## Calculus 10th Edition

$y=Ce^{-\frac{1}{2}x}$
First we must find the slope of the tangent line that runs between $(x,y)$ and $(x+2,0)$. $Slope=\frac{0-y}{x+2-x}=-\frac{y}{2}$ Therefore, the derivative of the function $f$ must be $-\frac{y}{2}$. $f'(x)=\frac{dy}{dx}=-\frac{y}{2}$ $\frac{1}{y}dy=-\frac{1}{2}dx$ $\int\frac{1}{y}dy=\int-\frac{1}{2}dx$ $\ln|y|=-\frac{1}{2}x+C_1$ $|y|=e^{-\frac{1}{2}x+C_1}$ $y=^+_-e^{-\frac{1}{2}x+C_1}$ $y=^+_-e^{C_1}e^{-\frac{1}{2}x}$ $y=Ce^{-\frac{1}{2}x}$, where $C=^+_-e^{C_1}$