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

Nov 10, 2015

$y = - \frac{4}{7} x + \frac{36}{7}$

#### Explanation:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({\left(3 {x}^{2} + 2 x\right)}^{\frac{1}{2}}\right)$

$= \frac{1}{2} {\left(3 {x}^{2} + 2 x\right)}^{- \frac{1}{2}} \left(6 x + 2\right)$

The tangent line should pass through the point $\left(2 , f \left(2\right)\right)$, and have a gradient of $- \frac{1}{f ' \left(2\right)}$.

$f \left(2\right) = 4$

$f ' \left(2\right) = \frac{7}{4}$

Equation of tangent line:

$\frac{y - f \left(2\right)}{x - 2} = - \frac{1}{f ' \left(2\right)}$

$\frac{y - 4}{x - 2} = - \frac{4}{7}$

$y = - \frac{4}{7} x + \frac{36}{7}$