Answer
The equation of the tangent line is $y=12x+16$.
Work Step by Step
The slope $m$ of the tangent line of the curve $f(x)$ at $A(a,b)$ is also the slope of the curve at that point, which is calculated by $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$
$$y=f(x)=x^3\hspace{1cm}A(-2,-8)$$
1) Find the slope $m$ of the tangent: $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$
Here $a=-2$ and $f(a)=b=-8$.
$$m=\lim_{h\to0}\frac{(h-2)^3-(-8)}{h}=\lim_{h\to0}\frac{(h^3-6h^2+12h-8)+8}{h}$$ $$m=\lim_{h\to0}\frac{h^3-6h^2+12h}{h}=\lim_{h\to0}(h^2-6h+12)$$ $$m=0^2-6\times0+12=12$$
2) Find the equation of the tangent line at $A(-2,-8)$:
The tangent line would have this form: $$y=12x+m$$
Substitute $A(-2,-8)$ here to find $m$: $$12\times(-2)+m=-8$$ $$-24+m=-8$$ $$m=16$$
So the equation of the tangent line is $y=12x+16$.
