Question

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

Answer 1

`y-23=1/30 (x+2)`

Explanation:

First find the slope of the tangent line at the given point. It would be f' (-2).

Thus find `f'(x) =-6x^2 +2x -2 -> f'(-2)= -30`
The slope of the normal would be `1/30`

At x=-2, f(x) would be 16+4+4-1 =23. The given point is this (-2, 23)

Equation of the normal line, in point slope form would be

`y-23=1/30 (x+2)`