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sketch the graph of the following function

$f(x) = e^{-2x}$ for $x \in \mathbb R$

this what i got,

y-intercept $x=0$ implies $y=\cfrac{1}{e^{2\times 0}}$ therefore $y=1$

and I have used the fact that $t= -2x$ $\rightarrow -\infty$ as $x \rightarrow \infty$

by the properties of the Natural Exponential Function:

$\displaystyle\lim_{x \to \infty} e^x=\infty$

$\displaystyle\lim_{x \to \infty} e^{-2x}=0$

$\displaystyle\lim_{x \to -\infty} e^{t}=0$

Therefore, I draw the curve graph with $y$-intercept coordinates $(0,1)$ and approaches zero.

but my prof said it's wrong. he commented = $e^x$ cannot be zero.

Could you please explain to me about this?...

BLAZE
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purugin
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    "Approaches zero when $x\to+\infty$" $\ne$ "Equals zero at some $x$". Did you make the graph of $y=e^{-x}$ actually meet the line $y=0$ at some point? – Did Sep 20 '15 at 07:52
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    Then we have to find a reason why your professor mentioned the fact that $e^x$ is never zero. – Did Sep 20 '15 at 08:03
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    A proof of what? The assertion of your professor is correct, the mystery, so far, is why they think it invalidates your solution. – Did Sep 20 '15 at 08:13
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    @purugin We all want to help you but please be clarify your objective: Is the goal simply "sketch the graph of the following function"? – BLAZE Sep 20 '15 at 08:41
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    yes just the graph . – purugin Sep 20 '15 at 08:42

1 Answers1

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Your professor is correct (as usual) he describes asymptotic behaviour - namely $y=0$ is a horizontal asymptote for $e^{-2x}$. Sketch the graph of $e^{-2x}$ to see this. Firstly you know how to sketch the graph of $e^x$ right? You then apply linear transformations which consist of a horizontal reflection in the $y$-axis and then a stretch of scale factor $1/2$ parallel to the $x$-axis. Does this make sense? If not let me know and I will explain further.

BLAZE
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