There are some rational functions that do a switcheroo when $x \to -\infty$ where (when graphed) they cross the horizontal asymptote in order to approach from the other side. It only seems to happen when the power is larger on bottom than the power on top.

I first came across this behavior in Algebra 2, and never found a satisfying answer to what was going on here. It's been nagging me since then as I went through Pre-Cal and AP Calculus, and never found any answers other than just a shrug and response of "that's just how it is."

Here's an example: $$y = \frac{x+1}{x^2}$$

When graphed, it looks like: Example Rational Function

Now that I've taken AP Calculus, I can also find that there's a relative minimum at $\left(-2, -\frac{1}{4}\right)$, but once again, it's unclear why. Can someone please shed some light on why these functions cross the horizontal asymptote?

  • $\begingroup$ Here is somewhat of a justification, beyond simply saying "horizontal asymptotes can be crossed". Horizontal asymptotes describe behavior at infinity, but this says nothing about the function in any window finite window, like $[-r,r]$. The function can do all sorts of things in any finite window, because it will have infinite room to get back to its horizontal asymptote. $\endgroup$ – SquirtleSquad Jun 10 '16 at 22:39
  • $\begingroup$ The function $$f(x) = \frac{(x + 2)(x - 1)}{(x + 1)(x - 2)}$$ has a horizontal asymptote crossing at $x = 0$, so it is not necessary that the degree of the denominator exceed that of the numerator. $\endgroup$ – N. F. Taussig Jun 10 '16 at 22:42
  • $\begingroup$ @N.F.Taussig I understand that horizontal asymptotes can be crossed. We learned that in Algebra 2 also, but I'm wondering if the "switcheroo" behavior is a different story, because my teachers always addressed my question like it was a "glitch" in rational functions. $\endgroup$ – 4castle Jun 10 '16 at 22:56
  • $\begingroup$ Here's the original function that gave me the head-scratcher: $$y = \frac{x+1}{x(x-3)}$$ $\endgroup$ – 4castle Jun 10 '16 at 23:20
  • $\begingroup$ What I was pointing out is that it is not necessary for the degree of the denominator to exceed that of the numerator in order to have a horizontal asymptote crossing, as you seem to be asserting in your first paragraph. Note, also, that the horizontal asymptote in the example I gave above is $y = 1$. In that example, while $f(x)$ is positive as $x \to \infty$ and as $x \to -\infty$, the graph approaches the asymptote from above as $x \to \infty$ and from below as $x \to -\infty$. $\endgroup$ – N. F. Taussig Jun 10 '16 at 23:37

The reason is because the horizontal asymptote is identically the $x$-axis for your function. Also, an asymptote only describes end behavior, so it's perfectly fine for the function to cross it somewhere near the origin. Since your function simultaneously has a root and has the $x$-axis as an asymptote, it will need to pull away from the axis, then slice through it, and then proceed eventually coming back to the axis.

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  • $\begingroup$ I never thought about the roots of the function. That actually makes a lot of intuitive sense. Thanks! I suppose though that I'm looking for a more general answer. One that doesn't only address the example I gave. $\endgroup$ – 4castle Jun 10 '16 at 22:48

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