Precalculus (6th Edition)

The given relation defines $y$ as a function of $x$. domain: $(-\infty, 3) \cup (3, +\infty)$ range: $(-\infty, 0) \cup (0, +\infty)$
The given equation above will give only one value of $y$ for every value of $x$ within its domain. This means that each $x$ is paired with only one value of $y$. Thus, the given relation defines $y$ as a function of $x$. The denominator is not allowed to be zero. Since $3$ will make the given expression's denominator equal to zero, then $x$ cannot be $3$. Thus, the domain is $(-\infty, 3) \cup (3, +\infty)$. Note that when $2$ is divided by any non-zero number, the quotient will never be equal to zero. Thus, the value of $y$ can be any real number except zero. In interval notation, the range is $(-\infty, 0) \cup (0, +\infty)$.