#### Answer

The $x-$intercept is $(-4/3,0)$ and the $y-$intercept is $(0,16)$.

#### Work Step by Step

$$y=x^3$$
1) First, find the derivative of $y$: $$y'=3x^2$$
We have $y'(-2)=3\times(-2)^2=3\times4=12$
2) At the point $(-2,-8)$, the tangent to the curve has the following equation: $$y=y'(-2)x+y_0$$ $$y=12x+y_0$$
To find $y_0$, we replace $x$ and $y$ with $-2$ and $-8$ respectively: $$-8=12\times(-2)+y_0$$ $$-8=-24+y_0$$ $$y_0=16$$
Therefore, the tangent to the curve at $(-2,-8)$ is $$y=12x+16$$
3) Find the $x-$intercept: for $y=0$: $$12x+16=0$$ $$x=-\frac{16}{12}=-\frac{4}{3}$$
Find the $y-$intercept: for $x=0$: $$y=12\times0+16=16$$
So the $x-$intercept is $(-4/3,0)$ and the $y-$intercept is $(0,16)$.