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Worksheet 1A
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Solutions
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(a)
\begin{align} y & = 5(x + 3)(x - 4) \phantom{000000} [\text{Minimum curve } \cup] \\ \\ \text{Let } & x = 0, \\ y & = 5(0 + 3)(0 - 4) \\ y & = -60 \phantom{000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = 5(x + 3)(x - 4) \\ 0 & = (x + 3)(x - 4) \\ \\ x & = -3 \phantom{0} \text{ or } x = 4 \phantom{00000000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } x & = {-3 + 4 \over 2} \\ x & = 0.5 \\ \\ \text{Substitute } & x = 0.5 \text{ into eqn of curve,} \\ y & = 5(0.5 + 3)(0.5 - 4) \\ y & = -61.25 \\ \\ \text{Turning point: } & (0.5, -61.25) \end{align}
$$ \text{Minimum value} = -61.25, \text{ when } x = 0.5 $$
(b)
\begin{align} y & = -{1 \over 6} (x - 12)(x - 9) \phantom{000000} [\text{Maximum curve } \cap] \\ \\ \text{Let } & x = 0, \\ y & = -{1 \over 6}(0 - 12)(0 - 9) \phantom{000000} [y \text{-intercept}] \\ y & = -18 \\ \\ \text{Let } & y = 0, \\ 0 & = -{1 \over 6}(x - 12)(x - 9) \\ 0 & = (x - 12)(x - 9) \\ \\ x & = 12 \phantom{0} \text{ or } x = 9 \phantom{000000000000} [x \text{-intercepts}] \\ \\ \text{Line of symmetry, } x & = {9 + 12 \over 2} \\ x & = 10.5 \\ \\ \text{Substitute } & x = 10.5 \text{ into eqn of curve,} \\ y & = -{1 \over 6}(10.5 - 12)(10.5 - 9) \\ y & = 0.375 \\ \\ \text{Turning point: } & (10.5, 0.375) \end{align}
$$ \text{Maximum value} = 0.375, \text{ when } x = 10.5 $$
(c)
\begin{align} y & = (2x + 7)(x + 8) \phantom{000000} [\text{Minimum curve } \cup] \\ \\ \text{Let } & x = 0, \\ y & = [2(0) + 7](0 + 8) \phantom{000000} [y \text{-intercept}] \\ y & = 56 \\ \\ \text{Let } & y = 0, \\ 0 & = (2x + 7)(x + 8) \\ \\ x & = -{7 \over 2} \phantom{0} \text{ or } x = -8 \phantom{00000000} [x \text{-intercepts}] \\ x & = -3.5 \\ \\ \text{Line of symmetry, } x & = {-3.5 + (-8) \over 2} \\ x & = -5.75 \\ \\ \text{Substitute } & x = -5.75 \text{ into eqn of curve,} \\ y & = [2(-5.75) + 7](-5.75 + 8) \\ y & = -10.125 \\ \\ \text{Turning point: } & (-5.75, -10.125) \end{align}
$$ \text{Minimum value} = -10.125, \text{ when } x = -5.75 $$
(d)
\begin{align} y & = (10 - x)(4x + 1) \\ & = 40x + 10 - 4x^2 -x \\ & = - 4x^2 + 39x + 10 \phantom{000000} [\text{Maximum curve } \cap] \\ \\ \text{Let } & x = 0, \\ y & = (10 - 0)[4(0) + 1] \\ y & = 10 \phantom{000000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = (10 - x)(4x + 1) \\ \\ x & = 10 \phantom{0} \text{ or } x = -{1 \over 4} \phantom{00000000} [x \text{-intercepts}] \\ & \phantom{00000000.} x = -0.25 \\ \\ \text{Line of symmetry, } x & = {10+ (-0.25) \over 2} \\ x & = 4.875 \\ \\ \text{Substitute } & x = 4.875 \text{ into eqn of curve,} \\ y & = (10 - 4.875)[4(4.875) + 1] \\ y & = 105{1 \over 16} \\ \\ \text{Turning point: } & \left( 4{7 \over 8}, 105{1 \over 16}\right) \end{align}
$$ \text{Maximum value} = 105{1 \over 16}, \text{ when } x = 4{7 \over 8} $$
(e)
\begin{align} y & = 4(x + 5)^2 \phantom{000000} [\text{Minimum curve } \cup] \\ y & = 4(x + 5)^2 + 0 \\ \\ \text{Turning point: } & (-5, 0) \\ \\ \text{Let } & x = 0, \\ y & = 4(0 + 5)^2 + 0 \\ y & = 100 \phantom{0000000000} [y \text{-intercept}] \end{align}
$$ \text{Minimum value} = 0, \text{ when } x = -5 $$
(f)
\begin{align} y & = 3(x - 2)^2 - 6 \phantom{000000} [\text{Minimum curve } \cup] \\ \\ \text{Turning point: } & (2, -6) \\ \\ \text{Let } & x = 0, \\ y & = 3(0 - 2)^2 - 6 \\ y & = 6 \phantom{0000000000000000} [y \text{-intercept}] \end{align}
$$ \text{Minimum value} = -6, \text{ when } x = 2 $$
(g)
\begin{align} y & = -0.1(x - 1)^2 \phantom{000000} [\text{Maximum curve } \cap] \\ y & = -0.1(x - 1)^2 + 0 \\ \\ \text{Turning point: } & (1, 0) \\ \\ \text{Let } & x = 0, \\ y & = -0.1(0 - 1)^2 \\ y & = -0.1 \phantom{0000000000000000} [y \text{-intercept}] \end{align}
$$ \text{Maximum value} = 0, \text{ when } x = 1 $$
(h)
\begin{align} y & = -2(x + 3)^2 + 4 \phantom{000000} [\text{Maximum curve } \cap] \\ \\ \text{Turning point: } & (-3, 4) \\ \\ \text{Let } & x = 0, \\ y & = -2(0 + 3)^2 + 4 \\ y & = -14 \phantom{0000000000000000} [y \text{-intercept}] \end{align}
$$ \text{Maximum value} = 4, \text{ when } x = -3 $$
(a)
\begin{align} x^2 - 8x & = x^2 - 8x + \left(8 \over 2\right)^2 - \left(8 \over 2\right)^2 \\ & = x^2 - 8x + 4^2 - 4^2 \\ & = (x - 4)^2 - 16 \end{align}
(b)
\begin{align} 2x^2 + 10x & = 2 (x^2 + 5x) \\ & = 2 \left[ x^2 + 5x + \left(5 \over 2\right)^2 - \left(5 \over 2\right)^2 \right] \\ & = 2 \left[ \left(x + {5 \over 2}\right)^2 - {25 \over 4} \right] \\ & = 2 \left(x + {5 \over 2}\right)^2 - {25 \over 2} \end{align}
(c)
\begin{align} {1 \over 3}x^2 + 4x & = {1 \over 3}(x^2 + 12x) \\ & = {1 \over 3} \left[ x^2 + 12x + \left(12 \over 2\right)^2 - \left(12 \over 2\right)^2 \right] \\ & = {1 \over 3} ( x^2 + 12x + 6^2 - 6^2 ) \\ & = {1 \over 3} [ (x + 6)^2 - 36 ] \\ & = {1 \over 3} (x + 6)^2 - 12 \end{align}
(d)
\begin{align} 3x^2 + 12x + 12 & = 3 (x^2 + 4x) + 12 \\ & = 3 \left[ x^2 + 4x + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 \right] + 12 \\ & = 3 ( x^2 + 4x + 2^2 - 2^2 ) + 12 \\ & = 3 [ (x + 2)^2 - 4] + 12 \\ & = 3(x + 2)^2 - 12 + 12 \\ & = 3(x + 2)^2 \end{align}
(e)
\begin{align} 4x^2 - 6x - 1 & = 4 \left(x^2 - {3 \over 2}x\right) - 1 \\ & = 4 \left[ x^2 - {3 \over 2}x + \left(3 \over 4\right)^2 - \left(3 \over 4\right)^2 \right] - 1 \\ & = 4 \left[ \left(x - {3 \over 4}\right)^2 - {9 \over 16} \right] - 1 \\ & = 4 \left(x - {3 \over 4}\right)^2 - {9 \over 4} - 1 \\ & = 4 \left(x - {3 \over 4}\right)^2 - {13 \over 4} \end{align}
(f)
\begin{align} -x^2 - 10x + 3 & = -(x^2 + 10x) + 3 \\ & = - \left[ x^2 + 10x + \left(10 \over 2\right)^2 - \left(10 \over 2\right)^2 \right] + 3 \\ & = - ( x^2 + 10x + 5^2 - 5^2) + 3 \\ & = - [ (x + 5)^2 - 25 ] + 3 \\ & = - (x + 5)^2 + 25 + 3 \\ & = - (x + 5)^2 + 28 \end{align}
(g)
\begin{align} 7 + 12x - 0.4x^2 & = - 0.4x^2 + 12x + 7 \\ & = - 0.4 (x^2 - 30x) + 7 \\ & = - 0.4 \left[ x^2 - 30x + \left(30 \over 2\right)^2 - \left(30 \over 2\right)^2 \right] + 7 \\ & = - 0.4 (x^2 - 30x + 15^2 - 15^2) + 7 \\ & = - 0.4 [ (x - 15)^2 - 225 ] + 7 \\ & = - 0.4 (x - 15)^2 + 90 + 7 \\ & = - 0.4 (x - 15)^2 + 97 \end{align}
(h)
\begin{align} (3x + 5)(2x - 1) & = 6x^2 - 3x + 10x - 5 \\ & = 6x^2 + 7x - 5 \\ & = 6 \left(x^2 + {7 \over 6}x\right) - 5 \\ & = 6 \left[ x^2 + {7 \over 6}x + \left(7 \over 12\right)^2 - \left(7 \over 12\right)^2 \right] - 5 \\ & = 6 \left[ \left(x + {7 \over 12}\right)^2 - {49 \over 144} \right] - 5 \\ & = 6 \left(x + {7 \over 12}\right)^2 - {49 \over 24} - 5 \\ & = 6 \left(x + {7 \over 12}\right)^2 - {169 \over 24} \end{align}
(i)
\begin{align} 5(x - 4)(x + 8) & = 5(x^2 + 8x - 4x - 32) \\ & = 5 (x^2 + 4x - 32) \\ & = 5 \left[ x^2 + 4x + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 - 32 \right] \\ & = 5 \left( x^2 + 4x + 2^2 - 2^2 - 32 \right) \\ & = 5 [ (x+ 2)^2 - 4 - 32 ] \\ & = 5 [ (x + 2)^2 - 36 ] \\ & = 5 (x + 2)^2 - 180 \end{align}
(j)
\begin{align} 1.2(x - 6)(x - 5) & = 1.2 (x^2 - 5x - 6x + 30) \\ & = 1.2 (x^2 - 11x + 30) \\ & = 1.2 \left[ x^2 - 11x + \left(11 \over 2\right)^2 - \left(11 \over 2\right)^2 + 30 \right] \\ & = 1.2 \left[ \left(x - {11 \over 2}\right)^2 - {121 \over 4} + 30 \right] \\ & = 1.2 \left[ \left(x - {11 \over 2}\right)^2 -{1 \over 4} \right] \\ & = 1.2 \left(x - {11 \over 2}\right)^2 - {3 \over 10} \end{align}
(i)
\begin{align} 4x^2 - 20x + 15 & = 4 (x^2 - 5x) + 15 \\ & = 4 \left[ x^2 - 5x + \left(5 \over 2\right)^2 - \left(5 \over 2\right)^2 \right] + 15 \\ & = 4 \left[ \left(x - {5 \over 2} \right)^2 - {25 \over 4} \right] + 15 \\ & = 4 \left(x - {5 \over 2}\right)^2 - 25 + 15 \\ & = 4 \left(x - {5 \over 2}\right)^2 - 10 \end{align}
(ii)
\begin{align} y & = 4x^2 - 20x + 15 \phantom{0000000000} [\text{Minimum curve } \cup] \\ y & = 4 \left(x - {5 \over 2}\right)^2 - 10 \\ \\ \text{Turning point: } & \left({5 \over 2}, -10\right) \\ \\ \text{Let } & x = 0, \\ y & = 4 \left(0 - {5 \over 2}\right)^2 - 10 \\ & = 15 \phantom{000000000000000000000} [y \text{-intercept}] \end{align}
(i)
\begin{align} -2x^2 - 10x - 3 & = -2(x^2 + 5x) - 3 \\ & = -2 \left[ x^2 + 5x + \left(5 \over 2\right)^2 - \left(5 \over 2\right)^2 \right] - 3 \\ & = -2 \left[ \left(x + {5 \over 2}\right)^2 - {25 \over 4} \right] - 3 \\ & = -2 \left(x + {5 \over 2}\right)^2 + {25 \over 2} - 3 \\ & = -2 \left(x + {5 \over 2}\right)^2 + {19 \over 2} \end{align}
(ii)
\begin{align} y & = -2x^2 - 10x - 3 \phantom{0000000000} [\text{Maximum curve } \cap] \\ y & = -2 \left(x + {5 \over 2}\right)^2 + {19 \over 2} \\ \\ \text{Turning point: } & \left(-{5 \over 2}, {19 \over 2}\right) \\ \\ \text{Let } & x = 0, \\ y & = -2 \left(0 + {5 \over 2}\right)^2 + {19 \over 2} \\ & = -3 \phantom{000000000000000000000} [y \text{-intercept}] \end{align}
$$ \text{Maximum value} = 9{1 \over 2}, \text{ when } x = -2{1 \over 2} $$
(iii)
\begin{align} y & = -2x^2 - 10x - 3 + p \phantom{0000000000} [\text{Maximum curve } \cap] \\ y & = -2 \left(x + {5 \over 2}\right)^2 + {19 \over 2} + p \\ \\ \text{Turning point: } & \left(-{5 \over 2}, {19 \over 2} + p\right) \\ \\ \text{For curve to not } & \text{intersect the } x \text{-axis, turning point is below } x \text{-axis} \\ \\ \implies {19 \over 2} + p & < 0 \\ p & < -{19 \over 2} \\ p & < -9{1 \over 2} \\ \\ \text{Possible value} & = - 10 \end{align}
(a)
\begin{align} f(x) & = x^2 - 4x - 1 \\ & = x^2 - 4x + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 - 1 \\ & = x^2 - 4x + 2^2 - 2^2 - 1 \\ & = (x - 2)^2 - 4 - 1 \\ & = (x - 2)^2 - 5 \\ \\ \\ y & = (x - 2)^2 - 5 \phantom{000000} [\text{Minimum curve } \cup] \\ \\ \text{Turning} & \text{ point: } (2, -5) \\ \\ \text{Let } & x = 0, \\ y & = (0 - 2)^2 - 5 \\ y & = -1 \phantom{0000000000000} [y \text{-intercept}] \end{align}
(b)
\begin{align} f(x) & = 2x^2 + 6x + 9 \\ & = 2(x^2 + 3x) + 9 \\ & = 2 \left[ x^2 + 3x + \left(3 \over 2\right)^2 - \left(3 \over 2\right)^2 \right] + 9 \\ & = 2 \left[ \left(x + {3 \over 2}\right)^2 - {9 \over 4} \right] + 9 \\ & = 2 \left(x + {3 \over 2}\right)^2 - {9 \over 2} + 9 \\ & = 2 \left(x + {3 \over 2}\right)^2 + {9 \over 2} \\ \\ \\ y & = 2 \left(x + {3 \over 2}\right)^2 + {9 \over 2} \phantom{000000} [\text{Minimum curve } \cup] \\ \\ \text{Turning} & \text{ point: } \left(-{3 \over 2}, {9 \over 2}\right) \\ \\ \text{Let } & x = 0, \\ y & = 2\left(0 + {3 \over 2}\right)^2 + {9 \over 2} \\ y & = 9 \phantom{0000000000000000000} [y \text{-intercept}] \end{align}
(c)
\begin{align} f(x) & = 5 + 3x - x^2 \\ & = - x^2 + 3x + 5 \\ & = - (x^2 - 3x) + 5 \\ & = - \left[ x^2 - 3x + \left(3 \over 2\right)^2 - \left(3 \over 2\right)^2 \right] + 5 \\ & = - \left[ \left(x - {3 \over 2}\right)^2 - {9 \over 4} \right] + 5 \\ & = - \left(x - {3 \over 2}\right)^2 + {9 \over 4} + 5 \\ & = - \left(x - {3 \over 2} \right)^2 + {29 \over 4} \\ \\ \\ y & = - \left(x - {3 \over 2} \right)^2 + {29 \over 4} \phantom{000000} [\text{Maximum curve } \cap] \\ \\ \text{Turning} & \text{ point: } \left({3 \over 2}, {29 \over 4}\right) \\ \\ \text{Let } & x = 0, \\ y & = - \left(0 - {3 \over 2}\right)^2 + {29 \over 4} \\ y & = 5 \phantom{0000000000000000000} [y \text{-intercept}] \end{align}
(d)
\begin{align} f(x) & = 7x - 8 - {1 \over 2}x^2 \\ & = -{1 \over 2}x^2 + 7x - 8 \\ & = - {1 \over 2} (x^2 - 14x) - 8 \\ & = - {1 \over 2} \left[ x^2 - 14x + \left(14 \over 2\right)^2 - \left(14 \over 2\right)^2 \right] - 8 \\ & = - {1 \over 2} ( x^2 - 14x + 7^2 - 7^2 ) - 8 \\ & = - {1 \over 2} [ (x - 7)^2 - 49 ] - 8 \\ & = - {1 \over 2}(x - 7)^2 + {49 \over 2} - 8 \\ & = - {1 \over 2}(x - 7)^2 + {33 \over 2} \\ \\ \\ y & = - {1 \over 2}(x - 7)^2 + {33 \over 2} \phantom{000000} [\text{Maximum curve } \cap] \\ \\ \text{Turning} & \text{ point: } \left(7, {33 \over 2}\right) \\ \\ \text{Let } & x = 0, \\ y & = - {1 \over 2} (0 - 7)^2 + {33 \over 2} \\ y & = -8 \phantom{0000000000000000000} [y \text{-intercept}] \end{align}
\begin{align} 10 + x - 4x^2 & = - 4x^2 + x + 10 \\ & = -4 \left(x^2 - {1 \over 4}x \right) + 10 \\ & = -4 \left[ x^2 - {1 \over 4}x + \left(1 \over 8\right)^2 - \left(1 \over 8\right)^2 \right] + 10 \\ & = -4 \left[ \left(x - {1 \over 8}\right)^2 - {1 \over 64} \right] + 10 \\ & = -4 \left(x - {1 \over 8}\right)^2 + {1 \over 16} + 10 \\ & = -4 \left(x - {1 \over 8}\right)^2 + {161 \over 16} \\ \\ \\ \text{Max. value} & = {161 \over 16}, \text{ when } x = {1 \over 8} \end{align}
\begin{align} {1 \over 10} (x^2 - x) + c & = {1 \over 10} \left[ x^2 - x + \left(1 \over 2\right)^2 - \left(1 \over 2\right)^2 \right] + c \\ & = {1 \over 10} \left[ \left(x - {1 \over 2}\right)^2 - {1 \over 4} \right] + c \\ & = {1 \over 10} \left(x - {1 \over 2}\right)^2 - {1 \over 40} + c \\ \\ \text{Max. value} & = -{1 \over 40} + c \\ -3 & = -{1 \over 40} + c \\ -c & = -{1 \over 40} + 3 \\ -c & = {119 \over 40} \\ c & = -{119 \over 40} \end{align}
(a)
\begin{align} px^2 - 6x + q & = p \left(x^2 - {6 \over p}x \right) + q \\ & = p \left[ x^2 - {6 \over p}x + \left(3 \over p\right)^2 - \left(3 \over p\right)^2 \right] + q \\ & = p \left[ \left(x - {3 \over p}\right)^2 - {9 \over p^2} \right] + q \\ & = p \left(x - {3 \over p}\right)^2 - {9 \over p} + q \\ \\ \text{For max. value, } & p < 0 \\ \\ \text{Max. value} & = -{9 \over p} + q \\ 8 & = -{9 \over p} + q \\ \\ \text{Let } & p = -1, \\ 8 & = -{9 \over -1} + q \\ 8 & = 9 + q \\ 8 - 9 & = q \\ -1 & = q \\ \\ \\ \therefore \text{Possible } & \text{values: } p = -1, q = -1 \end{align}
(b)
\begin{align} px^2 - 6x + q & = p \left(x - {3 \over p}\right)^2 - {9 \over p} + q \\ \\ \text{For min. value, } & p > 0 \\ \\ \text{Min. value} & = -{9 \over p} + q \\ 8 & = -{9 \over p} + q \\ \\ \text{Let } & p = 3, \\ 8 & = -{9 \over 3} + q \\ 8 & = -3 + q \\ 11 & = q \\ \\ \\ \therefore \text{Possible } & \text{values: } p = 3, q = 11 \end{align}
(a)
\begin{align} & 7(x + 3)^2 - 1 \\ \\ \text{Minimum value} & = -1, \text{ when } x = -3 \\ \\ \therefore \text{Statement is fal} & \text{se} \end{align}
(b)
\begin{align} & 7(x + 3)^2 - 1 \\ \\ \text{Minimum value} & = -1, \text{ when } x = -3 \\ \\ \therefore \text{Statement is tru} & \text{e} \end{align}
(c)
\begin{align} y & = 7(x + 3)^2 - 1 \phantom{000000} [\text{Minimum curve } \cup] \\ \\ \text{Turning} & \text{ point: } (-3, -1) \\ \\ \therefore & \text{Statement is false} \end{align}
(d)
\begin{align} y & = 7(x + 3)^2 - 1 \phantom{000000} [\text{Minimum curve } \cup] \\ \\ \text{Turning} & \text{ point: } (-3, -1) \text{ is below } x \text{-axis} \\ \\ \implies y & = 7(x + 3)^2 - 1 \text{ will cut } x \text{-axis twice} \\ \\ \\ \therefore & \text{Statement is true} \end{align}
\begin{align} \text{Curve 1: }y & = a(x - h)^2 + k \\ y & = (x - 2)^2 + 3 \phantom{000000} [\text{Minimum curve } \cup] \\ \\ \text{Turning} & \text{ point: } (2, 3) \\ \\ \text{Let } & x = 0, \\ y & = (0 - 2)^2 + 3 \\ y & = 7 \phantom{000000000000000} [y \text{-intercept}] \\ \\ \\ \\ \text{Curve 2: }y & = a(x - p)(x - q) \\ y & = -(x - 1)(x - 3) \phantom{000} \left[ {1 + 3 \over 2} = 2, \text{ maximum curve } \cap \right] \\ \\ \text{Let } & x = 0, \\ y & = -(0 - 1)(0 - 3) \\ y & = -3 \phantom{000000000000000} [y \text{-intercept}] \\ \\ \text{Let } & y = 0, \\ 0 & = -(x - 1)(x - 3) \\ 0 & = (x - 1)(x - 3) \\ \\ x & = 1 \phantom{0} \text{ or } \phantom{0} x = 3 \phantom{0000000} [x \text{-intercepts}] \\ \\ \\ \text{Line of } & \text{symmetry, } x = 2 \\ \\ \text{Let } & x = 2, \\ y & = - (2 - 1)(2 - 3) \\ y & = 1 \\ \\ \text{Turning} & \text{ point: } (2, 1) \end{align}
\begin{align} & \text{Curve 1 does not meet the } x \text{-axis while curve 2 meets } x \text{-axis twice} \\ \\ & \text{Both curves have the same line of symmetry, } x = 2 \\ \\ & \text{Both curves do not intersect} \\ \\ & \text{Both curves are reflections of each other about the line } y = 2 \end{align}