Answer
(a) The slope of the tangent of the curve at $P$ is $-4$.
(b) The equation of the tangent line of the curve at $P$ is $y=-4x+11$
Work Step by Step
$$y=7-x^2\hspace{1cm} P(2,3)$$
1) First, take $Q(2+h,y)$ to be a nearby point of $P$ on the graph of the function.
$$y=7-(2+h)^2=7-(4+4h+h^2)=7-4-4h-h^2=3-4h-h^2$$
So $Q(2+h,3-4h-h^2)$
2) Find the slope of the secant $PQ$:
$$\frac{\Delta y}{\Delta x}=\frac{3-4h-h^2-3}{2+h-2}=\frac{-4h-h^2}{h}=-4-h$$
3) Find out what happens if $Q$ approaches $P$
As $Q$ approaches $P$, $2+h$ will gradually approach $2$, while $3-4h-h^2$ approaches $3$. Both of these mean that $h$ will approach $0$ and hence secant slope will approach $-4$.
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 $b$:
$$-4\times2+b=3$$ $$-8+b=3$$ $$b=11$$
Therefore, the equation of the tangent line of the curve at $P$ is $y=-4x+11$
