# What is the equation of the normal line of f(x)= x^3-3x^2-2x+6 at x = 4?

Feb 3, 2016

Equation of normal:

$y = - \frac{1}{22} x + \frac{156}{11}$

#### Explanation:

$f \left(4\right) = 14$

The line of normal will pass through the point $\left(4 , 14\right)$.

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

$f ' \left(4\right) = 22$

From coordinate geometry, the gradient of the line of normal, $m$, will have a value of

$m = \frac{- 1}{22}$

The $y$-intercept, $c$, is given by

$c = y - m x$

$= 14 - \left(- \frac{1}{22}\right) \cdot \left(4\right)$

$= \frac{156}{11}$

Equation of normal:

$y = - \frac{1}{22} x + \frac{156}{11}$

Here is the graph