Answer
tangent line:
$y = -x+2$
normal line:
$y = x+2$
Work Step by Step
$y = (2+x)e^{-x}$
$y' = e^{-x}-(2+x)e^{-x}$
We can find $y'$ when $x = 0$:
$y' = e^{0}-(2+0)e^{0} = 1-2 = -1$
The tangent line has a slope of $-1$ at this point.
We can find the equation of the tangent line:
$y-2 = (-1)(x-0)$
$y = -x+2$
The normal line has a slope of $1$ at this point.
We can find the equation of the normal line:
$y-2 = (1)(x-0)$
$y = x+2$
