Question

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

Answer 1

`y=x/e`

Explanation:

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

By Product Rule,

`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:

1. Point: `(x_1,y_1)=(1,1/e)`
2. Slope: `m=y'(1)=(2-1)cdot1cdot e^{-1}=1/e`

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.