Calculus 8th Edition

$$y =-x+2$$
Given $$y=(x+2)e^{-x},\ \ \ (0,2)$$ Since $$y'= -(x+2)e^{-x}+e^{-x}=-( x+1)e^{-x}$$ Then $m=y'\bigg|_{(0,2)}=-1$, hence the tangent line given by \begin{align*} \frac{y-y_1}{x-x_1}&=m\\ \frac{y-2}{x-0}&=-1\\ y&=-x+2 \end{align*}