Question

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

Answer 1

`y=sqrt(3)/2x-sqrt(3)`

Explanation:

Given `f(x)=-sqrt((x+1)(x+3))`
we get `f(0)=-sqrt(3)`

Now we will compute the slope of the normal line:

`f'(x)=-((x+1)(x+3))^(-1/2)*(x+3+x+1)`so
`f'(x)=-(4+2x)/(2sqrt((x+1)(x+3))`

`f'(0)=-2/sqrt(3)` so the slope of our normal line is given by

`m_N=sqrt(3)/2`

`y=sqrt(3)/2x+n`

The Point is given by `(0,-sqrt(3))`

and we get `n=-sqrt(3)`

therefore our line is given by

`y=sqrt(3)/2x-sqrt(3)`