Question 1045715
I am a rational function having a vertical asymptote at 
the lines x = 3 and x = -3, and a horizontal asymptote y = 1. 
If my only x-intercept is 5, and y-intercept is -5/9, What 
function am I?
<pre>
Since it has vertical asymptotes at the lines x = 3 and x = -3, the
denominator is (x-3)(x+3) or x²-9

Since it has horizontal asymptote y = 1, the leading term of the numerator
must by x²

So it has form:

{{{y}}}{{{""=""}}}{{{(x^2+Bx+C)/(x^2-9)}}}, for some B and C

Since it has x-intercept 5, it passes through the point (5,0),
so we substitute (x,y) = (5,0)

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

{{{0}}}{{{""=""}}}{{{(25+5B+C)/(25-9)}}}

{{{0}}}{{{""=""}}}{{{(25+5B+C)/16}}}

Multiply both sides by 16

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

Since it has y-intercept -5/9, it passes through the point (0,-5/9),
so we substitute (x,y) = (0,-5/9)

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

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

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

Multiply through by -9

{{{5}}}{{{""=""}}}{{{C}}}

Substitute in

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

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

{{{0}}}{{{""=""}}}{{{30+5B}}}

{{{-30}}}{{{""=""}}}{{{5B}}}

{{{-6}}}{{{""=""}}}{{{B}}}

So 

{{{y}}}{{{""=""}}}{{{(x^2+Bx+C)/(x^2-9)}}}

becomes


{{{y}}}{{{""=""}}}{{{(x^2-6x+5)/(x^2-9)}}}

Edwin</pre>