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

##### 1 Answer

Jan 25, 2017

#### Explanation:

#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

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