# Chapter 2 - Section 2.1 - Rates of Change and Tangents to Curves - Exercises - Page 56: 8

(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 $y=-4x+11$

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