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

Apr 13, 2018

The equation of the normal line is $y = - \frac{1}{14} x - \frac{85}{12}$.

#### Explanation:

We have the function, $f \left(x\right) = {x}^{3} - {x}^{2}$.

At $x = - 2$, $f \left(x\right) = {\left(- 2\right)}^{3} - {\left(- 2\right)}^{2}$

$= - 8 - 4$

$= - 12$

So, the point is at $\left(- 2 , - 12\right)$.

First, we find the slope of the tangent line at $x = - 2$, and that's the derivative of the function.

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

So at $x = - 2$,

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

$= 3 \cdot 4 + 4$

$= 14$

Now, the normal line is perpendicular to the tangent line, and we know that the slope of the normal line is the negative reciprocal of the tangent line, as

${m}_{1} {m}_{2} = - 1$, where ${m}_{1} , {m}_{2}$ are the gradients of the tangent line and normal line, respectively.

And so, the slope becomes, $- \frac{1}{14}$.

The point-slope form states that,

$y - {y}_{0} = m \left(x - {x}_{0}\right)$

And so,

$y - \left(- 12\right) = - \frac{1}{14} \left(x - \left(- 2\right)\right)$

$y + 12 = - \frac{1}{14} \left(x + 2\right)$

$y + 12 = - \frac{1}{14} x - \frac{1}{7}$

$y = - \frac{1}{14} x - \frac{1}{7} - 12$

$= - \frac{1}{14} x - \frac{85}{12}$

Here is a graph of the line: