Determining functions from horizontal & vertical asymptotes

[!!!]

"Find a formula for a function that has vertical asymptotes x = 1 and x = 3 and horizontal asymptote y =1."

So far I have the function derived from the vertical asymptotes, which is:
1 / (x-1)(x-3)
But I don't know what to do with the horizontal asymptote.
I know the answer is x^2 / (x-1)(x-3), but how do you get that from the problem?

 

[pka]

Have you tried \(\displaystyle \frac{{e^x }}{{\left( {x - 1} \right)\left( {x - 3} \right)}} + 1\)?

 

[galactus]

If the power of the numerator and denomiantor are the same then the ratio of the leading coeffcients is the horizontal asymptote. Since you have a quadratic, a power of 2, in the denominator, then you must have a x^2 in the numerator. Also the ratio of leading coefficients must be 1. So x^2 would belong in the numerator.