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

Jan 7, 2017

$y = - \frac{1}{1.9} x + 4$

#### Explanation:

Given -

$y = \sqrt{3 {x}^{3} - 2 x}$

It slope at any point is given by its first derivative.

dy/dx=(9x^2-2)/(2* sqrt(3x^3-2x)

At $x = 2$ The slope is -

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{9 \cdot {2}^{2} - 2}{2 \cdot \sqrt{3 \cdot {2}^{3} - 2 \cdot 2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{36 - 2}{2 \cdot \sqrt{24 - 4}}$

dy/dx=34/(2*sqrt20))
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{34}{2 \cdot 4.5}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{34}{9} = 3.8$
${m}_{1} = 3.8$

At $x = 2$ the value of the function is

$y = \sqrt{3 \cdot {2}^{3} - 2 \cdot 2}$
$y = \sqrt{24 - 4} = \sqrt{20} = 4.5$

The normal is passing through the point $\left(2 , 4.5\right)$

The slope of the normal ${m}_{2} = \frac{- 1}{m} _ 1 = \frac{- 1}{3.8}$

To find the equation of the normal -

$y = m x + c$
$m x + c = y$
$\left(- \frac{1}{3.8}\right) \left(2\right) + c = 4.5$
$- \frac{2}{3.8} + c = 4.5$
$- \frac{1}{1.9} + c = 4.5$
$c = 4.5 - \frac{1}{1.9} = \frac{8.6 - 1}{1.9} = 4$
The equation is -
$y = - \frac{1}{1.9} x + 4$