What is the equation of the line tangent to # f(x)=(x^2)e^(x+2) # at # x=2 #?

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Answer 1 · 2016-11-27T15:15:26

#y=435.8x-655.2#. A near-origin graph and a contracted high-y graph are inserted to show that the tangent at ( 2, 218.4) is nearly vertical. .

At (x = 2, y = 4e^4=218.4, nearly.

So, the point of contact is P( 2, 218.4)

#y' =xe^(x+2)(2+x)=8e^2=436.8#, at P.

The tangent is inclined at #tan^(-1)436.8=89.9^o#, nearly to the x-

axis.

The equation of the tangent is

#y-218.4=436.8(x-2)#. Rearranging,

#y=436.8x-655.2#

graph{x^2e(x+2) [-606, 606, -303, 303]}
graph{x^2e(x+2) [-10, 10, -5, 5]}