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

(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 $y=-2x-1$
$$y=x^2-4x\hspace{1cm} P(1,-3)$$ 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 $y=-2x-1$