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

Mar 23, 2016

$y - 23 = \frac{1}{30} \left(x + 2\right)$

#### Explanation:

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

Thus find $f ' \left(x\right) = - 6 {x}^{2} + 2 x - 2 \to f ' \left(- 2\right) = - 30$
The slope of the normal would be $\frac{1}{30}$

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

$y - 23 = \frac{1}{30} \left(x + 2\right)$