# What is the equation of the normal line of f(x)=-x^4+2x^3-x^2-9x-3 at x=-1?

Nov 24, 2015

$y = - \frac{1}{3} x + \frac{5}{3}$

#### Explanation:

$f \left(- 1\right) = - 1 - 2 - 1 + 9 - 3 = 2$

$f ' \left(x\right) = - 4 {x}^{3} + 6 {x}^{2} - 2 x - 9$

so $f ' \left(- 1\right) = 4 + 6 + 2 - 9 = 3$.

The slope of the tangent line at $\left(- 1 , 2\right)$ is $3$, so the normal line has slope $- \frac{1}{3}$.

The line with slope $m = - \frac{1}{3}$ through the point $\left(- 1 , 2\right)$ is

$y - 2 = - \frac{1}{3} \left(x - \left(- 1\right)\right)$

$y - 2 = - \frac{1}{3} x - \frac{1}{3}$

$y = - \frac{1}{3} x + \frac{5}{3}$