## Precalculus (6th Edition) Blitzer

No vertical asymptote and $x=3\text{ is a hole}$ .
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 the numerator and $q\left( x \right)$ is the 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 $x-3$ from the numerator and denominator function then, $f\left( x \right)=x+3$. Denominator of the function $f\left( x \right)$ is constant (there are no values of $x$ for which the denominator is equal to zero). Hence, there is no 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 the numerator and $q\left( x \right)$ is the denominator then $x=a$ is called a hole if $x-a$ is a common factor of the numerator and denominator. There exists $x-3$, a common factor between the numerator and denominator. So, $x=3$ is a hole. Hence, there is no vertical asymptote and $x=3$.