Verified answer

Suppose f(x) = p(x)/q(x) and given:

f(0) = 0 ... is the only zero ("x-intercept") of f. That means that p(0) = 0 at x=0 and nowhere else. That means p(x) must be a power of x times a polynomial with no real roots. The simplest possible case of that is just x times a constant.

p(x) = ax

The only (vertical) asymptotes are at x=-6 and x=5, so those are the only real roots of q(x). That means q(x) is some power of (x+6) times some power of (x-5) times some polynomial that with no real roots. The simplest case of that is:

q(x) = (x + 6)(x - 5) = x² + x - 30

So f(x) = ax / (x² + x - 30) fits the given intercept and asymptote conditions for any nonzero value of a. Use the given point value to find out what a must be:

f(6) = -12

(a*6)/(6² + 6 - 30) = -12 .... substitute for f(6)

6a / 12 = -12

6a = -144

a = -24

So f(x) = -24x / (x² + x - 30)

-x^2 has only one x-intercept at the origin

dividing by (x+6)^2 and (x-5)^2 gives us negative asymptotes

this passes through (6,-1/4), so a possible formula is -48x^2 / ((x+6)^2 * (x-5)^2),

as might -288x/((x+6)^2 * (x-5)^2)

http://www.futhead.com/17/players/20781/lee-novak/