# What is the equation of the line normal to  f(x)=sqrt(x/(x+2)  at  x=-3?

The Normal Line: $y = - \sqrt{3} x - 2 \sqrt{3}$

#### Explanation:

Given $f \left(x\right) = \sqrt{\frac{x}{x + 2}}$

$f \left(- 3\right) = \sqrt{3}$

the point $\left(- 3 , \sqrt{3}\right)$

$f ' \left(- 3\right)$ = slope for tangent line

$- \frac{1}{f ' \left(- 3\right)}$ = slope for the Normal Line=$- \sqrt{3}$