# What is the equation of the normal line of f(x)=x^3-8x^2+9x-3 at x=-1?

Nov 22, 2016

$f \left(- 1\right) = - 21$

$f ' \left(x\right) = 3 {x}^{2} - 16 x + 9$

$f ' \left(- 1\right) = 3 + 16 + 9 = 28 = {m}_{\tan}$

The slope of the normal line at $x = - 1$ is the negative reciprocal of the slope of the tangent line $m = 28$ which would be $- \frac{1}{28}$.

So, using a slope formula:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

We get:

$y + 21 = - \frac{1}{28} \left(x + 1\right)$

$y = - \frac{1}{28} x - \frac{589}{28}$