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

Dec 18, 2016

$y = - \frac{x}{164} + \frac{27555}{164}$

#### Explanation:

Given -

$y = 6 {x}^{3} + 2 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 18 {x}^{2} + 2$

Slope, exactly at $x = 3$

At x=3; 18(3)^2+2=164

This is the slope of the tangent ${m}_{1} = 164$.
Slope of the normal ${m}_{2} = - \frac{1}{{m}_{1}} = - \frac{1}{164}$

At x=3; 6(3)^3+2(3)=162+6=168

We have to find the equation of the line passing through the point $\left(3 , 168\right)$ with the slope $- \frac{1}{164}$

$m x + c = y$
$- \frac{1}{164} \left(3\right) + c = 168$
$c = 168 + \frac{3}{164} = \frac{27552 + 3}{164} = \frac{27555}{164}$

$y = - \frac{x}{164} + \frac{27555}{164}$