In Problems 58, list the intercepts and test for symmetry.

$y=\frac{x^2-4}{2x}$

The given equation is$y=\frac{x^2-4}{2x}$.

For $x$-intercept, substitute 0 for $y$and then solve for $x$.

$\begin{array}{rcl}0& =& \frac{x^2-4}{2x}\\ x^2-4& =& 0\\ \left(x-2\right)\left(x+2\right)& =& 0\\ x& =& -2,2\end{array}$

For $y$-intercept, substitute 0 for $x$and then solve for $y$.

$\begin{array}{rcl}y& =& \frac{0^2-4}{2\left(0\right)}\\ & =& \mathrm{Undefined}\end{array}$

Substitute $y$for $-y$in the equation to check for symmetry with respect to $x$axis:

$\begin{array}{rcl}-y& =& \frac{x^2-4}{2x}\\ y& =& \frac{-x^2+4}{2x}\end{array}$

Notice that the right side of the equation after substitution is not equal to the right side of the initial equation.

Since the equations are not equivalent, the graph is not symmetric with respect to the $x$-axis.

Substitute $x$with $-x$in the equation to check for the symmetry with respect to $y$-axis:

$\begin{array}{rcl}y& =& \frac{{\left(-x\right)}^2-4}{2\left(-x\right)}\\ & =& \frac{-x^2+4}{2x}\end{array}$

Notice that the right side of the equation after substitution is not equal to the right side of the initial equation.

Since the equations are not equivalent, the equation is not symmetric with respect to the $y$-axis.

Substitute $x$with $-x$and $y$with $-y$in the equation to check for the symmetry with respect to the origin:

$\begin{array}{rcl}-y& =& \frac{{\left(-x\right)}^2-4}{2\left(-x\right)}\\ y& =& \frac{x^2-4}{2x}\end{array}$

Notice that the right side of the equation after substitution is equal to the right side of the initial equation.

Since the equations are equivalent, the equation is symmetric with respect to the origin.

There is no $y$-intercept.

The $x$-intercept are $\left(2,0\right)$and $\left(-2,0\right)$.

The given equation is symmetric about the origin.