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 $y=-2x-1$
Work Step by Step
$$y=x^2-4x\hspace{1cm} P(1,-3)$$
1) First, take $Q(1+h,y)$ to be a nearby point of $P$ on the graph of the function.
$$y=(1+h)^2-4(1+h)=1+2h+h^2-4-4h=h^2-2h-3$$
So $Q(1+h,h^2-2h-3)$
2) Find the slope of the secant $PQ$:
$$\frac{\Delta y}{\Delta x}=\frac{h^2-2h-3-(-3)}{1+h-1}=\frac{h^2-2h-3+3}{h}=\frac{h^2-2h}{h}=h-2$$
3) Find out what happens if $Q$ approaches $P$
As $Q$ approaches $P$, $1+h$ will gradually approach $1$, 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 $-2$ 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: $$y=-2x+b$$
Replace the coordinates of $P$ here to find $b$:
$$-2\times1+b=-3$$ $$-2+b=-3$$ $$b=-1$$
Therefore, the equation of the tangent line of the curve at $P$ is $y=-2x-1$
