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

##### 1 Answer
Nov 27, 2015

$y = - \frac{1}{3} x + \frac{2}{3}$

#### Explanation:

$f \left(x\right) = {x}^{2} - x - 2$ at $x = 2$ we get $y = f \left(2\right) = 0$

$f ' \left(x\right) = 2 x - 1$, so the slope of the tangent at $\left(2 , 0\right)$ is $f ' \left(2\right) = 3$.

Therefore the slope of the normal line is $- \frac{1}{3}$. And the equation of the line through $\left(2 , 0\right)$ with slope $- \frac{1}{3}$ is $y = - \frac{1}{3} x + \frac{2}{3}$.