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

Jun 25, 2018

$y = \frac{\sqrt{6}}{6} x$

#### Explanation:

We get by the product and chain rule

$f ' \left(x\right) = \sqrt{2} \left(- \sqrt{x + 1} + \left(2 - x\right) \cdot \left(\frac{1}{2}\right) \cdot {\left(x + 1\right)}^{- \frac{1}{2}}\right)$

we get $f ' \left(x\right) = \sqrt{2} \cdot \left(- \sqrt{3}\right) = - \sqrt{6}$
so we get the slope of the normal line

${m}_{n} = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6}$

Now we have $f \left(2\right) = 0$
so our equation is given by

$y = \frac{\sqrt{6}}{6} x$