# Multiple Asymptotes and Rational Functions

When sketching rational functions $(f\left(x\right)=p\left(x\right)/q\left(x\right)$ where $p\left(x\right)$ and $q\left(x\right)$ are polynomials) in MCV4U, there are many steps in the process when all is said and done. One of these steps is determining whether there are any asymptotes. There are a few different kinds:

• Vertical asymptotes (VA): The vertical line $x=c$ such that $q\left(c\right)=0$ but $p\left(c\right)\neq0$
• Horizontal asymptotes (HA): The horizontal line $y=c$ such that either ${\displaystyle \lim_{x\rightarrow\infty}f\left(x\right)=c}$ or ${\displaystyle \lim_{x\rightarrow-\infty}f\left(x\right)=c}$
• Oblique asymptotes (OA): When the degree of $p\left(x\right)$ is $1$ plus the degree of $q\left(x\right)$, the OA is the slanted line $y=mx+b$ which is the quotient after dividing $q\left(x\right)$ into $p\left(x\right)$ via polynomial division

Once when I was teaching these concepts in a MCV4U class, the following very interesting question was asked: “Can a function have multiple asymptotes?” It turns out that the answer is pretty interesting and enlightening, and it is considered in this post.

For VAs the answer is a fairly straightforward “yes.” For example, simply add factors to the denominator and keep a constant numerator:

${\displaystyle f\left(x\right)=\frac{1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}}$

This function has VAs $x=1$, $2$, and $3$. But, for HAs and OAs the answer is no longer so clear. Consider the following graph:

This function, as a fact, has two HAs: $y=\pm\pi/2$. Similarly, if one were to rotate the graph above while fixing the axis, a graph of a function would be created with two OAs.

With all this said, therefore it should be true that functions can have two HAs or OAs… right? In fact, despite this, it is not true for rational functions. That is, there are indeed functions that can satisfy this, but given any rational function in the world, there is not way it can have more than one HA or OA. Essentially, the reason for this is that the process in obtaining the asymptotes results in there only being one answer, so it is impossible for there to be anything more than that. Algebraically, the answer goes a bit beyond the curriculum, but not too far. These algebraic proofs is what follows.

## Rational functions have at most one HA

Let

${\displaystyle f\left(x\right)=\frac{a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}}{b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0}}}$

where $a_{n}\neq0\neq b_{m}$. Of course, we will assume that there actually is a HA. For this to occur, it must be that $m\geq n$. This is because otherwise $m, so that when we do the usual process for finding HAs of

${\displaystyle \frac{a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}}{b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0}}\times\frac{1/x^{n}}{1/x^{n}}}={\displaystyle \frac{a_{n}+a_{n-1}/x+\cdots+a_{1}/x^{n-1}+a_{0}/x^{n}}{b_{m}x^{m-n}+b_{m-1}x^{m-n-1}+\cdots+b_{1}x^{1-n}+b_{0}x^{-n}}}$

and take $x\rightarrow\pm\infty$, we get everything in the denominator going to $0$, contradicting there being a HA:

${\displaystyle \lim_{x\rightarrow\pm\infty}f\left(x\right)=\frac{a_{n}+0+\cdots+0+0}{0+0+\cdots+0+0}=\frac{a_{n}}{0}}$

So, indeed, we must have that $m\geq n$. Consequently, the process of finding a HA instead leads us to dividing by $x^{m}$ and obtaining

${\displaystyle \lim_{x\rightarrow\pm\infty}f\left(x\right)=\lim_{x\rightarrow\pm\infty}{\displaystyle \frac{a_{n}x^{n-m}+a_{n-1}x^{n-m-1}+\cdots+a_{1}x^{1-m}+a_{0}x^{-m}}{b_{m}+\frac{b_{m-1}}{x}+\cdots+\frac{b_{1}}{x^{m-1}}+\frac{b_{0}}{x^{m}}}}}$

If $n=m$, then the limit is $a_{n}/b_{m}$ (computed like above). Otherwise, $m>n\Longrightarrow n-m<0$ so that the limit is $0/b_{m}=0$. This means that the HA is either $y=a_{m}/b_{m}$ or $y=0$, depending on $n$ and $m$. However, the answer for the HA does not depend on whether the limit is $x\rightarrow\infty$ or $x\rightarrow-\infty$. Consequently, the HA is either one of these, but not both. This completes the proof

## Rational functions have at most one OA

Let $f\left(x\right)=p\left(x\right)/q\left(x\right)$. Since there can only be an OA if the degree of $p\left(x\right)$ is $1$ more than that of $q\left(x\right)$, let the degree of $p\left(x\right)$ be $n+1$ and that of $q\left(x\right)$ be $n$. Euclidean division of polynomials, a theorem of dividing polynomials, then says that for dividing $p\left(x\right)$ by $q\left(x\right)$, there are unique polynomials $s\left(x\right)$ (the quotient) and $r\left(x\right)$ (the remainder) such that

$p\left(x\right)=s\left(x\right)q\left(x\right)+r\left(x\right)$

where the degree of $r\left(x\right)$ is less than the degree of $q\left(x\right)$, which is $n$. (Unique here means that given $p\left(x\right)$ and $q\left(x\right)$, the only quotient and remainder you can get by dividing $p\left(x\right)$ by $q\left(x\right)$ is $s\left(x\right)$ and $r\left(x\right)$–nothing else.) By dividing the equation by $q\left(x\right)$ we get the alternate expression

${\displaystyle f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}=s\left(x\right)+\frac{r\left(x\right)}{q\left(x\right)}}$

According to the procedure for finding an OA, the OA should therefore be $y=s\left(x\right)$. However, we need to know that $s\left(x\right)$ is linear, i.e. it has degree $1$. But, since $p\left(x\right)$ has degree $n+1$, $q\left(x\right)$ has degree $n$, and $r\left(x\right)$ has degree less than $n$, then according to the equation

$p\left(x\right)=s\left(x\right)q\left(x\right)+r\left(x\right)$

it must be that $s\left(x\right)$ has in fact degree $1$. So, indeed $y=s\left(x\right)=mx+b$ is the OA. Moreover, it is the only answer because it is unique–that is, there cannot be any other OAs. This completes the proof.