Answer
(a) The slope of the tangent line of the curve at $P$ is $2$.
(b) The equation of the tangent line of the curve at $P$ is $y=2x-7$
Work Step by Step
$$y=x^2-2x-3\hspace{1cm} P(2,-3)$$
1) First, take $Q(2+h,y)$ to be a nearby point of $P$ on the graph of the function.
$$y=(2+h)^2-2(2+h)-3=4+4h+h^2-4-2h-3$$ $$y=2h+h^2-3=h^2+2h-3$$
So $Q(2+h,h^2+2h-3)$
2) Find the slope of the secant $PQ$:
$$\frac{\Delta y}{\Delta x}=\frac{h^2+2h-3-(-3)}{2+h-2}=\frac{h^2+2h-3+3}{h}=\frac{h^2+2h}{h}=h+2$$
3) Find out what happens if $Q$ approaches $P$
As $Q$ approaches $P$, $2+h$ will gradually approach $2$, while $h^2+2h-3$ approaches $-3$. Both of these mean that $h$ will approach $0$ and hence secant slope $h+2$ will approach $2$.
So we take $2$ the slope of the tangent line of the curve at $P$.
4) The equation of the tangent line at $P$ would have this form: $$y=2x+b$$
Replace the coordinates of $P$ here to find $b$:
$$2\times2+b=-3$$ $$4+b=-3$$ $$b=-7$$
Therefore, the equation of the tangent line of the curve at $P$ is $y=2x-7$
