Answer
$\frac{4}{9}$
Work Step by Step
Given: $x=t^{3}+6t+1$ and $y=2t-t^{2}$
$\frac{dx}{dt}=3t^{2}+6$
$\frac{dy}{dt}=2-2t$
$\frac{dy}{dx}=\frac{{dy}/{dt}}{{dx}/{dt}}$
$=\frac{2-2t}{3t^{2}+6}$
Let $m$ be the slope of the tangent line to the given curve.
$m=\frac{dy}{dx}|_{t=-1}=\frac{2-2t}{3t^{2}+6}|_{t=-1}$
$=\frac{2-2\times (-1)}{3(-1)^{2}+6}$
Hence, $m=\frac{4}{9}$
