Answer
See graph and explanations.
Work Step by Step
Step 1. Given the function $f(x)=\frac{2x^2}{x^2-1}=\frac{2x^2}{(x+1)(x-1)}$, we can identify two vertical asymptotes as $x=\pm1$
Step 2. We can find a horizontal asymptote as $y=2$
Step 3. We can find the x-intercept as $x=0$ and y-intercept as $y=0$
Step 4. The function is even because $f(-x)=f(x)$
Step 5. The signs of the function when crossing the vertical asymptotes can be found as
$...(+)...(-1)...(-)...(1)...(+)...$
Step 6. We can graph the function as shown in the figure.
