Answer
$y=Ce^{-\frac{1}{2}x}$
Work Step by Step
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 $C=^+_-e^{C_1}$
