Finding a function with given conditions I want to find the equation of a function $f(x)$ such that the following conditions are met:
$(i)\ f(3) = 0, f(0) =1$.
$(ii)\ f(x)$ is an even function.
$(iii)\ f(x)$ has vertical asymptodes at $x=\pm 4$.
$(iv)\ f(x)$ has horizontal asymptode at $ y=2$.
Now since the function is even, I get $x=-3$ as another root and then I can graph the function. But how do I find its equation? Any hints? Thank you.
In general is there some method or software to find the equation of a function given its graph?

I am assuming you want to find a rather than all functions that satisfy that property. You can approach this with a sort of an ad hoc method.
First lets find some function $f(x)$ that satisfies $(i)$. This should be easy enough, the polynomial $x(x-3)$ does the job. (This i incorrect as I misread the condition to be $f(0) = 0$, this is fixed in the edit below)
Now for $(ii)$, we want $f(x) = f(-x)$. Notice squaring a number makes its sign redundant so we try incorporating that into our function to get $x^2(x^2-9)$. (always check back to see you havent lost $(i)$)
$(iii)$ requires us to divide by $0$ when $x = \pm 4$. So let $$f(x) = \frac{x^2(x^2-9)}{(x-4)(x+4)} = \frac{x^2(x^2-9)}{(x^2-16)}$$
Notice how this still has properties $(i)(ii)$
Finally, we need the horizontal asymptotic condition. Notice right now that $\lim_{x \rightarrow \infty} f(x) = \infty$ since the numerator is of degree $4$ while the denominator is of degree $2$. We can first let $f(x)$ converge when $x$ goes to infinity simply by squaring the denominator. This results in 
$$f(x) = \frac{x^2(x^2-9)}{(x^2-16)^2}$$
This gives us a horizontal asymptote of $y=1$. Multiplying our function by $2$ then fixes this
$$f(x) = 2\frac{x^2(x^2-9)}{(x^2-16)^2}$$
we are done.
EDIT (Sorry for the error)
we remove the $x^2$ and the square on the denominator to get $\frac{2(x^2-9)}{x^2 - 16}$ Notice at $x = 0$ this evaluates to $9/8$. We want to add a term that disappears at $x = \pm 3$ and as $x \rightarrow \infty $ and evaluates to $-1/8$ when $x = 0$. Following similar line of reasoning as above we get $\frac{(x^2 - 9)16^2}{9(8)(x^2 - 16)^2}$  so our function becomes
$$f(x) = \frac{2(x^2-9)}{x^2 - 16} + \frac{(x^2 - 9)16^2}{9(8)(x^2 - 16)^2} = (\frac{x^2-9}{x^2 -16})(2+ \frac{32}{9(x^2 -16)})$$
