Determining the positioning of rational functions without plotting points

Question

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?

Answer 1

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.

Answer 2

In this particular case, you can look at the asymptotic behavior of the rational function. In particular, $\frac{2x}{x^2-1}$ is always positive for $x > 1$. This rules out the plot on the left.

More generally, rational functions go like ratios of the leading term of the numerator and denominator, so if your plot extends far enough, you can use that.

Answer 3

You may consider my answer to be essentially plugging in values and checking (because it is), but I'll provide it anyway.

The function $$ \frac{2x}{x^2-1} $$ has vertical asymptotes at $-1$ and $1$, so I think about points near those.

An input slightly smaller than $-1$ will give a negative numerator and a positive denominator, so the function value is negative overall. (This is what I mean by "essentially plugging in values", though I'm not actually reaching for my calculator.)

An input slightly bigger than $-1$ will give a negative numerator and a negative denominator, so the function value is positive overall.

An input slightly smaller than $1$ will give a positive numerator and a negative denominator, so the function value is negative overall.

An input slightly bigger than $1$ will give a positive numerator and a positive denominator, so the function value is positive overall.

All this information tells me that the second picture is the correct one.