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

Feb 24, 2017

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

#### Explanation:

Equation of normal at $x = {x}_{0}$ on the curve $y = f \left(x\right)$ is perpendicular to the tangent at $x = {x}_{0}$ at $f \left(x\right)$.

As slope of tangent at $x = {x}_{0}$ on the curve $f \left(x\right)$, is given by $f ' \left({x}_{0}\right)$, slope of normal is $- \frac{1}{f \left({x}_{0}\right)}$ and equation of normal is

$y = - \frac{1}{f \left({x}_{0}\right)} \left(x - {x}_{0}\right) + f \left({x}_{0}\right)$

Here, we have $y = f \left(x\right) = - \sqrt{\left(x + 1\right) \left(x + 3\right)}$

and $f ' \left(x\right) = \frac{\mathrm{df}}{\mathrm{dx}}$ - using chain rule

= $- 1 \times \frac{1}{2 \sqrt{\left(x + 1\right) \left(x + 3\right)}} \left[1 \times \left(x + 3\right) + 1 \times \left(x + 1\right)\right]$

= $- \frac{1}{2 \sqrt{\left(x + 1\right) \left(x + 3\right)}} \left(2 x + 4\right) = - \frac{x + 2}{\sqrt{\left(x + 1\right) \left(x + 3\right)}}$

and while $f \left(0\right) = - \sqrt{3}$, $f ' \left(0\right) = - \frac{2}{\sqrt{3}}$

as such slope of normal at $x = 0$ is $\frac{\sqrt{3}}{2}$ and

equation of normal is $y = \frac{\sqrt{3}}{2} x - \sqrt{3}$

graph{(y-sqrt3/2x+sqrt3)(y+sqrt((x+1)(x+3)))=0 [-12.92, 7.08, -8.52, 1.48]}
The curve or normal is not appearing in the interval $\left(- 1 , - 3\right)$ as $f \left(x\right) = - \sqrt{\left(x + 1\right) \left(x + 3\right)}$ is not defined in the interval $\left(- 1 , - 3\right)$.