Answer
The slope of the tangent line of the curve at $P$ is $-9$.
The equation of the tangent line of the curve at $P$ is $y=-9x-2$
Work Step by Step
$$y=x^3-12x\hspace{1cm} P(1,-11)$$
1) First, take $Q(1+h,y)$ to be a nearby point of $P$ on the graph of the function.
$$y=(1+h)^3-12(1+h)=1+3h+3h^2+h^3-12-12h$$ $$y=h^3+3h^2-9h-11$$
So $Q(1+h,h^3+3h^2-9h-11)$
2) Find the slope of the secant $PQ$:
$$\frac{\Delta y}{\Delta x}=\frac{h^3+3h^2-9h-11-(-11)}{1+h-1}=\frac{h^3+3h^2-9h}{h}=h^2+3h-9$$
3) Find out what happens if $Q$ approaches $P$
As $Q$ approaches $P$, $1+h$ will gradually approach $1$, while $h^3+3h^2-9h-11$ approaches $-11$. Both of these mean that $h$ will approach $0$ and hence secant slope $h^2+3h-9$ will approach $-9$.
So we take $-9$ 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: $$y=-9x+b$$
Replace the coordinates of $P$ here to find $b$:
$$(-9)\times1+b=-11$$ $$-9+b=-11$$ $$b=-2$$
Therefore, the equation of the tangent line of the curve at $P$ is $y=-9x-2$
