Question

What is the equation of the line normal to `f(x)=sqrt(x/(x+2)` at `x=-3`?

Answer 1

The Normal Line: `y=-sqrt(3)x-2 sqrt(3)`

Explanation:

Given `f(x) = sqrt (x/(x+2))`

`f(-3)=sqrt(3)`

the point `(-3, sqrt(3))`

`f' (-3)` = slope for tangent line

`-1/(f' (-3))` = slope for the Normal Line=`-sqrt(3)`