Answer
$y=12x+16$, as shown in the figure.
Work Step by Step
Given the curve as $y=x^3$ and a point $(-2,-8)$ on the curve, we can find the slope of the tangent line at this point as:
$m=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}==\lim_{h\to0}\frac{(-2+h)^3-(-2)^3}{h}=\lim_{h\to0}\frac{12h-6h^2+h^3}{h}=\lim_{h\to0}(12-6h+h^2)=12$
The equation for the tangent line at $(-2,-8)$ can then be written as $y+8=12(x+2)$ or $y=12x+16$, as shown in the figure.