Question

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

Answer 1

`y=sqrt3/2x-sqrt3`

Explanation:

Equation of normal at `x=x_0` on the curve `y=f(x)` is perpendicular to the tangent at `x=x_0` at `f(x)`.

As slope of tangent at `x=x_0` on the curve `f(x)`, is given by `f'(x_0)`, slope of normal is `-1/(f(x_0))` and equation of normal is

`y=-1/(f(x_0))(x-x_0)+f(x_0)`

Here, we have `y=f(x)=-sqrt((x+1)(x+3))`

and `f'(x)=(df)/(dx)` - using chain rule

= `-1xx1/(2sqrt((x+1)(x+3)))[1xx(x+3)+1xx(x+1)]`

= `-1/(2sqrt((x+1)(x+3)))(2x+4)=-(x+2)/sqrt((x+1)(x+3))`

and while `f(0)=-sqrt3`, `f'(0)=-2/sqrt3`

as such slope of normal at `x=0` is `sqrt3/2` and

equation of normal is `y=sqrt3/2x-sqrt3`

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 `(-1,-3)` as `f(x)=-sqrt((x+1)(x+3))` is not defined in the interval `(-1,-3)`.