Question

What is the equation of the line normal to `f(x)=x^2-x-2` at `x=2`?

Answer 1

`y = -1/3x+2/3`

Explanation:

`f(x)=x^2-x-2` at `x=2` we get `y=f(2) = 0`

`f'(x) = 2x-1`, so the slope of the tangent at `(2,0)` is `f'(2)=3`.

Therefore the slope of the normal line is `-1/3`. And the equation of the line through `(2,0)` with slope `-1/3` is `y = -1/3x+2/3`.