Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises - Page 352: 17

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Work Step by Step

$e\approx2.718$ 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 $e^{-0} =e^{0}=1$

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