Precalculus, Chapter 1, 1.2, Section 1.2, Problem 68

Find the intercepts of the equation $\displaystyle y = \frac{x^2 - 4}{2x}$ and test for symmetry.

$x$-intercepts:


$
\begin{equation}
\begin{aligned}

y =& \frac{x^2 - 4}{2x}
&& \text{Given equation}
\\
0 =& \frac{x^2 - 4}{2x}
&& \text{To find the $x$-intercept, we let } y = 0
\\
0 =& x^2 - 4
&&
\\
4 =& x^2
&&
\\
\pm 2 =& x
&&

\end{aligned}
\end{equation}
$


The $x$-intercepts are $(-2,0)$ and $(2,0)$

$y$-intercepts:


$
\begin{equation}
\begin{aligned}

y =& \frac{x^2 - 4}{2x}
&& \text{Given equation}
\\
y =& \frac{(0)^2 - 4}{2(0)}
&& \text{To find the $y$-intercept, we let } x = 0
\\
y =& \frac{-4}{0}
&&

\end{aligned}
\end{equation}
$


There is no real solutions for $y$

Test for symmetry

$x$-axis:


$
\begin{equation}
\begin{aligned}

y =& \frac{x^2 - 4}{2x}
&& \text{Given equation}
\\
-y =& \frac{x^2 - 4}{2x}
&& \text{To test for $x$-axis symmetry, replace $y$ by $-y$ and see if the equation is still the same}

\end{aligned}
\end{equation}
$


The equation changes so it is not symmetric to the $x$-axis

$y$-axis:


$
\begin{equation}
\begin{aligned}

y =& \frac{x^2 - 4}{2x}
&& \text{Given equation}
\\
y =& \frac{(-x)^2 - 4}{2(-x)}
&& \text{To test for $y$-axis symmetry, replace$ x$ by $-x$ and see if the equation is still the same}
\\
y =& - \frac{x^2 - 4}{2x}
&&

\end{aligned}
\end{equation}
$


The equation changes so it is not symmetric to the $x$-axis

Origin:


$
\begin{equation}
\begin{aligned}

y =& \frac{x^2 - 4}{2x}
&& \text{Given equation}
\\
-y =& \frac{(-x)^2 - 4}{2(-x)}
&& \text{To test for origin symmetry, replace both $x$ by $-x$ and y by $-y$ and see if the equation is still the same}
\\
-y =& - \frac{x^2 -4}{2x}
\\
y =& \frac{x^2 -4}{2x}
&&

\end{aligned}
\end{equation}
$


The equation is still the same so it is symmetric to the origin.

Therefore, the equation $\displaystyle y = \frac{x^2 - 4}{2x}$ has an intercepts $(-2,0)$ and $(2,0)$ and it is symmetry to the origin.