Answer
(a) The slope of the tangent line of the curve at $P$ is $4$.
(b) The equation of the tangent line of the curve at $P$ is $y=4x-9$
Work Step by Step
$$y=x^2-5\hspace{1cm} P(2,-1)$$
1) First, take $Q(2+h,y)$ to be a nearby point of $P$ on the graph of the function.
$$y=(2+h)^2-5=4+h^2+4h-5=h^2+4h-1$$
So $Q(2+h,h^2+4h-1)$
2) Find the slope of the secant $PQ$:
$$\frac{\Delta y}{\Delta x}=\frac{h^2+4h-1-(-1)}{2+h-2}=\frac{h^2+4h}{h}=h+4$$
3) Find out what happens if $Q$ approaches $P$
As $Q$ approaches $P$, $2+h$ will gradually approach $2$, while $h^2+4h-1$ approaches $-1$, meaning $h$ approaches $0$ and secant slope approaches $4$.
So we take $4$ the slope of the tangent of the curve at $P$.
4) The equation of the tangent line at $P$ would have this form: $$y=4x+b$$
Replace the coordinates of $P$ here to find $b$:
$$4\times2+b=-1$$ $$8+b=-1$$ $$b=-9$$
Therefore, the equation of the tangent line of the curve at $P$ is $y=4x-9$
