Find a rational function $f(x)$ with H. asymptote of y=2, V. asymptotes at x=-3, x=3 and a y-intercept at $\frac{-2}{3}$ 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.  
 A: 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.)
A: 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.
A: 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, $f(x)=\frac{8/3}2g(x)-\frac23.$
