Question 1185294
<pre>
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:

{{{y}}}{{{""=""}}}{{{(x^2+Ax+B)/((x-3)(x+3))}}}

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

{{{0}}}{{{""=""}}}{{{(5^2+A(5)+B)/((5-3)(5+3))}}}

{{{0}}}{{{""=""}}}{{{(25+5A+B)/((2)(8))}}}

Multiply through by 16

{{{0}}}{{{""=""}}}{{{25+5A+B}}}

{{{5A+B}}}{{{""=""}}}{{{-25}}}

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

{{{-5/9}}}{{{""=""}}}{{{(0^2+A(0)+B)/((0-3)(0+3))}}}

{{{-5/9}}}{{{""=""}}}{{{B/(-9)}}}

Multiply both sides by -9

{{{5}}}{{{""=""}}}{{{B}}}

Substituting in

{{{5A+B}}}{{{""=""}}}{{{-25}}}
{{{5A+5}}}{{{""=""}}}{{{-25}}}
{{{5A}}}{{{""=""}}}{{{-30}}}
{{{A}}}{{{""=""}}}{{{-6}}}

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

{{{y}}}{{{""=""}}}{{{(x^2+Ax+B)/((x-3)(x+3))}}}

{{{y}}}{{{""=""}}}{{{(x^2-6x+5)/((x-3)(x+3))}}}

Let's graph it and see what we have:

{{{drawing(400,400,-10,10,-10,10,
green(line(-3,-11,-3,11),line(3,-11,3,11),line(-11,1,11,1)),
graph(400,400,-10,10,-10,10,(x^2-6x+5)/((x-3)(x+3))))}}}

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:

{{{y}}}{{{""=""}}}{{{(x^2-6x+5)(x-1)/((x-3)(x+3)(x-1))}}}

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

{{{drawing(400,400,-10,10,-10,10,circle(1,0,.4),
green(line(-3,-11,-3,11),line(3,-11,3,11),line(-11,1,11,1)),
graph(400,400,-10,10,-10,10,((x^2-6x+5)(x-1))/((x-1)(x-3)(x+3))))}}}

The un-factored form of the rational function is:

{{{y}}}{{{""=""}}}{{{(x^3-7x^2+11x-5)/(x^3-x^2-9x+9)}}}

Edwin</pre>