What is the equation of the line that is normal to #f(x)=(x+4)^2e^x# at # x=-3 #?

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Answer 1 · 2017-02-08T03:16:27

#1/3e^3x+y-e^(-3)+e^3=0#. See normal-inclusive Socratic graph

graph{(e^x(x+4)^2-y)(1/3e^3x+y-e^(-3)+e^3)=0 [-4, -2, -0.5, 0.5]}

y > 0.

As #xto=oo, y to 0#, revealing that #y = 0 larr# is the asymptote.

At #x =-3, y = e^(-3)#.

So, the foot of the normal is #P(-3, e^(-3))#

#y'=e^x((x+4)^2+2(x+4))=3e^(-3)#, at #x = -3#.

The slope of the normal is #-1/(y')=-1/3e^3#

And so, the equation to the normal at P is

#y-e^(-3)=-1/3e^3(x+3)#, giving

#1/3e^3x+y-e^(-3)+e^3=0#