# What is the equation of the normal line of f(x)=12x^3-4x^2-5x at x=-2?

Jul 21, 2016

$y = - \frac{1}{155} x - \frac{15812}{155}$

#### Explanation:

Normal line will be perpendicular to the tangent line. As we know that the product of perpendicular gradients is always $- 1$ we can find the gradient of the normal from the gradient of the tangent.

We compute the slope of the tangent by evaluating the first derivative:

$f ' \left(x\right) = 36 {x}^{2} - 8 x - 5$

$f ' \left(- 2\right) = 144 + 16 - 5 = 155$

So the slope of the normal (${m}_{n}$) can be found by:

${m}_{n} \cdot 155 = - 1 \implies {m}_{n} = - \frac{1}{155}$

To calculate the equation of the normal line we use

$y - b = m \left(x - a\right)$

To use this, we need a point on the line. We know that x = -2 is on the line, so we evaluate the original function at this point to get :

$f \left(- 2\right) = - 102$ hence:

$y - \left(- 102\right) = - \frac{1}{155} \left(x - \left(- 2\right)\right)$

$y + 102 = - \frac{1}{155} x - \frac{2}{155}$

$y = - \frac{1}{155} x - \frac{15812}{155}$