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

Nov 25, 2015

$y = \frac{1}{108} x - \frac{3135}{56}$

#### Explanation:

The normal line to a tangent is perpendicular to the tangent. We can find the slope of the tangent line using the derivative of the original function, then take its opposite reciprocal to find the slope of the normal line at the same point.

$f \left(x\right) = 3 {x}^{4} - {x}^{3}$

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

$f ' \left(- 2\right) = 12 {\left(- 2\right)}^{3} - 3 {\left(- 2\right)}^{2} = 12 \left(- 8\right) - 3 \left(4\right) = - 108$

If $- 108$ is the slope of the tangent line, the slope of the normal line is $\frac{1}{108}$.

The point on $f \left(x\right)$ that the normal line will intersect is $\left(- 2 , - 56\right)$.

We can write the equation of the normal line in point-slope form:

$y + 56 = \frac{1}{108} \left(x + 2\right)$

In slope-intercept form:

$y = \frac{1}{108} x - \frac{3135}{56}$