# What is the equation of the line normal to  f(x)=-sqrt((x+1)(x+3) at  x=0?

##### 1 Answer
Jul 30, 2018

$y = \frac{\sqrt{3}}{2} x - \sqrt{3}$

#### Explanation:

Given $f \left(x\right) = - \sqrt{\left(x + 1\right) \left(x + 3\right)}$
we get $f \left(0\right) = - \sqrt{3}$

Now we will compute the slope of the normal line:

$f ' \left(x\right) = - {\left(\left(x + 1\right) \left(x + 3\right)\right)}^{- \frac{1}{2}} \cdot \left(x + 3 + x + 1\right)$so
f'(x)=-(4+2x)/(2sqrt((x+1)(x+3))

$f ' \left(0\right) = - \frac{2}{\sqrt{3}}$ so the slope of our normal line is given by

${m}_{N} = \frac{\sqrt{3}}{2}$

$y = \frac{\sqrt{3}}{2} x + n$

The Point is given by $\left(0 , - \sqrt{3}\right)$

and we get $n = - \sqrt{3}$

therefore our line is given by

$y = \frac{\sqrt{3}}{2} x - \sqrt{3}$