# What is the equation of the normal line of f(x)=x^3-11x^2-5x-2 at x=0?

Nov 22, 2016

$y = \frac{1}{5} x - 2$

#### Explanation:

Since $f \left(0\right) = - 2$, the normal line passes through the point $\left(0 , - 2\right)$.

To find the slope of the normal line, first find the slope of the tangent line. Since the tangent line and normal line are perpendicular, their slopes will be the opposite reciprocals of one another.

The slope of the tangent line at $x = 0$ is $f ' \left(0\right)$, so the slope of the normal line is $- \frac{1}{f ' \left(0\right)}$.

First, through the power rule, find the function's derivative: $f ' \left(x\right) = 3 {x}^{2} - 22 x - 5$. Thus $f ' \left(0\right) = - 5$ and $- \frac{1}{f ' \left(0\right)} = \frac{1}{5}$.

So, the normal line has slope $\frac{1}{5}$ and passes through the point $\left(0 , - 2\right)$. We could write this using point-slope form, but we already know the slope $m = \frac{1}{5}$ and $y$-intercept $b = - 2$ for a linear equation in $y = m x + b$ form: $y = \frac{1}{5} x - 2$.