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

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 symmetery, } 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 symmetery, } 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 symmetery, } 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 symmetery, } 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 symmetery, } 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 symmetery, } 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{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 } -0.58113 \\ 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 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*}