Answer
The slope of the tangent line of the curve at $P$ is $0$.
The $Ox$ line is the tangent line of the curve at $P$
Work Step by Step
$$y=x^3-3x^2+4\hspace{1cm} P(2,0)$$
1) First, take $Q(2+h,y)$ to be a nearby point of $P$ on the graph of the function.
$$y=(2+h)^3-3(2+h)^2+4$$ $$y=(8+3\times4h+3\times2h^2+h^3)-3(4+4h+h^2)+4$$ $$y=8+12h+6h^2+h^3-12-12h-3h^2+4$$ $$y=3h^2+h^3$$
So $Q(2+h,3h^2+h^3)$
2) Find the slope of the secant $PQ$:
$$\frac{\Delta y}{\Delta x}=\frac{3h^2+h^3-0}{2+h-2}=\frac{3h^2+h^3}{h}=3h+h^2$$
3) Find out what happens if $Q$ approaches $P$
As $Q$ approaches $P$, $2+h$ will gradually approach $2$, while $3h^2+h^3$ approaches $0$. Both of these mean that $h$ will approach $0$ and hence secant slope $3h+h^2$ will approach $0$ as well.
So we take $0$ 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: $$y=0x+b$$ $$y=b$$
Replace the coordinates of $P$ here to find $b$:
$$b=0$$
Therefore, the equation of the tangent line of the curve at $P$ is $y=0$, which is the $Ox$ line.
