Vertical Asymptotes of Rational Functions (College Board AP® Precalculus): Revision Note

Vertical asymptotes of rational functions

What is a vertical asymptote of a rational function?

• A vertical asymptote is often represented by a dashed vertical line on a graph

• For a rational function, it corresponds to an input value for which the function is not defined

  • However the graph of the function approaches closer and closer to the line from either side

  • It is a boundary for the graph of the function, rather than a part of the graph itself

Where do vertical asymptotes occur in rational functions?

• Let $r\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$ be a rational function

  • i.e. where $p\left(x\right)$ and $q\left(x\right)$ are both polynomial functions

• If $a$ is a real zero of $q$, i.e. if $q\left(a\right)=0$

  • and if $a$ is not a real zero of $p$, i.e. if $p\left(a\right)\ne 0$

  • then the graph of $r$ has a vertical asymptote at $\mathit{x}\mathbf{=}\mathit{a}$

    • E.g. $\frac{x+4}{x-3}$ has a vertical asymptote at $x=3$

• If $a$ is also a real zero of $p$, there can still be a vertical asymptote at $x=a$

  • if the multiplicity of $a$ as a real zero of $q$ is greater than its multiplicity as a real zero of $p$

    • E.g. $\frac{\left(x+4\right){\left(x-3\right)}^5}{{\left(x-3\right)}^6}$ has a vertical asymptote at $x=3$

      • because the multiplicity of $x=3$ in the denominator (6) is greater than its multiplicity in the numerator (5)

      • Note that the expression $\frac{\left(x+4\right){\left(x-3\right)}^5}{{\left(x-3\right)}^6}$ can be simplified to give $\frac{\left(x+4\right)\cancel{{\left(x-3\right)}^5}}{\left(x-3\right)\cancel{{\left(x-3\right)}^5}}=\frac{\left(x+4\right)}{\left(x-3\right)}$

  • If the multiplicity of $a$ as a real zero of $q$ is less than or equal to its multiplicity as a real zero of $p$, then there is a hole at $x=a$

How do I express the behavior of rational functions near vertical asymptotes using limit notation?

• If $x=a$ is a vertical asymptote of a rational function $r$

  • then near the vertical asymptote the values of the function's denominator become arbitrarily close to zero

  • This means that near the vertical asymptote the values of $r$ increase or decrease without bound

• E.g. consider the behavior of $\frac{x+4}{x-3}$ close to $x=3$

• For input values near $\mathit{a}$ and greater than $\mathit{a}$, the corresponding limit notation is

$\underset{x\to a^{+}}{\lim }r\left(x\right)=\infty$or $\underset{x\to a^{+}}{\lim }r\left(x\right)=-\infty$

• For input values near $\mathit{a}$ and less than $\mathit{a}$, the corresponding limit notation is

$\underset{x\to a^{-}}{\lim }r\left(x\right)=\infty$or $\underset{x\to a^{-}}{\lim }r\left(x\right)=-\infty$

• $\underset{x\to a^{+}}{\lim }$ can be read as "the limit as $x$ approaches $a$ from above (or from the right)"

  • and $\underset{x\to a^{-}}{\lim }$ can be read as "the limit as $x$ approaches $a$ from below (or from the left)"

• Whether a particular limit is $\infty$ or $-\infty$ will depend on the precise nature of the function $r$

How do I decide whether a limit goes to ∞ or -∞?

• To decide whether the limit is $\infty$ or $-\infty$, look at values of the function near to the vertical asymptote

  • E.g. $s\left(x\right)=\frac{x+4}{x-3}$ has a vertical asymptote at $x=3$

    • At $x=2$, $s\left(2\right)=\frac{2+4}{2-3}=\frac{6}{-1}=-6$

      • As $x$ increases towards 3, the numerator stays positive

      • and the denominator stays negative (while getting closer and closer to 0)

      • so $\underset{x\to 3^{-}}{\lim }s\left(x\right)=-\infty$

    • At $x=4$, $s\left(4\right)=\frac{4+4}{4-3}=\frac{8}{1}=8$

      • As $x$ decreases towards 3, the numerator stays positive

      • and the denominator stays positive (while getting closer and closer to 0)

      • so $\underset{x\to 3^{+}}{\lim }s\left(x\right)=\infty$