## Precalculus (6th Edition) Blitzer

The vertical asymptote is $x=3$ and $x=0$ is a hole.
If there is a rational function $f\left( x \right)=\frac{p\left( x \right)}{q\left( x \right)}$ where $p\left( x \right)$ is a numerator and $q\left( x \right)$ is a denominator then $x=a$ is a vertical asymptote of the function $f\left( x \right)$ if $x=a$ is a zero of the denominator $q\left( x \right)$. After cancelling the common factor $x$ from the numerator and denominator of the function, we get $h\left( x \right)=\frac{1}{x-3}$ For the vertical asymptote, equate the denominator to zero. \begin{align} & x-3=0 \\ & x=3 \end{align} Thus, $x=3$ is a vertical asymptote. If there is a rational function $f\left( x \right)=\frac{p\left( x \right)}{q\left( x \right)}$ where $p\left( x \right)$ is a numerator and $q\left( x \right)$ is a denominator then $x=a$ is called a hole if $x-a$ is a common factor of the numerator and denominator. The common factor between $x$ and $x\left( x-3 \right)$ is $x$. So, $x=0$ is a hole. Hence, $x=3$ is a vertical asymptote and $x=0\text{ is a hole}$ .