Greenestamps didn't think of the sneaky trick of eliminating an x-intercept by
putting an extra factor in the numerator and denominator to create a hole in the
curve instead of having an x-intercept.  I'll use that trick.

To have those vertical asymptotes, the denominator, when set = 0, must have
roots 3 and -3, so the denominator must contain (x-3)(x+3), which has degree 2
and leading coefficient 1.  That might be enough for the denominator.  If so,
the numerator must also be of degree 2 and have leading coefficient 1 in order
to have horizontal asymptote y = 1.  So let's try this version:



To have x-intercept 5, it must go through (5,0).  So we substitute 





Multiply through by 16





To have y-intercept -5/9, it must go through (0,-5/9).  So we substitute 





Multiply both sides by -9



Substituting in






So let's substitute those values for A and B in





Let's graph it and see what we have:



Oh darn! That has an extra x-intercept at (1,0).  Aha, but I can play the trick!
I'll put a hole in the curve at (1,0) by putting in a factor of (x-1) in
the numerator and the denominator:



Then the graph has a hole at (1,0) instead of an x-intercept there.



The un-factored form of the rational function is:



Edwin