How do you find the equation of the tangent line to the curve #y=(x^2)e^(x+2)# at x=2?

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Answer 1 · 2015-06-18T12:32:36

Find the slope of the tangent line using the derivative, find the #y# value of the point where #x=2#, and find the equation of the line.

#y=(x^2)e^(x+2)#

at x=2, we get #y=4e^4#, So our line goes through the point: #(2, 4e^4)#

To find the slope, find #y'#. To find #y'# use the product rule and the derivative of exponentials (and the chain rule).

#y' = 2xe^(x+2) + x^2 e^(x+2) (1)# -- I included the #(1)# to remind us of the chain rule.)

At #x=2#, #m = y' |_2 = 4e^4+4e^3 = 8e^4#

The line with #m=8e^4# through #(2, 4e^4)# has equation(s):

#y-4e^4 = 8e^4(x-2)#

#y-4e^4 = 8e^4x-16e^4#

#y = 8e^4x - 12 e^4#