Finding a polynomial equation based off its graph

Hello, DasRabbit!

Quote:

http://i46.tinypic.com/vrdoo5.png

There are vertical asymptotes at $x = 3$ and $x = \text{-}3.$
$f(x)$ has $(x-3)(x+3)$ in the denominator.

With the "usual" vertical asymptote, the graph goes to $+\infty$ on one side
. . and to $\text{-}\infty$ on the other side (as it does at $x = \text{-}3$ ).

At $x = 3$ , the graph goes to $+\infty$ on both sides.
This indicates that the factor $(x-3)$ in the denominator is to an even power.
. . Hence, the denominator has: . $(x-3)^2(x+3)$

Since there is a horizontal asymptote $(y = 0)$ , the numerator has a degree less than 3.
. . Assume (for simplicity) that the numerator is linear.

We have: . $f(x) \:=\:\frac{ax+b}{(x-3)^2(x+3)}$

We want: $f(0) = \text{-}1\!:\;\frac{b}{(-3)^2(3)} \:=\:\text{-}1 \quad\Rightarrow\quad b \,=\,\text{-}27$

We want: $f(1) = 0\!:\;\frac{a+b}{(\text{-}2)^2(4)} \:=\:0 \quad\Rightarrow\quad\frac{a-27}{16} \:=\:0 \quad\Rightarrow\quad a \:=\:27$

Therefore: . $f(x) \:=\:\frac{27x - 27}{(x-3)^2(x+3)} \quad\Rightarrow\quad f(x) \:=\:\frac{27(x-1)}{(x-3)^2(x+3)}$

Test a few points to ensure that graph is on the correct sides of the x-axis.