I'm having a little bit of a difficulty trying to decide the denominator. Its power definitely has to be bigger than 2, but then since there are no vertical asymptotes mentioned, do I take it for granted that there is no vertical asymptote in this function?
On
On $how$ to solve it: Since $f(x)$ is continuous we cannot have a vertical asymptote, so $f(x)=p(x)/q(x)$ where $p(x),q(x)$ are polynomials with no common polynomial (non-constant) factor, and with $q(x)$ never $0$.
We have $p(4)=p(-4)=0,$ so try $p(x)=x^2-16.$
Then $p(0)=-16$ and $f(0)=-7$ so $q(0)=16/7$. Now deg($q$)>deg($p$) in order that $\lim_{x\to \infty}f(x)=0.$ So let $q(x)=x^4+16/7,$ recalling that $q$ is never $0.$ Now all the conditions are met.
There are infinitely many other $f$ that will meet all the conditions. For example, with $p(x)=x^2-16$ we can let $q(x)=x^{2n+2}+16/7$ for any $n\in \mathbb N.$
Since it is continuous, you can't have vertical asymptotes, so the denominator cannot have real roots. You can have a quadratic on top and quartic on bottom to fit these conditions. Let $$f(x)=\frac{ax^2+b}{x^4+1}$$ You can use the first 2 conditions to solve for $a$ and $b$, while satisfying the other conditions.