\begin{align*} y & = x^3 - 2x^2 + ax + c \\ \\ {dy \over dx} & = 3x^2 - 2(2)x + a \\ & = 3x^2 - 4x + a \\ \\ \text{When } & x = 4 \text{ and } {dy \over dx} = 0, \phantom{000000} [(4, -91) \text{ is a stationary point}] \\ 0 & = 3(4)^2 - 4(4) + a \\ 0 & = 48 - 16 + a \\ 0 & = 32 + a \\ -32 & = a \\ \\ y & = x^3 - 2x^2 - 32x + c \\ \\ \text{Using } & (4, -91), \\ -91 & = (4)^3 - 2(4)^2 - 32(4) + c \\ -91 & = 64 - 32 - 128 + c \\ -91 & = -96 + c \\ 5 & = c \\ \\ \therefore a & = -32, c = 5 \end{align*}

\begin{align*} y & = x^3 - 3x^2 + 12x + 2 \\ \\ {dy \over dx} & = 3x^2 - 3(2)x + 12 \\ & = 3x^2 - 6x + 12 \\ & = 3(x^2 - 2x) + 12 \\ & = 3 \left[ x^2 - 2x + \left(2 \over 2\right)^2 - \left(2 \over 2\right) \right] + 12 \phantom{000000} [\text{Complete the square}] \\ & = 3 [ (x - 1)^2 - 1 ] + 12 \\ & = 3(x - 1)^2 - 3 + 12 \\ & = 3(x - 1)^2 + 9 \end{align*}
\begin{align*} \text{For all values} & \text{ of } x , \\ (x - 1)^2 & \ge 0 \\ 3(x - 1)^2 & \ge 0 \\ 3(x - 1)^2 + 9 & \ge 9 \\ \\ \implies {dy \over dx} \ge 9 & \text{ and } {dy \over dx} \ne 0 \\ \\ \therefore \text{Curve has } & \text{no stationary point} \end{align*}

\begin{align*} y & = {1 \over 2x + 3} \\ y & = (2x + 3)^{-1} \\ \\ {dy \over dx} & = (-1)(2x + 3)^{-2} (2) \phantom{000000} [\text{Chain rule}] \\ & = -2 (2x + 3)^{-2} \\ & = - {2 \over (2x +3)^2 } \end{align*} \begin{align*} \text{For } x > -1.5 &, \\ (2x + 3)^2 & > 0 \\ {2 \over (2x + 3)^2} & > 0 \\ -{2 \over (2x + 3)^2} & < 0 \\ \\ \implies {dy \over dx} < 0 & \text{ and } {dy \over dx} \ne 0 \\ \\ \therefore \text{Curve has } & \text{no stationary point} \end{align*}