## University Calculus: Early Transcendentals (3rd Edition)

(a) The slope of the tangent line of the curve at $P$ is $12$. (b) The equation of the tangent line of the curve at $P$ is $y=12x-16$.
$$y=x^3\hspace{1cm} P(2,8)$$ 1) First, take $Q(2+h,y)$ to be a nearby point of $P$ on the graph of the function. $$y=(2+h)^3=8+3\times2^2\times h+3\times 2h^2+h^3$$ $$y=8+12h+6h^2+h^3$$ So $Q(2+h,8+12h+6h^2+h^3)$ 2) Find the slope of the secant $PQ$: $$\frac{\Delta y}{\Delta x}=\frac{8+12h+6h^2+h^3-8}{2+h-2}=\frac{12h+6h^2+h^3}{h}=12+6h+h^2$$ 3) Find out what happens if $Q$ approaches $P$ As $Q$ approaches $P$, $2+h$ will gradually approach $2$, while $8+12h+6h^2+h^3$ approaches $8$. Both of these mean that $h$ will approach $0$ and hence secant slope $12+6h+h^2$ will approach $12$. So we take $12$ 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: $$y=12x+b$$ Replace the coordinates of $P$ here to find $b$: $$12\times2+b=8$$ $$24+b=8$$ $$b=-16$$ Therefore, the equation of the tangent line of the curve at $P$ is $y=12x-16$