Chapter 2 - Section 2.1 - Rates of Change and Tangents to Curves - Exercises - Page 56: 7

(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 $y=4x-9$

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