algebra-precalculusgraphing-functionsrational-functions
When graphing rational functions, how do I determine the orientation of a rational function around the asymptotes without plotting points? For example, is it possible to determine which one of these is the correct graph of $\frac{2x}{x^2-1}$ without plugging in points and checking?
Basically, how would I check which one of these is correct, once I've found the asymptotes, or is plotting points necessary?
Let $f(x)=\frac{x-2}{x^2-4}$. Let us prove that there is an asymptote at $x = -2$:
$$\lim_{x\to -2^+}f(x)=\lim_{x\to -2^+}\frac{x-2}{x^2-4} = \lim_{x\to -2^+}\frac{1}{x+2} = +\infty,$$
$$\lim_{x\to -2^-}f(x)=\lim_{x\to -2^-}\frac{x-2}{x^2-4} = \lim_{x\to -2^-}\frac{1}{x+2} = -\infty.$$
However, there is actually no asymptote at $x = 2$. Your mistake is that you didn't check the limit:
$$\lim_{x\to 2}f(x)=\lim_{x\to 2}\frac{x-2}{x^2-4} = \lim_{x\to 2}\frac{1}{x+2} = \frac 14.$$
As you can see, the limit is not $\pm\infty$, which would be needed for it to be an asymptote. Actually, $f$ can be extended continuously:
$$g(x):=\begin{cases} f(x),& x\neq 2\\ \lim_{t\to 2}f(t),& x= 2\\ \end{cases}$$ and immediately it follows that $g(x) = \frac{1}{x+2}$.
This explains why the graph of $f$ looks like the graph of $g$; the only difference is that one point must be erased from the graph: $(2,\frac 14)$. If you want to emphasize it, this would be a way to do it:
If you want similar example, plot function $x\mapsto \frac{\sin x}x$ and observe that there is no asymptote at $x = 0$.
I believe the issue here is the definition of negative numbers raised to rational powers that's used by Wolfram or your graphing software. A robust definition of exponentials that's used by most software because it generalises to complex numbers is: $$a^b := \exp({b\ln a}),$$ where $\exp x=e^x$, and in the case where $a$ is complex, $\ln$ is the principal logarithm. The issue now, is when you try to evaluate something like $(-1)^{2/3}$. If we use the definition we have $$(-1)^{2/3}=\exp\left(\frac23\ln(-1)\right),$$ which clearly makes no sense in $\mathbb R$ (the $\ln$ of a negative number is undefined). So to answer your question, the reason wolfram thinks $(-1)^{2/3}$ is undefined is because of the robust definition for exponentiation, which gives $(-1)^{2/3}$ as a complex number.
Best Answer
Since the rational expression factors as $\dfrac{2x}{(x-1)(x+1)}$, and the power of each factor of the numerator/denominator is odd, the sign definitely changes at the zeros of the numerator and the denominator (in contrast to a situation like $x^4$ or $1/x^2$). Therefore, one of the two pictures is correct (as opposed to pictures where it approaches $\infty$ on both sides of $x=1$, say). Since extremely large $x$ makes all the factors of the numerator and denominator positive, it must be the picture on the right.
The point of the above is that looking at whether you're dealing with odd/even powers saves you from looking at every interval as in Austin Mohr's answer, and this is a general technique.