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

Answer 1

Ezra Adams

#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#

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Answer 2

Scarlett Allison

The equation of the line that is normal to f(x)=(x+4)^2e^x at x=-3 is y = -3e^(-3) + 5.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.