Answer
x-intercept $a=-4/3$,
y-intercept $b=16$
Work Step by Step
Step 1. Given $y=x^3$, we have $y'=3x^2$, which gives the slope of tangent lines to the curve.
Step 2. At $(-2,-8)$, the slope $m=y'=3(-2)^2=12$ and the tangent line equation is $y+8=12(x+2)$ or $y=12x+16$
Step 3. To find the x- and y-intercepts, first let $x=0$; we get the y-intercept as $b=16$. Second, we let $y=0$ and we can find the x-intercept as $a=-16/12=-4/3$
