Question

What is the equation of the line normal to `f(x)=-x^2+3x - 1` at `x=-1`?

Answer 1

`y+5=-1/5 (x+1)`

Explanation:

`f'(x)=-2x+3`
slope of tangent line when x = -1 is `5` so slope of normal is `-1/5`.
When `x =-1, y=-5`
Equation of normal: `y+5=-1/5 (x+1)`