# What is the equation of the line normal to f(x)=x^3-6x^2-2  at x=1?

Dec 30, 2015

$y + 7 = \frac{1}{9} \left(x - 1\right)$

#### Explanation:

Find the slope of the tangent line, to which the normal line is perpendicular.

$f \left(1\right) = - 7$

The line will pass through the point $\left(1 , - 7\right)$.

$f ' \left(x\right) = 3 {x}^{2} - 12 x$

$f ' \left(1\right) = - 9$

If the slope of the tangent line is $- 9$, the slope of the normal line will be the opposite reciprocal of $- 9$ since the lines are perpendicular. This means the slope is $\frac{1}{9}$.

Write the equation in point-slope form:

$y + 7 = \frac{1}{9} \left(x - 1\right)$