The maximum slope of the curve $y=-x^{3}+3 x^{2}+2 x-27$ is

$y=-x^{3}+3 x^{2}+2 x-27$ On differentiating w.r.t. $x$, we get $\frac{d y}{d x}=-3 x^{2}+6 x+2$ Slope of the curve $=-3 x^{2}+6 x+2$ Let $m=-3 x^{2}+6 x+2$ $\therefore \quad \frac{d m}{d x}=-6 x+6$ Put $\frac{d m}{d x}=0$ to find the critical points, $\begin{aligned} \Rightarrow &-6 x+6 &=0 \\ \Rightarrow \quad x &=1 \\ &\left(\frac{d^{2} m}{d x^{2}}\right)_{x=1}=-6 < 0 \end{aligned}$ Slope is maximum at $x=1$ The maximum slope, $\begin{aligned} m_{\max } &=-3(1)^{2}+6(1)+2 \\ &=-3+6+2=5 \end{aligned}$