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

$y = - \frac{x}{6} - \frac{11}{3}$ or $x + 6 y = - 22$

#### Explanation:

the given: $f \left(x\right) = {x}^{3} - 6 x$ at $x = 2$

Find the value of the function at $x = 2$

$f \left(2\right) = {\left(2\right)}^{3} - 6 \left(2\right) = 8 - 12 = - 4$

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

The point on the curve is $\left(2 , - 4\right)$

Find first derivative $f ' \left(x\right)$ then evaluate at $x = 2$

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

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

$m = 6$ to be used for Tangent line.

$m$(perpendicular)$= - \frac{1}{6}$ to be used for the Normal Line

Use now the $m = - \frac{1}{6}$ and the point $\left(2 , - 4\right)$

The Normal Line:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - \left(- 4\right) = \left(- \frac{1}{6}\right) \left(x - 2\right)$

$y + 4 = \left(- \frac{1}{6}\right) \left(x - 2\right)$

$6 y + 24 = - x + 2$

$6 y = - x - 22$

$y = - \frac{x}{6} - \frac{11}{3}$ the required Normal Line