(a)

\begin{align} y & = 3(x - 1)^2 - 1 \\ \\ \text{Comparing } \text{with } y & = a(x - h)^2 + k, \phantom{0} a = 3, h = 1 \text{ and } k = -1. \\ \\ \implies & \text{Minimum curve } (\cup) \text{ since } a > 0 \\ \\ \implies & \text{Turning point is } (1, -1) \\ \\ \\ \text{When } & x = 0, \\ y & = 3(0 - 1)^2 - 1 \\ & = 2 \\ \\ \implies & y \text{-intercept is } 2 \end{align}

(b)

\begin{align} y & = -2(x + 1)^2 - 3 \\ & = -2[x - (-1)]^2 - 3 \\ \\ \text{Comparing with } y & = a(x - h)^2 + k, \phantom{0} a = -2, h = -1 \text{ and } k = - 3. \\ \\ \implies & \text{Maximum curve } (\cap) \text{ since } a < 0 \\ \\ \implies & \text{Turning point is } (-1, -3) \\ \\ \\ \text{When } & x = 0, \\ y & = -2(0 + 1)^2 - 3 \\ & = -5 \\ \\ \implies & y\text{-intercept is } -5 \end{align}

(c)

\begin{align} y & = {1 \over 4}(x + 2)^2 + 1 \\ & = {1 \over 4}[x - (-2)]^2 + 1 \\ \\ \text{Comparing with } y & = a(x - h)^2 + k, \phantom{0} a = {1 \over 4}, h = -2 \text{ and } k = 1. \\ \\ \implies & \text{Minimum curve } (\cup) \text{ since } a > 0 \\ \\ \implies & \text{Turning point is } (-2, 1) \\ \\ \\ \text{When } & x = 0, \\ y & = {1 \over 4}(0 + 2)^2 + 1 \\ & = 2 \\ \\ \implies & y \text{-intercept is } 2 \end{align}

(d)

\begin{align} y & = -3(x - 2)^2 \\ & = -3(x - 2)^2 + 0 \\ \\ \text{Comparing with } y & = a(x - h)^2 + k, \phantom{0} a = -3, h = 2 \text{ and } k = 0. \\ \\ \implies & \text{Maximum curve } (\cap) \text{ since } a < 0 \\ \\ \implies & \text{Turning point is } (2, 0) \\ \\ \\ \text{When } & x = 0, \\ y & = -3(0 - 2)^2 \\ & = -12 \\ \\ \implies & y \text{-intercept is } -12 \end{align}

(a)

\begin{align} 5x^2 - x & - 2 = 0 \\ \\ [a = 5, b & = -1, c = -2] \\ \\ b^2 - 4ac & = (-1)^2 - 4(5)(-2) \\ & = 41 \\ \\ \text{Since } b^2 - 4ac > 0, & \text{ equation has 2 distinct real roots}. \end{align}

(b)

\begin{align} 9x^2 + 6x & + 1 = 0 \\ \\ [a = 9, b & = 6, c = 1] \\ \\ b^2 - 4ac & = (6)^2 - 4(9)(1) \\ & = 0 \\ \\ \text{Since } b^2 - 4ac = 0, & \text{ equation has 2 equal real roots}. \end{align}

(c)

\begin{align} x^2 + x + 1 & = -2x^2 \\ x^2 + 2x^2 + x + 1 & = 0 \\ 3x^2 + x + 1 & = 0 \\ \\ [a = 3, b & = 1, c = 1] \\ \\ b^2 - 4ac & = (1)^2 - 4(3)(1) \\ & = -11 \\ \\ \text{Since } b^2 - 4ac < 0, & \text{ equation has no real roots}. \end{align}

(d)

\begin{align} (x - 2)^2 & = -6 \\ (x)^2 - 2(x)(2) + (2)^2 & = -6 \\ x^2 - 4x + 4 & = - 6 \\ x^2 - 4x + 10 & = 0 \\ \\ [a = 1, b & = -4, c = 10] \\ \\ b^2 - 4ac & = (-4)^2 - 4(1)(10) \\ & = -24 \\ \\ \text{Since } b^2 - 4ac < 0, & \text{ equation has no 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)

\begin{align} y & = 4x^2 - 4x - p \\ \\ [a = 4, b & = -4, c = -p] \\ \\ b^2 - 4ac & > 0 \phantom{00000} [\text{Two } x \text{-intercepts}] \\ (-4)^2 - 4(4)(-p) & > 0 \\ 16 + 16p & > 0 \\ 16p & > -16 \\ p & > {-16 \over 16} \\ p & > -1 \end{align}

(b)

\begin{align} y & = 9x^2 - px + 1 \\ \\ [a = 9, b & = -p, c = 1] \\ \\ b^2 - 4ac & = 0 \phantom{00000} [\text{Only 1 } x \text{-intercept}] \\ (-p)^2 - 4(9)(1) & = 0 \\ p^2 - 36 & = 0 \\ p^2 & = 36 \\ p & = \pm \sqrt{36} \\ & = \pm 6 \end{align}

(c)

\begin{align} y & = px^2 - 2x + 3 \\ \\ [a = p, b & = -2, c = 3] \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No } x \text{-intercepts}] \\ (-2)^2 - 4(p)(3) & < 0 \\ 4 - 12p & < 0 \\ -12p & < - 4 \\ p & > {-4 \over -12} \\ p & > {1 \over 3} \end{align}

(a)

\begin{align} 2x^2 + 2x & + k \\ \\ [a = 2, b & = 2, c = k] \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No } x \text{-intercepts}] \\ (2)^2 - 4(2)(k) & < 0 \\ 4 - 8k & < 0 \\ -8k & < - 4 \\ k & > {-4 \over -8} \\ k & > {1 \over 2} \end{align}

(b)

\begin{align} -3x^2 + 6x + k - 1 \\ \\ [a = -3, b & = 6, c = k - 1] \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No } x \text{-intercepts}] \\ (6)^2 - 4(-3)(k - 1) & < 0 \\ 36 + 12(k - 1) & < 0 \\ 36 + 12k - 12 & < 0 \\ 12k + 24 & < 0 \\ 12k & < - 24 \\ k & < {-24 \over 12} \\ k & < -2 \end{align}

(c)

\begin{align} y & = 2x^2 + x - 2k \\ \\ [a = 2, b & = 1, c = -2k] \\ \\ b^2 - 4ac & < 0 \phantom{00000} [\text{No } x \text{-intercepts}] \\ (1)^2 - 4(2)(-2k) & < 0 \\ 1 - 8(-2k) & < 0 \\ 1 + 16k & < 0 \\ 16k & < -1 \\ k & < -{1 \over 16} \end{align}

Since it is not explicitly stated how many times the curve meets the x-axis, the curve can meet the x-axis once or twice. This suggests b² - 4ac ≥ 0.

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

(a)

\begin{align} f(x) & = -2x^2 + 4x + 3 \\ & = -2(x^2 - 2x) + 3 \\ & = -2\left[ x^2 - 2x + \left(2 \over 2\right)^2 - \left(2 \over 2\right)^2 \right] + 3 \\ & = -2 [ x^2 - 2x + 1^2 - 1^2 ] + 3 \\ & = -2 [ (x - 1)^2 - 1 ] + 3 \\ & = -2(x - 1)^2 + 2 + 3 \\ & = -2(x - 1)^2 + 5 \\ \\ \\ \text{Comparing with } f(x) & = a(x - h)^2 + k, \phantom{0} a = -2, h = 1 \text{ and } k = 5. \\ \\ \implies & \text{Maximum curve } (\cap) \text{ since } a < 0 \\ \\ \implies & \text{Turning point is } (1, 5) \end{align}
$$ \text{Discriminant is positive}$$

(b)

\begin{align} f(x) & = 3x^2 - 12x + 13 \\ & = 3(x^2 - 4x) + 13 \\ & = 3\left[ x^2 - 4x + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 \right] + 13 \\ & = 3( x^2 - 4x + 2^2 - 2^2 ) + 13 \\ & = 3[ (x - 2)^2 - 4 ] + 13 \\ & = 3(x - 2)^2 - 12 + 13 \\ & = 3(x - 2)^2 + 1 \\ \\ \\ \text{Comparing with } f(x) & = a(x - h)^2 + k, \phantom{0} a = 3, h = 2 \text{ and } k = 1. \\ \\ \implies & \text{Minimum curve } (\cup) \text{ since } a > 0 \\ \\ \implies & \text{Turning point is } (2, 1) \end{align}
$$ \text{Discriminant is negative}$$

(c)

\begin{align} f(x) & = {1 \over 2}x^2 - 4x + 8 \\ & = {1 \over 2}(x^2 - 8x) + 8 \\ & = {1 \over 2}\left[ x^2 - 8x + \left(8 \over 2\right)^2 - \left(8 \over 2\right)^2 \right] + 8 \\ & = {1 \over 2} (x^2 - 8x + 4^2 - 4^2) + 8 \\ & = {1 \over 2} [ (x - 4)^2 - 16] + 8 \\ & = {1 \over 2}(x - 4)^2 - 8 + 8 \\ & = {1 \over 2}(x - 4)^2 + 0 \\ \\ \\ \text{Comparing with } f(x) & = a(x - h)^2 + k, \phantom{0} a = {1 \over 2}, h = 4 \text{ and } k = 0. \\ \\ \implies & \text{Minimum curve } (\cup) \text{ since } a > 0 \\ \\ \implies & \text{Turning point is } (4, 0) \end{align}
$$ \text{Discriminant is equals to zero}$$

(d)

\begin{align} f(x) & = -4x^2 - 8x - 5 \\ & = -4(x^2 + 2x) - 5 \\ & = -4\left[ x^2 + 2x + \left(2 \over 2\right)^2 - \left(2 \over 2\right)^2 \right] - 5 \\ & = -4 (x^2 + 2x + 1^2 - 1^2) - 5 \\ & = -4 [ (x + 1)^2 - 1 ] - 5 \\ & = -4(x + 1)^2 + 4 - 5 \\ & = -4(x + 1)^2 - 1 \\ & = -4[x - (-1)]^2 - 1 \\ \\ \\ \text{Comparing with } f(x) & = a(x - h)^2 + k, \phantom{0} a = -4, h = -1 \text{ and } k = -1. \\ \\ \implies & \text{Maximum curve } (\cap) \text{ since } a < 0 \\ \\ \implies & \text{Turning point is } (-1, -1) \end{align}
$$ \text{Discriminant is negative}$$

(a)

\begin{align} 3x^2 & = 2x + p - 1 \\ 3x^2 - 2x + 1 - p & = 0 \\ \\ [a = 3, b & = -2, c = 1 - p] \\ \\ b^2 - 4ac & > 0 \\ (-2)^2 - 4(3)(1 - p) & > 0 \\ 4 -12(1 - p) & > 0 \\ 4 - 12 + 12p & > 0 \\ -8 + 12p & > 0 \\ 12p & > 8 \\ p & > {8 \over 12} \\ p & > {2 \over 3} \end{align}

(b)

\begin{align} x^2 + p^2 & = 3px - 5 \\ x^2 - 3px + p^2 + 5 & = 0 \\ \\ [a = 1, b & = -3p, c = p^2 + 5] \\ \\ b^2 - 4ac & = 0 \\ (-3p)^2 - 4(1)(p^2 + 5) & = 0 \\ 9p^2 - 4(p^2 + 5) & = 0 \\ 9p^2 - 4p^2 - 20 & = 0 \\ 5p^2 - 20 & = 0 \\ 5p^2 & = 20 \\ p^2 & = {20 \over 5} \\ & = 4 \\ p & = \pm \sqrt{4} \\ & = \pm 2 \end{align}

(c)

\begin{align} (x + 1)(2x - 1) & = p - 2 \\ 2x^2 - x + 2x - 1 & = p - 2 \\ 2x^2 + x - 1 & = p - 2 \\ 2x^2 + x - 1 + 2 - p & = 0 \\ 2x^2 + x + 1 - p & = 0 \\ \\ [a = 2, b & = 1, c = 1 - p] \\ \\ b^2 - 4ac & > 0 \\ (1)^2 - 4(2)(1 - p) & > 0 \\ 1 - 8(1 - p) & > 0 \\ 1 - 8 + 8p & > 0 \\ -7 + 8p & > 0 \\ 8p & > 7 \\ p & > {7 \over 8} \end{align}

(d)

\begin{align} 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 - 3p + 4p + 2 & = 0 \\ px^2 + (-2p - 1)x + p + 2 & = 0 \\ \\ [a = p, b & = -2p - 1, c = p + 2] \\ \\ b^2 - 4ac & < 0 \\ (-2p - 1)^2 - 4(p)(p + 2) & < 0 \\ (-2p)^2 - 2(-2p)(1) + (1)^2 - 4p(p + 2) & < 0 \\ 4p^2 + 4p + 1 - 4p^2 - 8p & < 0 \\ 1 - 4p & < 0 \\ -4p & < -1 \\ 4p & > 1 \\ p & > {1 \over 4} \end{align}

\begin{align} x^2 - 2kx + k^2 & = 3 + x \\ x^2 - 2kx - x + k^2 - 3 & = 0 \\ x^2 + (-2k - 1)x + k^2 - 3 & = 0 \\ \\ [a = 1, b & = -2k - 1, c = k^2 - 3] \\ \\ b^2 - 4ac & \ge 0 \phantom{00000} [\text{Two distinct or equal real roots}] \\ (-2k - 1)^2 - 4(1)(k^2 - 3) & \ge 0 \\ (-2k)^2 - 2(-2k)(1) + (1)^2 - 4(k^2 - 3) & \ge 0 \\ 4k^2 + 4k + 1 - 4k^2 + 12 & \ge 0 \\ 4k + 13 & \ge 0 \\ 4k & \ge -13 \\ k & \ge -{13 \over 4} \\ \\ \therefore \text{Least value of } k & = -{13 \over 4} \end{align}

\begin{align} 2x^2 + p & = 2(x - 1) \\ 2x^2 + p & = 2x - 2 \\ 2x^2 - 2x + p + 2 & = 0 \\ \\ [a = 2, b & = -2, c = p + 2] \\ \\ b^2 - 4ac & = (-2)^2 - 4(2)(p + 2) \\ & = 4 - 8(p + 2) \\ & = 4 - 8p - 16 \\ & = -8p - 12 \\ & = -{24 \over 3}p - 12 \\ & = -12 \left( {2 \over 3}p + 1 \right) \\ \\ \text{For } p > -{3 \over 2} &, \\ {2 \over 3}p + 1 & > 0 \\ -12 \left({2 \over 3}p + 1\right) & < 0 \\ \\ \therefore b^2 - 4ac & < 0 \\ \\ \implies \text{Equation has } & \text{no real roots.} \end{align}

\begin{align} px^2 + 3px + p + q & = 0 \\ \\ [a = p, b & = 3p, c = p + q] \\ \\ b^2 - 4ac & = 0 \\ (3p)^2 - 4(p)(p + q) & = 0 \\ 9p^2 - 4p(p + q) & = 0 \\ 9p^2 - 4p^2 - 4pq & = 0 \\ 5p^2 - 4pq & = 0 \\ p(5p - 4q) & = 0 \\ \\ p = 0 \text{ (Reject)} \phantom{000}&\text{or}\phantom{000} 5p - 4q = 0 \\ & \phantom{or000-4q} 5p = 4q \\ & \phantom{or000-4q5} p = {4 \over 5}q \text{ (Shown)} \end{align}

(a)

\begin{align} 3x^2 - 3x & > x + k \\ 3x^2 - 3x - x - k & > 0 \\ 3x^2 - 4x - k & > 0 \\ \\ [a = 3, b & = -4, c = -k] \\ \\ \text{For equation to be always positive }& (>0), \text{ equation has no real roots.} \\ \\ b^2 - 4ac & < 0 \\ (-4)^2 - 4(3)(-k) & < 0 \\ 16 - 12(-k) & < 0 \\ 16 + 12k & < 0 \\ 12k & < - 16 \\ k & < -{16 \over 12} \\ k & < -{4 \over 3} \end{align}

(b)

\begin{align} kx^2 + 1 & > 2kx - k \\ kx^2 - 2kx + 1 + k & > 0 \\ \\ [a = k, b & = -2k, c = 1 + k] \\ \\ \text{For equation to be always positive }& (>0), \text{ equation has no real roots.} \\ \\ b^2 - 4ac & < 0 \\ (-2k)^2 - 4(k)(1 + k) & < 0 \\ 4k^2 - 4k(1 + k) & < 0 \\ 4k^2 - 4k - 4k^2 & < 0 \\ -4k & < 0 \\ 4k & > 0 \\ k & > 0 \end{align}

(a)

\begin{align} y & = kx - 5 \phantom{000} \text{ --- (1)} \\ \\ x^2 & = 2y + 1 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 & = 2(kx - 5) + 1 \\ x^2 & = 2kx - 10 + 1 \\ x^2 & = 2kx - 9 \\ x^2 - 2kx + 9 & = 0 \\ \\ [a = 1, b & = -2k, c = 9] \\ \\ b^2 - 4ac & = 0 \\ (-2k)^2 - 4(1)(9) & = 0 \\ 4k^2 - 36 & = 0 \\ 4k^2 & = 36 \\ k^2 & = {36 \over 4} \\ & = 9 \\ k & = \pm \sqrt{9} \\ & = \pm 3 \end{align}

(b)

\begin{align} x + 3y & = k - 1 \\ 3y & = k - 1 - x \\ y & = {k - 1 - x \over 3} \phantom{000} \text{ --- (1)} \\ \\ y^2 & = 2x + 5 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ \left(k - 1 - x \over 3\right)^2 & = 2x + 5 \\ {(k - 1 - x)^2 \over 9} & = 2x + 5 \\ (k - 1 - x)^2 & = 18x + 45 \\ (k - 1 - x)(k - 1 - x) & = 18x + 45 \\ k^2 - k - kx -k + 1 + x - kx + x + x^2 & = 18x + 45 \\ x^2 + 2x - 2kx + k^2 - 2k + 1 & = 18x + 45 \\ x^2 - 16x - 2kx + k^2 - 2k - 44 & = 0 \\ x^2 + (-16 - 2k)x + (k^2 - 2k - 44) & = 0 \\ \\ [a = 1, b & = -16 - 2k, c = k^2 - 2k - 44] \\ \\ b^2 - 4ac & \ge 0 \phantom{00000} [\text{1 or 2 points of intersection}] \\ (-16 - 2k)^2 - 4(1)(k^2 - 2k - 44) & \ge 0 \\ (-16)^2 - 2(-16)(2k) + (2k)^2 - 4(k^2 - 2k - 44) & \ge 0 \\ 256 + 64k + 4k^2 - 4k^2 + 8k + 176 & \ge 0 \\ 72k + 432 & \ge 0 \\ 72k & \ge -432 \\ k & \ge {-432 \over 72} \\ k & \ge -6 \end{align}

(c)

\begin{align} y & = kx + 2 \phantom{000} \text{ --- (1)} \\ \\ y^2 & = 8x - x^2 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (kx + 2)^2 & = 8x - x^2 \\ (kx)^2 + 2(kx)(2) + (2)^2 & = 8x - x^2 \\ k^2 x^2 + 4kx + 4 & = 8x - x^2 \\ k^2 x^2 + x^2 + 4kx - 8x + 4 & = 0 \\ (k^2 + 1)x^2 + (4k - 8)x + 4 & = 0 \\ \\ [a = k^2 + 1, b & = 4k - 8, c = 4] \\ \\ b^2 - 4ac & > 0 \\ (4k - 8)^2 - 4(k^2 + 1)(4) & > 0 \\ (4k)^2 - 2(4k)(8) + (8)^2 - 16(k^2 + 1) & > 0 \\ 16k^2 - 64k + 64 - 16k^2 - 16 & > 0 \\ -64k + 48 & > 0 \\ -64k & > -48 \\ 64k & < 48 \\ k & < {48 \over 64} \\ k & < {3 \over 4} \end{align}

(d)

\begin{align} y & = x + k - 1 \phantom{000} \text{ --- (1)} \\ \\ (y - 1)^2 & = 4x \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (x + k - 1 - 1)^2 & = 4x \\ (x + k - 2)^2 & = 4x \\ (x + k - 2)(x + k - 2) & = 4x \\ x^2 + kx - 2x + kx + k^2 - 2k - 2x -2k + 4 & = 4x \\ x^2 + 2kx - 4x + k^2 - 4k + 4 & = 4x \\ x^2 + 2kx - 8x + k^2 - 4k + 4 & = 0 \\ x^2 + (2k - 8)x + (k^2 - 4k + 4) & = 0 \\ \\ [a = 1, b & = 2k - 8, c = k^2 - 4k + 4] \\ \\ b^2 - 4ac & < 0 \\ (2k - 8)^2 - 4(1)(k^2 - 4k + 4) & < 0 \\ (2k)^2 - 2(2k)(8) + (8)^2 - 4(k^2 - 4k + 4) & < 0 \\ 4k^2 - 32k + 64 - 4k^2 + 16k - 16 & < 0 \\ -16k + 48 & < 0 \\ -16k & < - 48 \\ 16k & > 48 \\ k & > {48 \over 16} \\ k & > 3 \end{align}

(i)

\begin{align} y & = kx(x + 2) \phantom{000} \text{ --- (1)} \\ \\ y & = x - k \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ kx(x + 2) & = x - k \\ kx^2 + 2kx & = x - k \\ kx^2 + 2kx - x + k & = 0 \\ kx^2 + (2k - 1)x + k & = 0 \\ \\ [a = k, b & = 2k - 1, c = k] \\ \\ b^2 - 4ac & \ge 0 \phantom{00000} [\text{1 or 2 points of intersections}] \\ (2k - 1)^2 - 4(k)(k) & \ge 0 \\ (2k)^2 - 2(2k)(1) + (1)^2 - 4k^2 & \ge 0 \\ 4k^2 - 4k + 1 - 4k^2 & \ge 0 \\ -4k + 1 & \ge 0 \\ -4k & \ge - 1 \\ 4k & \le 1 \\ k & \le {1 \over 4} \end{align}

(ii) If line is tangent to the curve, there is only 1 point of intersection, which implies b² - 4ac = 0

$$ \therefore k = {1 \over 4} $$

(iii)

\begin{align} b^2 - 4ac & = -4k + 1 \\ \\ \text{When } k & = 1, \\ b^2 - 4ac & = -4(1) + 1 \\ & = -3 \\ \\ \text{Since } b^2 - 4ac < 0, & \text{ there are no points of intersection.} \end{align}

(i)

\begin{align} \text{When } y & = 50, \\ 50 & = -5x^2 + 20x + c \\ 0 & = -5x^2 + 20x + c - 50 \\ \\ [a = -5, b & = 20, c = c - 50] \\ \\ b^2 - 4ac & < 0 \\ (20)^2 - 4(-5)(c - 50) & < 0 \\ 400 + 20(c - 50) & < 0 \\ 400 + 20c - 1000 & < 0 \\ 20c - 600 & < 0 \\ 20c & < 600 \\ c & < {600 \over 20} \\ c & < 30 \end{align}

(ii)

\begin{align} y & = -5x^2 + 20x + 20 \\ & = -5(x^2 - 4x) + 20 \\ & = -5\left[ x^2 - 4x + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 \right] + 20 \\ & = -5 (x^2 - 4x + 2^2 - 2^2) + 20 \\ & = -5 [ (x - 2)^2 - 4] + 20 \\ & = -5(x - 2)^2 + 20 + 20 \\ & = -5(x - 2)^2 + 40 \\ \\ \\ \text{Comparing with } y & = a(x - h)^2 + k, \phantom{0} a = -5, h = 2 \text{ and } k = 40. \\ \\ \implies & \text{Maxmium curve } (\cap) \text{ since } a < 0 \\ \\ \implies & \text{Turning point is } (2, 40) \\ \\ \\ \text{When } & x = 0, \\ y & = -5(0)^2 + 20(0) + 20 \\ & = 20 \end{align}

(iii)

\begin{align} & \text{From (i), the baseball does not reach a height of 50 m when } c < 30 \\ & \text{Referring to graph from (ii), the ball only reaches a maximum height of 40 m} \end{align}

(i)

\begin{align} k^2 - 4k + 12 & = k^2 - 4k + \left(4 \over 2\right)^2 - \left(4 \over 2\right)^2 + 12 \\ & = k^2 - 4k + 2^2 - 2^2 + 12 \\ & = (k - 2)^2 - 4 + 12 \\ & = (k - 2)^2 + 8 \text{ (Shown)} \end{align}

(ii)

\begin{align} x^2 + kx & = 3 - k \\ x^2 + kx + k - 3 & = 0 \\ \\ [a = 1, b & = k, c = k - 3] \\ \\ b^2 - 4ac & = (k)^2 - 4(1)(k - 3) \\ & = k^2 - 4(k - 3) \\ & = k^2 - 4k + 12 \\ & = (k - 2)^2 + 8 \phantom{00000} [\text{Apply result from part (i)]} \\ \\ \\ \text{For all real } & \text{values of } k, \\ (k - 2)^2 & \ge 0 \\ (k - 2)^2 + 8 & > 8 \\ \\ \therefore b^2 - 4ac & > 0 \\ \\ \implies \text{Equation has real } & \text{solutions for all real values of } k. \end{align}

(i)

\begin{align} \text{When } & p = 4, \\ y & = x^2 + 4x - 4 + 3 \\ y & = x^2 + 4x - 1 \\ \\ [a = 1, b & = 4, c = -1] \\ \\ b^2 - 4ac & = (4)^2 - 4(1)(-1) \\ & = 16 + 4 \\ & = 20 > 0 \\ \\ \text{Since } b^2 - 4ac > 0, & \phantom{0} \text{there are 2 points of intersection}. \end{align}

(ii)

\begin{align} \text{When } & p = -2, \\ y & = x^2 + (-2)x - (-2) + 3 \\ & = x^2 - 2x + 2 + 3 \\ & = x^2 - 2x + 5 \\ \\ [a = 1, b & = -2, c = 5] \\ \\ b^2 - 4ac & = (-2)^2 - 4(1)(5) \\ & = 4 - 20 \\ & = -16 < 0 \\ \\ \text{Since } b^2 - 4ac < 0, & \phantom{0} \text{there no points of intersection with the } x\text{-axis}. \\ \\ \text{Since } a > 0, \text{ it is a } &\text{minimum curve/function} (\cup). \\ \\ \therefore \text{Value of } y & \text{ is always positive}. \end{align}

(iii)

\begin{align} y & = x^2 + px - p + 3 \\ & = x^2 + px + 3 - p \\ \\ [a = 1, b & = p, c = 3 - p] \\ \\ b^2 - 4ac & = 0 \\ (p)^2 - 4(1)(3 - p) & = 0 \\ p^2 - 4(3 - p) & = 0 \\ p^2 - 12 + 4p & = 0 \\ p^2 + 4p - 12 & = 0 \\ (p - 2)(p + 6) & = 0 \\ \\ p - 2 = 0 \phantom{00}&\text{or}\phantom{000} p + 6 = 0 \\ p = 2 \phantom{00}&\phantom{or000+6} p = - 6 \\ \\ \\ \text{For } & p = 2, \\ y & = x^2 + 2x - 2 + 3 \\ & = x^2 + 2x + 1 \\ \\ \text{When } & y = 0, \\ 0 & = x^2 + 2x + 1 \\ 0 & = (x + 1)^2 \\ 0 & = x + 1 \\ - 1 & = x \\ \\ \therefore \text{When } p = 2, & \phantom{0} \text{coordinate is } (-1, 0). \\ \\ \\ \text{When } & p = -6, \\ y & = x^2 + (-6)x - (-6) + 3 \\ & = x^2 - 6x + 6 + 3 \\ & = x^2 - 6x + 9 \\ \\ \text{When } & y = 0, \\ 0 & = x^2 - 6x + 9 \\ 0 & = (x - 3)^2 \\ 0 & = x - 3 \\ 3 & = x \\ \\ \therefore \text{When } p = -6, &\phantom{0} \text{coordinate is } (3, 0). \end{align}

(i)

\begin{align} y & = x^2 + 6x - 5 \phantom{000} \text{ --- (1)} \\ \\ y & = 8x + c \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ x^2 + 6x - 5 & = 8x + c \\ x^2 - 2x - 5 - c & = 0 \\ \\ [a = 1, b & = -2, c = -5 - c] \\ \\ \\ b^2 - 4ac & \ge 0 \phantom{00000} [\text{1 or 2 points of intersections}] \\ (-2)^2 - 4(1)(-5 - c) & \ge 0 \\ 4 - 4(- 5 - c) & \ge 0 \\ 4 + 20 + 4c & \ge 0 \\ 24 + 4c & \ge 0 \\ 4c & \ge -24 \\ c & \ge {-24 \over 4} \\ c & \ge -6 \end{align}

(ii) For f(x) to be always greater than g(x), there are no points of intersection between the two graphs

\begin{align} \text{From part (i), } a = 1, b & = -2, c = -5 - c \\ \\ \\ b^2 - 4ac & < 0 \\ (-2)^2 - 4(1)(-5 - c) & < 0 \\ 4 - 4(- 5 - c) & < 0 \\ 4 + 20 + 4c & < 0 \\ 24 + 4c & < 0 \\ 4c & < -24 \\ c & < {-24 \over 4} \\ c & < -6 \end{align}

(i)

\begin{align} \text{When } & C = 10, \\ 10 & = 1.2n^2 - 14.4n + 53.7 \\ 0 & = 1.2n^2 -14.4n + 43.7 \\ \\ [a = 1.2, b & = -14.4, c = 43.7] \\ \\ b^2 - 4ac & = (-14.4)^2 - 4(1.2)(43.7) \\ & = -2.4 < 0 \\ \\ \text{Since } b^2 - 4ac < 0, & \phantom{0} \text{equation has no real solutions}. \\ \\ \therefore \text{It is not possible to have } & \text{a cost of production of 10 thousand dollars.} \end{align}

(ii)

\begin{align} \text{When } & C = 80, \\ 80 & = 1.2n^2 - 14.4n + 53.7 \\ 0 & = 1.2n^2 -14.4n - 26.3 \\ \\ [a = 1.2, b & = -14.4, c = -26.3] \\ \\ b^2 - 4ac & = (-14.4)^2 - 4(1.2)(-26.3) \\ & = 333.6 > 0 \\ \\ \text{Since } b^2 - 4ac > 0, & \phantom{0} \text{equation has 2 real solutions}. \\ \\ \therefore \text{It is possible to have } & \text{the cost of production to reach 80 thousand dollars.} \end{align}

(iii) For the cost of production to be always more than x thousand dollars, the curve does not intersect the line C = x.

\begin{align} \text{When } & C = x, \\ x & = 1.2n^2 - 14.4n + 53.7 \\ 0 & = 1.2n^2 -14.4n + 53.7 - x \\ \\ [a = 1.2, b & = -14.4, c = 53.7 - x] \\ \\ \\ b^2 - 4ac & < 0 \\ (-14.4)^2 - 4(1.2)(53.7 - x) & < 0 \\ 207.36 - 4.8(53.7 - x) & < 0 \\ 207.36 - 257.76 + 4.8x & < 0 \\ -50.4 + 4.8x & < 0 \\ 4.8x & < 50.4 \\ x & < {50.4 \over 4.8} \\ x & < 10.5 \\ \\ \text{Since cost of production is } & \text{more than } \$ 0, \phantom{.} 0 < x < 10.5 \end{align}

(i)

\begin{align} y & = -0.5t^2 + 7t + k \\ \\ \text{When } t = 10 & \text{ and } y = 26, \\ 26 & = -0.5(10)^2 + 7(10) + k \\ 26 & = -50 + 70 + k \\ 26 + 50 - 70 & = k \\ 6 & = k \end{align}

(ii)

\begin{align} y & = -0.5t^2 + 7t + 6 \\ \\ \text{When } & t = 0, \\ y & = -0.5(0)^2 + 7(0) + 6 \\ & = 6 \\ \\ \therefore \text{He} & \text{ight} = 6 \text{ m} \end{align}

(iii)

\begin{align} y & = -0.5t^2 + 7t + 6 \\ \\ \text{When } & y = 30, \\ 30 & = -0.5t^2 + 7t + 6 \\ 0 & = -0.5t^2 + 7t - 24 \\ 0 & = -t^2 + 14t - 48 \\ 0 & = t^2 - 14t + 48 \\ 0 & = (t - 6)(t - 8) \\ \\ t - 6 = 0 \phantom{00}&\text{or}\phantom{000} t - 8 = 0 \\ t = 6 \phantom{00}&\phantom{or000-8} t = 8 \\ \\ \text{Since the coefficient of } t^2 \text{ is negative}, & \phantom{0} \text{the curve is a maximum curve} (\cap). \\ \\ \text{Thus the ball is above 30 m} & \text{ between 6 and 8 seconds.} \\ \\ \therefore 6 < \phantom{.}& t < 8 \end{align}

(iv)

\begin{align} t & = {6 + 8 \over 2} \\ & = 7 \\ \\ \text{When } & t = 7, \\ y & = -0.5(7)^2 + 7(7) + 6 \\ & = 30.5 \\ \\ \therefore \text{Maximum height} & = 30.5 \text{ m} \end{align}

(v)(a)

\begin{align} y & = -0.5t^2 + 7t + 6 \\ & = -0.5(t^2 - 14t) + 6 \\ & = -0.5 \left[ t^2 - 14t + \left(14 \over 2\right)^2 - \left(14 \over 2\right)^2 \right] + 6 \\ & = -0.5 (t^2 - 14t + 7^2 - 7^2) + 6 \\ & = -0.5 [ (t - 7)^2 - 49 ] + 6 \\ & = -0.5(t - 7)^2 + 24.5 + 6 \\ & = -0.5(t - 7)^2 + 30.5 \\ \\ \\ \text{Comparing with } y & = a(t - p)^2 + q, \phantom{0} a = -0.5, p = 7 \text{ and } q = 30.5. \\ \\ \implies & \text{Maximum curve } (\cap) \text{ since } a < 0 \\ \\ \implies & \text{Turning point is } (7, 30.5) \end{align}

(v)(b)

\begin{align} & \text{No.} \\ & \text{For } y = -0.5t^2 + 7t + k \text{, the graph is restricted to } t \ge 0 \\ & \text{On the other hand, for } y = -0.5x^2 + 7x + k \text{, there are no restrictions on the values of } x \end{align}

\begin{align} \text{If } m = 0, \phantom{.} (0)x^2 + 6x + 7 & = 0 \\ 6x + 7 & = 0 \\ 6x & = - 7 \\ x & = -{7 \over 6} \\ \\ \text{For } m = 0, \text{ equation only } & \text{has 1 real root} \\ \\ \text{Thus, answer is incorrect } & \end{align}

(i)

\begin{align} \alpha + \beta & = {-b + \sqrt{b^2 - 4ac} \over 2a} + {-b - \sqrt{b^2 - 4ac} \over 2a} \\ & = {-b + \sqrt{b^2 - 4ac} - b - \sqrt{b^2 - 4ac} \over 2a} \\ & = {-2b \over 2a} \\ & = -{b \over a} \end{align}

(ii)

\begin{align} \alpha \beta & = \left({-b + \sqrt{b^2 - 4ac} \over 2a}\right) \left({-b - \sqrt{b^2 - 4ac} \over 2a}\right) \\ & = { (-b)^2 - (\sqrt{b^2 - 4ac})^2 \over (2a)^2} \\ & = { b^2 - (b^2 - 4ac) \over 4a^2} \\ & = { b^2 - b^2 + 4ac \over 4a^2} \\ & = { 4ac \over 4a^2} \\ & = { c \over a} \phantom{00} \text{ (Shown)} \end{align}

Further question

\begin{align} \text{Yes, the results are valid since } \alpha + \beta \text{ and } \alpha \beta \text{ can be complex numbers} \end{align}

Parts (i) to (iii)

\begin{align} A: y & = - 0.3x^2 + 3 &&& B: y & = -0.2x^2 + 1.8 &&& C: y & = -0.08x^2 + 2.4 \\ \\ \text{Let } & y = 0, &&& \text{Let } & y = 0, &&& \text{Let } & y = 0, \\ 0 & = -0.3x^2 + 3 &&& 0 & = -0.2x^2 + 1.8 &&& 0 & = -0.08x^2 + 2.4 \\ 0.3x^2 & = 3 &&& 0.2x^2 & = 1.8 &&& 0.08x^2 & = 2.4 \\ x^2 & = 10 &&& x^2 & = 9 &&& x^2 & = 30 \\ x & = \pm \sqrt{10} &&& x^2 & = \pm 3 &&& x^2 & = \pm \sqrt{30} \end{align}
\begin{align} & \text{(i) Water jet C since it has the furthest } x \text{-intercept } (\sqrt{30}, 0) \\ \\ & \text{(ii) Water jet A since it has the highest turning point } (0, 3) \\ \\ & \text{(iii) Water jet B since it has the lowest turning point } (0, 1.8) \text{ and the closest } x \text{-intercept } (3, 0) \end{align}

Further question

\begin{align} \text{Max. height} & \ge 10 \text{ m} \\ y & \ge 10 \\ -0.5x^2 + 2x + k & \ge 10 \\ -0.5x^2 + 2x + (k - 10) & \ge 0 \\ \\ b^2 - 4ac & = (2)^2 - 4(-0.5)(k - 10) \\ & = 4 + 2(k - 10) \\ & = 4 + 2k - 20 \\ & = 2k - 16 \\ \\ b^2 - 4ac & \ge 0 \\ 2k - 16 & \ge 0 \\ 2k & \ge 16 \\ k & \ge 8 \end{align}