Answer
$y=-x$
Work Step by Step
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$Find the point at which $t = -1$
$x|_{t=-1}=(-1)^3+1=0$
$y|_{t=-1}=(-1)^4+(-1)=0$
Point: $(0,0)$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$Now find $\displaystyle \frac{dy}{dx}$
$dx=3t^2$
$dy=4t^3+1$
$\displaystyle \frac{dy}{dx}=\frac{4t^3+1}{3t^2}$
$\qquad\qquad\qquad\qquad\qquad$Now plug $t=-1$ in for $t$ and find the slope of the line:
$\displaystyle \frac{dy}{dx}|_{t=-1}=\frac{-4+1}{3}=\frac{-3}{3}=-1$
$\qquad\qquad\qquad\qquad\qquad$Now use the point-slope form of a line to
$\qquad\qquad\qquad\qquad\qquad$find the equation tangent to the parametric curve.
$y-y_1=m(x_1-x)\quad$
$y-0=(-1)(x-0)$
$y=-x$
