So first I multiplied the V. Asymptotes like so, $$(x+3)(x-3)$$ to get $$(x^2-9)$$
And knowing that because the horizontal Asymptote is a non-negative number, that the leading coefficient in the numerator has to be double that of the leading coefficient in the denominator. Meaning it will be $2$ along with the same degree value as the denominator.
So up till now I have $$f(x)=\frac{2x^2}{x^2-9}$$
This is where I get stuck because I am not sure how to incorporate the y-intercept.
On
You are on the right track. Note that when you have $x=0$ in your current function, the value of the function is also $0$ hence we can just have $$f(x)=-\frac23+\frac{8x^2}{3(x^2-9)}$$ Because the constant term gives the $y$-intercept, but the factor ahead of the rational term must also be changed to keep the same horizontal asymptote.
On
Your current function -- call it $g(x)$ -- has $g(0)=0$ and horizontal asymptote $y=2$.
To get a function with $f(0)=-\frac23$ and keep the horizontal asymptote $y=2$,
you can stretch $g(x)$ in the vertical direction so a distance of $2$ becomes $2+\frac23$,
and then translate down by $-\frac23$ so $f(0)=-\frac23$. That is, $f(x)=\frac{8/3}2g(x)-\frac23.$
You can use $f(x)=\frac{2x^2+c}{x^2-9}$ and then use $x=0$ to get $$f(0)=\frac{c}{-9}=\frac{-c}{9}=\frac{-2}{3}$$ or $c=6$.
So you get $f(x)=\frac{2x^2+6}{x^2-9}$
More generally, you can use:
$$f(x)=\frac{2x^2+bx+6}{x^2-9}$$ where $b\neq \pm 8.$ (When $b=\pm 8$ you lose one of the vertical asymptotes.)