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Is the graph of $(-a)^x$ possible? If yes, then what is it? If no, why not? This question came into my mind while studying transformations of graphs.

Here $a$ is any arbitrary constant.

Abhishek Verma
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    Are you trying to graph an exponential function with a negative base? – imranfat Sep 28 '13 at 01:48
  • If your a is a negative Real constant, then the graph (and,of, course, the function) is defined everywhere; if a>0 , then $a^x$ is defined for $x \geq 0$ but it is not defined for all x ; e.g., the square root of (-2) is not defined. – DBFdalwayse Sep 28 '13 at 01:49
  • Yes, it is possible depending on what one means by a graph. how do you draw the graph of $4^{2} = {2,-2}$ ? – jimjim Sep 28 '13 at 02:17
  • @DBF: if $a>0$, then it's not defined for all $x\geq 0$, it's defined for all integer $x$. (or all rational $x$ for which the denominator is not divisible by 2). – Glen O Sep 28 '13 at 03:13
  • Related. – goblin GONE Sep 28 '13 at 03:23
  • Supposing you mean that $a$ is a positive real number, so that $-a$ is a negative real, then $(-a)^x$ is not even (uniquely) defined in general, unless $x$ is integer. Thus there is nothing to make a graph of. – Marc van Leeuwen Sep 28 '13 at 03:49
  • @Glen O: you're right; I replied too quickly. – DBFdalwayse Sep 28 '13 at 03:52

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For $a>0$, if your graph is in the real numbers (and we take the principal root where possible), then the graph will only have values at rational points with the denominator being odd.

At these points, it will evaluate to $-(a^x)$. And as these points are dense within the reals, such a graph will appear to be the same as $-(a^x)$.

If one allows complex numbers, and take $x$ to be real, then one must be more specific as to the argument of $-a$. If we take $-1=e^{\pi i}$, then we get the graph of $$a^xe^{\pi xi}=a^x[\cos(\pi x)+i\sin(\pi x)]$$

We can represent this as two graphs - the real and imaginary components - in which case the real part will be $a^x\cos(\pi x)$.

Glen O
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