} \right) \\ \\ \therefore \text{Minimum value} & = -{529 \over 32} \end{align*}

(d)

\begin{align*} y & = (7 - 2x)(x + 3) \phantom{000} [\text{Maximum curve } (\cap)] \\ \\ \text{Let } & y = 0, \\ 0 & = (7 - 2x)(x + 3) \\ \\ 7 - 2x = 0 \phantom{0.} & \text{ or } \phantom{0} x + 3 = 0 \\ 7 = 2x \phantom{.} & \phantom{0000000} x = -3 \\ {7 \over 2} = x \phantom{0.} & \\ \\ \text{Line of symmetry, } x & = {{7 \over 2} + (-3) \over 2} \\ x & = {1 \over 4} \\ \\ \text{When } & x = {1 \over 4}, \\ y & = \left[ 7 - 2\left(1 \over 4\right) \right] \left( {1 \over 4} + 3 \right) \\ y & = {169 \over 8} \\ \implies & \text{Coordinates of maximum point is } \left({1 \over 4}, {169 \over 8} \right) \\ \\ \therefore \text{Maximum value} & = {169 \over 8} \end{align*}

You can use the formula below to complete the square: $$ x^2 \pm bx + c = \left(x \pm {b \over 2}\right)^2 - \left(b \over 2\right)^2 + c $$

(a)

\begin{align*} x^2 + 10x & = \left(x + {10 \over 2}\right)^2 - \left(10 \over 2\right)^2 \\ & = (x + 5)^2 - 25 \end{align*}

(b)

\begin{align*} x^2 + 15x & = \left(x + {15 \over 2}\right)^2 - \left(15 \over 2\right)^2 \\ & = \left(x + {15 \over 2} \right)^2 - {225 \over 4} \end{align*}

(c)

\begin{align*} x^2 - 8x & = \left(x - {8 \over 2}\right)^2 - \left(8 \over 2\right)^2 \\ & = (x - 4)^2 - 16 \end{align*}

(d)

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

\begin{align*} \text{For all real} & \text{ of values of } x, \\ (x - 1)^2 & \ge 0 \\ 4(x - 1)^2 & \ge 0 \\ 4(x - 1)^2 - 25 & \ge -25 \end{align*}

\begin{align*} \text{For all real} & \text{ of values of } x, \\ (x + 6)^2 & \ge 0 \\ -{2 \over 3}(x + 6)^2 & \le 0 \\ -{2 \over 3}(x + 6)^2 + 10 & \le 10 \end{align*}

Quadratic function in complete the square form: $$ y = a(x - h)^2 + k $$

In this form, the coordinates of the maximum/minimum point is $(h, k)$. Thus the maximum/minimum value is $k$

(a)

\begin{align*} \text{Coordinates of} & \text{ minimum point is } (-1, -8) \\ \\ \text{Minimum value} & = -8 \end{align*}

(b)

\begin{align*} \text{Coordinates of} & \text{ maximum point is } (4, 10) \\ \\ \text{Maximum value} & = 10 \end{align*}

(c)

\begin{align*} y & = {1 \over 4}(2 - x)^2 - 7 \\ y & = {1 \over 4}[-(x - 2)]^2 - 7 \\ y & = {1 \over 4}(-1)^2(x - 2)^2 - 7 \\ y & = {1 \over 4}(x - 2)^2 - 7 \\ \\ \text{Coordinates of} & \text{ minimum point is } (2, -7) \\ \\ \text{Minimum value} & = -7 \end{align*}

(d)

\begin{align*} \text{Coordinates of} & \text{ minimum point is } (0.6, 1.5) \\ \\ \text{Minimum value} & = 1.5 \end{align*}

Quadratic function in complete the square form: $$ y = a(x - h)^2 + k $$

In this form, the coordinates of the maximum/minimum point is $(h, k)$. Thus the maximum/minimum value is $k$

(a)

\begin{align*} \text{Coordinates of} & \text{ minimum point is } (5, 14) \\ \\ \text{Minimum value} & = 14 \end{align*}

(b)

\begin{align*} \text{Coordinates of} & \text{ maximum point is } (-3, -21) \\ \\ \text{Maximum value} & = -21 \end{align*}

(c)

\begin{align*} \text{Coordinates of} & \text{ minimum point is } \left( -7, {9 \over 16} \right) \\ \\ \text{Minimum value} & = {9 \over 16} \end{align*}

(d)

\begin{align*} f(x) & = -0.8 - 5(1 - x)^2 \\ f(x) & = -5(1 - x)^2 - 0.8 \\ f(x) & = -5[ -(x - 1)]^2 - 0.8 \\ f(x) & = -5(-1)^2 (x - 1)^2 - 0.8 \\ f(x) & = -5(x - 1)^2 - 0.8 \\ \\ \text{Coordinates of} & \text{ maximum point is }(1, -0.8) \\ \\ \text{Maximum value} & = -0.8 \end{align*}

\begin{align*} y & = -0.125(x - 4)^2 + 2 \\ \\ \text{Coordinates of} & \text{ maximum point is } (4, 2) \\ \\ \therefore \text{Greatest height} & = 2 \text{ m} \\ \text{Corres. horizontal distance} & = 4 \text{ m} \end{align*}

(a)

\begin{align*} y & = 6x^2 - x - 15 \\ y & = 6 \left( x^2 - {1 \over 6}x \right) - 15 \\ y & = 6 \left[ \left( x - {1 \over 12} \right)^2 - \left(1 \over 12\right)^2 \right] - 15 \\ y & = 6 \left[ \left(x - {1 \over 12} \right)^2 - {1 \over 144} \right] - 15 \\ y & = 6 \left(x - {1 \over 12}\right)^2 - {1 \over 24} - 15 \\ y & = 6 \left(x - {1 \over 12}\right)^2 - {361 \over 24} \\ \implies & \text{Coordinates of minimum point is } \left({1 \over 12}, -{361 \over 24} \right) \\ \\ \therefore \text{Minimum value} & = -{361 \over 24} \end{align*}

(b)

\begin{align*} y & = 12 - 40x - 7x^2 \\ y & = -7x^2 - 40x + 12 \\ y & = -7 \left( x^2 - {40 \over 7}x \right) + 12 \\ y & = -7 \left[ \left(x - {20 \over 7}\right)^2 - \left(20 \over 7\right)^2 \right] + 12 \\ y & = -7 \left[ \left(x - {20 \over 7}\right)^2 - {400 \over 49} \right] + 12 \\ y & = -7 \left(x - {20 \over 7}\right)^2 + {400 \over 7} + 12 \\ y & = -7 \left(x - {20 \over 7}\right)^2 + {484 \over 7} \\ \implies & \text{Coordinates of maximum point is } \left({20 \over 7}, {484 \over 7} \right) \\ \\ \therefore \text{Maximum value} & = {484 \over 7} \end{align*}

(a)

\begin{align*} x^2 - 6x + 5 & = \left(x - {6 \over 2} \right)^2 - \left(6 \over 2\right)^2 + 5 \\ & = (x - 3)^2 - 9 + 5 \\ & = (x - 3)^2 - 4 \end{align*}

(b)

\begin{align*} x^2 - 3x - 1.5 & = \left(x - {3 \over 2}\right)^2 - \left(3 \over 2\right)^2 - 1.5 \\ & = (x - 1.5)^2 - 2.25 - 1.5 \\ & = (x - 1.5)^2 - 3.75 \end{align*}

(c)

\begin{align*} 3x^2 + 4x - 12 & = 3 \left(x^2 + {4 \over 3}x \right) - 12 \\ & = 3 \left[ \left( x + {2 \over 3} \right)^2 - \left(2 \over 3\right)^2 \right] - 12 \\ & = 3 \left[ \left( x + {2 \over 3} \right)^2 - {4 \over 9} \right] - 12 \\ & = 3 \left(x + {2 \over 3}\right)^2 - {4 \over 3} - 12 \\ & = 3 \left(x + {2 \over 3}\right)^2 - {40 \over 3} \end{align*}

(d)

\begin{align*} -4x^2 - x + 9 & = -4 \left( x^2 + {1 \over 4}x \right)+ 9 \\ & = -4 \left[ \left(x + {1 \over 8}\right)^2 - \left(1 \over 8\right)^2 \right] + 9 \\ & = -4 \left[ \left(x + {1 \over 8}\right)^2 - {1 \over 64} \right] + 9 \\ & = -4 \left(x + {1 \over 8}\right)^2 + {1 \over 16} + 9 \\ & = -4 \left(x + {1 \over 8}\right)^2 + {145 \over 16} \end{align*}

\begin{align*} 5 - 4x - x^2 & = -x^2 - 4x + 5 \\ & = -(x^2 + 4x) + 5 \\ & = - \left[ \left(x + {4 \over 2}\right)^2 - \left(4 \over 2\right)^2 \right] + 5 \\ & = - [ (x + 2)^2 - 4 ] + 5 \\ & = -(x + 2)^2 + 4 + 5 \\ & = -(x + 2)^2 + 9 \\ \\ \text{For all real values} & \text{ of } x, \\ (x + 2)^2 & \ge 0 \\ -(x + 2)^2 & \le 0 \\ -(x + 2)^2 + 9 & \le 9 \\ \\ \therefore \text{Maximum value} & = 9, \text{ when } x = -2 \end{align*}

(a)

\begin{align*} f(x) & = 3x^2 - 18x + 14 \phantom{00000} [\text{Minimum curve } (\cup)] \\ & = 3(x^2 - 6x) + 14 \\ & = 3 \left[ \left(x - {6 \over 2}\right)^2 - \left(6 \over 2\right)^2 \right] + 14 \\ & = 3 [ (x - 3)^2 - 9 ] + 14 \\ & = 3(x - 3)^2 - 27 + 14 \\ & = 3(x - 3)^2 - 13 \\ \implies & \text{Coordinates of minimum point is } (3, -13) \\ \\ \therefore \text{Minimum value} & = -13 \end{align*}

(b)

\begin{align*} y & = 16 + 24x - 5x^2 \phantom{00000} [\text{Maximum curve } (\cap)] \\ & = -5x^2 + 24x + 16 \\ & = -5(x^2 - 4.8x) + 16 \\ & = -5 \left[ \left(x - {4.8 \over 2}\right)^2 - \left(4.8 \over 2\right)^2 \right] + 16 \\ & = -5 [ (x - 2.4)^2 - 5.76 ] + 16 \\ & = -5(x - 2.4)^2 + 28.8 + 16 \\ & = -5(x - 2.4)^2 + 44.8 \\ \implies & \text{Coordinates of maximum point is } (2.4, 44.8) \\ \\ \therefore \text{Maximum value} & = 44.8 \end{align*}

(a)

\begin{align*} y & = 6x - 13 - 3x^2 \phantom{00000} [\text{Maximum curve } (\cap)] \\ & = -3x^2 + 6x - 13 \\ & = -3(x^2 - 2x) - 13 \\ & = -3 \left[ \left(x - {2 \over 2}\right)^2 - \left(2 \over 2\right)^2 \right] - 13 \\ & = -3 [ (x - 1)^2 - 1 ] - 13 \\ & = -3(x - 1)^2 + 3 - 13 \\ & = -3(x - 1)^2 - 10 \\ \implies & \text{Coordinates of maximum point is } (1, -10) \\ \\ \therefore \text{Maximum value} & = -10 \end{align*}

(b)

\begin{align*} f(x) & = 5x^2 + 9x + 3\pi \phantom{00000} [\text{Minimum curve } (\cup)] \\ & = 5(x^2 + 1.8x) + 3\pi \\ & = 5 \left[ \left(x + {1.8 \over 2} \right)^2 - \left(1.8 \over 2\right)^2 \right] + 3\pi \\ & = 5 [ (x + 0.9)^2 - 0.81 ] + 3\pi \\ & = 5(x + 0.9)^2 - 4.05 + 3\pi \\ & = 5(x + 0.9)^2 + (3\pi - 4.05) \\ \implies & \text{Coordinates of minimum point is } (-0.9, 3\pi - 4.05) \\ \\ \therefore \text{Minimum value} & = 3\pi - 4.05 \end{align*}

(a)

\begin{align*} f(x) & = 4x^2 - 24x + 41 \phantom{00000} [\text{Minimum curve } (\cup)] \\ & = 4(x^2 - 6x) + 41 \\ & = 4 \left[ \left(x - {6 \over 2}\right)^2 - \left(6 \over 2\right)^2 \right] + 41 \\ & = 4 [ (x - 3)^2 - 9 ] + 41 \\ & = 4(x - 3)^2 - 36 + 41 \\ & = 4(x - 3)^2 + 5 \\ \implies & \text{Coordinates of minimum point is } (3, 5) \\ \\ \text{Since minimum } & \text{point is above the } x \text{-axis, the curve lies above the } x \text{ -axis} \end{align*}

(b)

\begin{align*} f(x) & = -{1 \over 3}x^2 + 2x - 9 \phantom{00000} [\text{Maximum curve } (\cap)] \\ & = -{1 \over 3} (x^2 - 6x) - 9 \\ & = -{1 \over 3} \left[ \left(x - {6 \over 2}\right)^2 - \left(6 \over 2\right)^2 \right] - 9 \\ & = -{1 \over 3} [ (x - 3)^2 - 9] - 9 \\ & = -{1 \over 3}(x - 3)^2 + 3 - 9 \\ & = -{1 \over 3}(x - 3)^2 - 6 \\ \implies & \text{Coordinates of maximum point is } (3, -6) \\ \\ \text{Since maximum } & \text{point is below the } x \text{-axis, the curve lies below the } x \text{ -axis} \end{align*}

(i)

\begin{align*} y & = -{1 \over 10}x^2 + 4x + {9 \over 5} \\ y & = -{1 \over 10}(x^2 - 40x) + {9 \over 5} \\ y & = -{1 \over 10} \left[ \left(x - {40 \over 2} \right)^2 - \left(40 \over 2\right)^2 \right] + {9 \over 5} \\ y & = -{1 \over 10} [ (x - 20)^2 - 400 ] + {9 \over 5} \\ y & = -{1 \over 10}(x - 20)^2 + 40 + {9 \over 5} \\ y & = -{1 \over 10}(x - 20)^2 + {209 \over 5} \\ y & = -0.1(x - 20)^2 + 41.8 \\ \implies & \text{Coordinates of maximum point is } (20, 41.8) \\ \\ \therefore \text{Greatest height} & = 41.8 \text{ m} \text{ when horizontal distance} = 20 \text{ m} \end{align*}

(ii) The object is at the ground when the height (y) is equals to 0

\begin{align*} y & = -0.1(x - 20)^2 + 41.8 \\ \\ \text{Let } & y = 0, \\ 0 & = -0.1(x - 20)^2 + 41.8 \\ 0.1(x - 20)^2 & = 41.8 \\ (x - 20)^2 & = {41.8 \over 0.1} \\ (x - 20)^2 & = 418 \\ x - 20 & = \pm \sqrt{418} \\ x & = \sqrt{418} + 20 \text{ or } -\sqrt{418} + 20 \\ x & = 40.445 \text{ or } -0.44504 \\ x & \approx 40.4 \text{ or } -0.445 \\ \\ \therefore \text{Horizontal distance} & = 40.4 \text{ m} \end{align*}

(i)

\begin{align*} y & = -{1 \over 20}x^2 + 75x - 600 \\ \\ \text{Let } & x = 0, \\ y & = -{1 \over 20}(0)^2 + 75(0) - 600 \\ y & = -600 \\ \\ "-600" \text{ refers} & \text{ to the fixed cost (\$600) paid by the company} \end{align*}
(Real-life examples of fixed costs: rent/lease, interest expenses, depreciation, etc)

(ii)

\begin{align*} y & = -{1 \over 20}x^2 + 75x - 600 \\ y & = -{1 \over 20}(x^2 - 1500x) - 600 \\ y & = -{1 \over 20} \left[ \left(x - {1500 \over 2} \right)^2 - \left(1500 \over 2\right)^2 \right] - 600 \\ y & = -{1 \over 20} [ (x - 750)^2 - 5625200 ] - 600 \\ y & = -{1 \over 20}(x - 750)^2 + 28125 - 600 \\ y & = -{1 \over 20}(x - 750)^2 + 27525 \\ \implies & \text{Coordinates of maximum point is } (750, 27525) \\ \\ \therefore \text{No. of goods sold} & = 750 \\ \text{Maximum profit} & = \$27,525 \end{align*}

(i)

\begin{align*} {9 \over 80}x^2 - {13 \over 5}x & = {9 \over 80} \left(x^2 - {208 \over 9}x \right) \\ & = {9 \over 80} \left[ \left(x - {104 \over 9}\right)^2 - \left(104 \over 9\right)^2 \right] \\ & = {9 \over 80} \left[ \left(x - {104 \over 9}\right)^2 - {10816 \over 81} \right] \\ & = {9 \over 80} \left(x - {104 \over 9}\right)^2 - {676 \over 45} \end{align*}

(ii)

\begin{align*} y & = {9 \over 80}x^2 - {13 \over 5}x \\ y & = {9 \over 80} \left(x - {104 \over 9}\right)^2 - {676 \over 45} \\ \implies & \text{Coordinates of minimum point is } \left({104 \over 9}, -{676 \over 45}\right) \\ \\ \therefore \text{Horizontal distance} & = {104 \over 9} \\ & = 11{5 \over 9} \text{ m} \end{align*}

(iii) The 'right shoulder' of the ramp is vertically 7 m below point O. Thus the y-coordinate of the 'right shoulder' is -7

\begin{align*} y & = {9 \over 80} \left(x - {104 \over 9}\right)^2 - {676 \over 45} \\ \\ \text{When } & y = -7, \\ -7 & = {9 \over 80} \left(x - {104 \over 9}\right)^2 - {676 \over 45} \\ -7 + {676 \over 45} & = {9 \over 80} \left(x - {104 \over 9}\right)^2 \\ {361 \over 45} & = {9 \over 80} \left(x - {104 \over 9}\right)^2 \\ {{361 \over 45} \over {9 \over 80}} & = \left(x - {104 \over 9}\right)^2 \\ {16576 \over 81} & = \left(x - {104 \over 9}\right)^2 \\ \pm \sqrt{5776 \over 81} & = x - {104 \over 9} \\ \pm {76 \over 9} & = x - {104 \over 9} \\ \\ x & = {76 \over 9} + {104 \over 9} \text{ or } -{76 \over 9} + {104 \over 9} \\ x & = 20 \text{ or } 3{1 \over 9} \\ \\ \therefore p & = 20 \end{align*}