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

Jan 25, 2017

$x - 2 y + 7 = 0$. See the normal-inclusive Socratic graph.

#### Explanation:

$f \left(- 1\right) = - 2 + 3 + 2 = 3$. So, the foot of the normal P is $\left(- 1 , 3\right)$.

f'=6x^2+6x-2=6-6-2=-2.

Slope of the normal = -1/f'=1/2, at P(-1, 3).

So, the equation to the normal at P is

$y - 3 = \frac{1}{2} \left(x + 1\right)$, giving

$x - 2 y + 7 = 0$

graph{(2x^3+3x^2-2x-y)(x-2y+7)((x+1)^2+(y-3)^2-.01)=0 [-10, 10, -5, 5]}