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

\begin{align} {1 \over 4}x^2 - 2x - 10 & = {1 \over 4} (x^2 - 8x) - 10 \phantom{0000000} [\text{must be } x^2 \text{ before completing the square}] \\ & = {1 \over 4} \left[ x^2 - 8x + \left(8 \over 2\right)^2 - \left(8 \over 2\right)^2 \right] - 10 \\ & = {1 \over 4} ( x^2 - 8x + 4^2 - 4^2 ) - 10 \\ & = {1 \over 4} [ (x - 4)^2 - 16 ] - 10 \\ & = {1 \over 4} (x - 4)^2 - 4 - 10 \\ & = {1 \over 4} (x - 4)^2 - 14 \end{align}

\begin{align} 9 + 3x - 2x^2 & = - 2x^2 + 3x + 9 \\ & = -2 \left(x^2 - {3 \over 2}x \right) + 9 \\ & = -2 \left[ x^2 - {3 \over 2}x + \left(3 \over 4\right)^2 - \left(3 \over 4\right)^2 \right] + 9 \\ & = -2 \left[ \left(x - {3 \over 4}\right)^2 - {9 \over 16} \right] + 9 \\ & = -2 \left(x - {3 \over 4}\right)^2 + {9 \over 8} + 9 \\ & = -2 \left(x - {3 \over 4}\right)^2 + {81 \over 8} \\ & = {81 \over 8} - 2 \left(x - {3 \over 4}\right)^2 \end{align}

\begin{align} -3x^2 + 6x - 8 & = -3 (x^2 - 2x) - 8 \\ & = -3 \left[ x^2 - 2x + \left(2 \over 2\right)^2 - \left(2 \over 2\right)^2 \right] - 8 \\ & = -3 (x^2 - 2x + 1^2 - 1^2) - 8 \\ & = -3 [ (x - 1)^2 - 1] - 8 \\ & = -3 (x - 1)^2 + 3 - 8 \\ & = -3(x - 1)^2 - 5 \\ \\ \text{For all real } & \text{values of } x, \\ (x - 1)^2 & \ge 0 \\ -3(x - 1)^2 & \le 0 \\ -3(x - 1)^2 -5 & \le -5 \\ \\ \therefore \text{Expression is always} & \text{ negative for all real values of } x \end{align}

\begin{align} y & = ax^2 + 2x + c \\ \\ \text{For curve to lie } & \text{below } x \text{-axis, curve must be a maximum curve } (\cap) \\ \\ \implies & \text{Condition #1: } a < 0 \\ \\ \\ y & = ax^2 + 2x + c \\ & = a \left(x^2 + {2 \over a}x\right) + c \\ & = a \left[ x^2 + {2 \over a}x + \left(1 \over a\right)^2 - \left(1 \over a\right)^2 \right] + c \\ & = a \left[ \left(x + {1 \over a}\right)^2 - {1 \over a^2} \right] + c \\ & = a \left(x + {1 \over a}\right)^2 - {a \over a^2} + c \\ & = a \left(x + {1 \over a}\right)^2 - {1 \over a} + c \\ \\ \text{Turning point: } & \left(-{1 \over a}, -{1 \over a} + c \right) \phantom{000000} [\text{Must be below } x \text{-axis}] \\ \\ -{1 \over a} + c & < 0 \\ \text{Condition #2: } c & < {1 \over a} \\ \\ \\ \text{Let } & a = -1, \phantom{000000} [\text{to satisfy condition #1}] \\ c & < {1 \over -1} \\ c & < -1 \\ \\ \\ \text{Possible values: } & a = -1, c = -2 \end{align}

\begin{align*} y & = {5 \over 4}x^2 - 20x + 340 \\ \\ \text{When } & x = 0, \\ y & = {5 \over 4}(0)^2 - 20(0) + 340 \\ & = 340 \\ \\ 340 \text{ is } & \text{the fixed cost of assembly} \end{align*}

\begin{align*} y & = {5 \over 4}x^2 - 20x + 340 \\ & = {5 \over 4}(x^2 - 16x) + 340 \\ & = {5 \over 4} \left[ x^2 - 16x + \left(16 \over 2\right)^2 - \left(16 \over 2\right)^2 \right] + 340 \\ & = {5 \over 4} ( x^2 - 16x + 8^2 - 8^2 ) + 340 \\ & = {5 \over 4} [ (x - 8)^2 - 64 ] + 340 \\ & = {5 \over 4} (x - 8)^2 - 80 +34 0\\ & = {5 \over 4} (x - 8)^2 + 260 \\ \\ \text{Minimum cost } (y) & = 260 \text{ when } x = 8 \\ \\ \therefore \text{Not possible } & \text{for cost to be } \$ 250 \end{align*}