# What is the equation of the normal line of f(x)= (1+2x)*e^(x^2) at x=-1/2?

May 26, 2018

The tangent line has equation y-y_0=f´(x_0)(x-x_0) and normal line y-y_0=-1/(f´(x_0))(x-x_0)

Then, f´(x)=2e^(x^2)+(1+2x)e^(x^2)·2x=

$= 2 {e}^{{x}^{2}} + 2 x {e}^{{x}^{2}} + 4 {x}^{2} {e}^{{x}^{2}}$

f´(-1/2)=2root(4)e-root(4)e+root(4)e=2root(4)e

Normal line equation is $y = - \frac{1}{2 \sqrt[4]{e}} \left(x + \frac{1}{2}\right)$

Because ${y}_{0} = 0$