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

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Answer 1 · 2017-01-25T11:24:01

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

#f(-1)=-2+3+2=3#. So, the foot of the normal P is #(-1, 3)#.

#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=1/2(x+1)#, giving

#x-2y+7=0#

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