# What is the equation of the line normal to f(x)=xe^x at x=7?

Feb 2, 2016

Equation of normal line

$y = \frac{7 \left(8 {e}^{14} + 1\right) - x}{8 {e}^{7}}$

#### Explanation:

$f \left(7\right) = 7 {e}^{7}$

The normal line has to pass through the point $\left(7 , f \left(7\right)\right)$.

$f ' \left(x\right) = \left(x + 1\right) {e}^{x}$

$f ' \left(7\right) = 8 {e}^{7}$

Since the normal line is perpendicular to the tangent line, we have the gradient of the normal line, $m$, given by

$m = \frac{- 1}{f ' \left(7\right)} = - \frac{1}{8} {e}^{- 7}$.

The $y$-intercept of the line is given by

$c = y - m x$

$= 7 {e}^{7} - \left(- \frac{1}{8} {e}^{- 7}\right) \left(7\right)$

$= 7 {e}^{- 7} \left({e}^{14} + \frac{1}{8}\right)$

Equation of normal line in slope-intercept form

$y = - \frac{1}{8} {e}^{- 7} x + 7 {e}^{- 7} \left({e}^{14} + \frac{1}{8}\right)$