# What is the equation of the normal line of f(x)=-2x^3+18x^2-25x at x=-1/3?

Oct 30, 2017

$y = - \frac{3}{113} \cdot x + \frac{31726}{3051}$

#### Explanation:

I took diferential both sides,

$f ' \left(x\right) = - 6 {x}^{2} + 36 x - 25$

$f ' \left(- \frac{1}{3}\right) = - \frac{2}{3} - 12 - 25 = - \frac{113}{3}$

Hence I found slope of normal line: $- \frac{3}{113}$

I found a point, which passes from it:

$f \left(- \frac{1}{3}\right) = \frac{2}{27} + \frac{18}{9} + \frac{25}{3} = \frac{281}{27}$

Thus,

$y - \frac{281}{27} = - \frac{3}{113} \cdot \left(x + \frac{1}{3}\right)$

$y - \frac{281}{27} = - \frac{3}{113} \cdot x - \frac{1}{113}$

$y = - \frac{3}{113} \cdot x + \frac{31726}{3051}$