How do you find the equation of the tangent line to the curve #y=x^2e^-x# at (1, 1/e)?

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Answer 1 · 2017-04-20T07:10:28

#y=x/e#

#y=x^2e^{-x}#

#y'=2x cdot e^{-x}+x^2 cdot(-e^{-x})=(2-x)xe^{-x}#

To find an equation of a line, we need two things:

By Point-Slope Form: #y-y_1=m(x-x_1)#,

#y-1/e=1/e(x-1)=x/e-1/e#

By adding #1/e#to both sides,

#y=x/e#

I hope that this was clear.