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**topsquark** Okay, so you have the slope of the tangent line at the point where x = 1. What is the y value of this point?

$\displaystyle y = 1^2e^{-1} = \frac{1}{e}$

Thus the tangent line at x = 1 has a slope of $\displaystyle \frac{1}{e}$ and touches the curve at the point $\displaystyle \left ( 1, \frac{1}{e} \right )$. (There are a couple of 1/e's floating around. Make sure you keep track of them.)

So your line equation, $\displaystyle y = mx + b$ becomes

$\displaystyle y = \frac{1}{e} \cdot x + b$

at the point $\displaystyle \left ( 1, \frac{1}{e} \right )$.

Plug this x and value into the line equation and solve for b.

-Dan