Answer
$$ y=12x+16$$
Work Step by Step
Given $$f(x)=x^{3}, \quad a=-2$$
Since $f(-2)=-8$ and
$$ f^{\prime}(a) =\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$$
Then
\begin{aligned}
f^{\prime}(-2) & =\lim _{h \rightarrow 0}\frac{(-2+h)^{3}-(-2)^{3}}{h} \\
&=\lim _{h \rightarrow 0} \frac{-8+12 h-6 h^{2}+h^{3}+8}{h}\\
&=\lim _{h \rightarrow 0}(12-6h+h^2)\\
&=12
\end{aligned}
Then the tangent line at $x=-2$ is given by
\begin{align*}
\frac{y-y_1}{x-x_1}&=f'(a)\\
\frac{y+8}{x+2}&=12\\
y+8&= 12(x+2)\\
\end{align*}
Hence $$ y=12x+16$$
