Answer
(a) The slope of the tangent line of the curve at $P$ is $12$.
(b) The equation of the tangent line of the curve at $P$ is $y=12x-16$.
Work Step by Step
$$y=x^3\hspace{1cm} P(2,8)$$
1) First, take $Q(2+h,y)$ to be a nearby point of $P$ on the graph of the function.
$$y=(2+h)^3=8+3\times2^2\times h+3\times 2h^2+h^3$$ $$y=8+12h+6h^2+h^3$$
So $Q(2+h,8+12h+6h^2+h^3)$
2) Find the slope of the secant $PQ$:
$$\frac{\Delta y}{\Delta x}=\frac{8+12h+6h^2+h^3-8}{2+h-2}=\frac{12h+6h^2+h^3}{h}=12+6h+h^2$$
3) Find out what happens if $Q$ approaches $P$
As $Q$ approaches $P$, $2+h$ will gradually approach $2$, while $8+12h+6h^2+h^3$ approaches $8$. Both of these mean that $h$ will approach $0$ and hence secant slope $12+6h+h^2$ will approach $12$.
So we take $12$ to be the slope of the tangent line of the curve at $P$.
4) The equation of the tangent line at $P$ would have this form: $$y=12x+b$$
Replace the coordinates of $P$ here to find $b$:
$$12\times2+b=8$$ $$24+b=8$$ $$b=-16$$
Therefore, the equation of the tangent line of the curve at $P$ is $y=12x-16$
