Answer
$y-1=-4(x+1)$ or $y=-4x-3$
Work Step by Step
Let the given point be $P(-1,1)$.
Let the slope of the secant line from $P$ to a distinct point $Q(x,x^4)$ be $m$.
$\implies m=\dfrac{x^4-1}{x+1}$
Now simplify $ m=\dfrac{x^4-1}{x+1}$.
We get, $m=\dfrac{(x^2-1)(x^2+1)}{x+1}=\dfrac{(x-1)(x+1)(x^2+1)}{x+1}=(x-1)(x^2+1)$
If point $Q$ moves toward the point $P$ on the curve.
Then, $x$ approaches $-1$.
That means $m$ approaches to $-2\times2=-4$
We know that if $Q$ gets closer to $P$, then the slope of secant gets closer to the slope of the tangent at $P$.
Hence, the slope of the tangent at $P$ is $-4$.
Thus, the equation of the tangent line at $P$ can be written using point-slope form of a line as follows:
$y-1=-4(x+1)$ or $y=-4x-3$
