## Calculus 10th Edition

$e\approx2.718$ is a number greater than one. So the value of $e^{-x}$ will go on decreasing, and the graph is asymptotic to the x-axis grows arbitrarily. For negative values, f $e^{-x}$ grows without bounds as we walk leftwards on the x-axis towards greater and greater negative values. The y-intercept is 1 because $e^{-0} =e^{0}=1$