Answer
$ (\pm2,0)$,
symmetric with respect to the origin.
Work Step by Step
Step 1. For x-intercept(s), let $y=0$, we have $x^2=4\longrightarrow (\pm2,0)$, no y-intercept(s) as we can not let $x=0$,
Step 2. To test for x-axis symmetry, replace $(x,y)$ with $(x,-y)$, we have $-y=\frac{(x)^2-4}{2(x)}$ which is different from the original equation, thus it is not symmetric with respect to the x-axis,
Step 3. To test for y-axis symmetry, replace $(x,y)$ with $(-x,y)$, we have $y=\frac{(-x)^2-4}{2(-x)}$ which is different from the original equation, thus it is not symmetric with respect to the y-axis,
Step 4. To test for origin symmetry, replace $(x,y)$ with $(-x,-y)$, we have $-y=\frac{(-x)^2-4}{2(-x)}$ which is the same as the original equation, thus it is symmetric with respect to the origin.
