Tangent and normal to an exponential curve

Statement

Given the curve $y=e^{-x}+2x$, find the equation of the tangent line at the point with $x=0$ and the equation of the normal line at the same point.

Solution

At the point: $y(0)=e^{0}+0=1$, so $P(0,1)$.

Derivative: $y'=-e^{-x}+2$, hence $y'(0)=-1+2=1$: the tangent has slope $m=1$.

Tangent: $y=1+1\cdot(x-0)=x+1$, i.e. slope $1$ and intercept $1$.

Normal: slope $-\dfrac{1}{m}=-1$, so $y=1-x$.

$\boxed{\text{tangent } y=x+1;\qquad \text{normal } y=-x+1}$