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.