Answer
$f(x)=\frac{2(x^2-9)}{x^2-4}$
Work Step by Step
Step 1. With the two vertical asymptotes, we can set the denominator as
$(x+2)(x-2)=x^2-4$
Step 2. With a horizontal asymptote of $y=2$, we can write
$f(x)=\frac{2(x^2+a)}{x^2-4}$
Step 3. The x-intercepts happens when $y=0$; we have $x^2+a=0$ when $x=\pm3$, which gives $a=-9$
Step 4 Check $f(x)=\frac{2(x^2-9)}{x^2-4}$ for the y-intercept by letting $x=0$, which gives $y=\frac{9}{2}$
Step 5. Check for y-symmetry:
$f(-x)=\frac{2((-x)^2-9)}{(-x)^2-4}=\frac{2(x^2-9)}{x^2-4}=f(x)$
Step 6. We conclude that one of the rational functions is
$f(x)=\frac{2(x^2-9)}{x^2-4}$
