Answer
$ y=-8x-7$
Work Step by Step
We have: $y(x)=(2x^2-3)^{-2}; x=-1$
Now, $y(-1)=[(2(-1)^2-3)^{-2}]=-1$
We differentiate both sides with respect to $x$.
$y^{\prime}(x)=-2 (2x^2-3)^{-3}(4x)=-8x (2x^2-3)^{-3}$
The equation of a tangent line at $(-1,1)$ is:
$y-y_1=m(x-x_1)\\ y-1=-8 (x+1) \\ y=-8x-7$
