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What happens in a graph when you modify the value of $a$ in the following equation:

$ae^x$ where e is the natural number.

At first it seemed like it was translating the graph horizontally but when $a$ is negative it inverses the graph. I then thought it was dilating the graph however it didn't look like the curve between the horizontal and vertical asymptote was changing, it was just moving horizontally. So how would i describe this change?

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    $\begingroup$ It is shrinking/expanding it vertically. If you go with $a$ from $1$ to $0$ it gets shrunk to the $OX$ axis. And when you go from $0$ to $-1$ it is expanded, but below the $OX$. A fast switch from $1$ to $-1$ looks like inversion but actually it is a process of shrinking and expanding. $\endgroup$ – savick01 Aug 7 '13 at 7:54
  • $\begingroup$ hmm i didnt think of that, it explains it perfectly, thanks. I also noticed that $a$ became the $y$ intercept if there were no other variables in the equation, is this a cause of the shrinking/expanding vertically $\endgroup$ – VikeStep Aug 7 '13 at 7:57
  • $\begingroup$ What do you mean by "$a$ became the $y$"? If $x=0$ then clearly $y=ae^x=ae^0=a$. One can say that it is a result of shrinking/expanding vertically a point with $y$-coordinate equal to $1$. $\endgroup$ – savick01 Aug 7 '13 at 8:00
  • $\begingroup$ that was what I meant, but I forgot that trait, thanks for all answers $\endgroup$ – VikeStep Aug 7 '13 at 8:03
  • $\begingroup$ Note that this function has no vertical asymptote. $\endgroup$ – Henning Makholm Aug 7 '13 at 8:35
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The fun thing about the exponential function is that stretching the graph vertically and translating it horizontally yield the same result. To be precise: $$a\cdot e^x=e^{x+b}\quad\text{where $a=e^b$.}$$ On the left: Stretch the graph vertically by a factor $a$. On the right: Translate it left a distance $b$.

Exercise: Figure out the analogous phenomenon on the graph of the logarithm.

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