# What is the equation of the line that is normal to f(x)= e^(2x) sqrt( 2x+2)  at  x=1 ?

May 29, 2016

$y - 2 {e}^{2} = \left(- \frac{2}{9 {e}^{2}}\right) \left(x - 1\right)$

#### Explanation:

At x=1, f(x)= $2 {e}^{2}$. The normal is required at point (1, $2 {e}^{2}$)

For slope of tangent to f(x), f'(x) = 2e^(2x) sqrt(2x+2)+ e^(2x) / sqrt(2x+2 .

At x=1, the slope of tangent =$4 {e}^{2} + {e}^{2} / 2$ = $\frac{9 {e}^{2}}{2}$

Slope of normal would thus be $- \frac{2}{9 {e}^{2}}$

Equation of normal line would be

$y - 2 {e}^{2} = \left(- \frac{2}{9 {e}^{2}}\right) \left(x - 1\right)$