Explaining Multiplicity of factors when writing Polynomial function from a graph 
The equation of this graph is:
$y = \frac{2(x+2)^2(x-5)}{(x+5)(x-2)^2}$
My question is about the exponents/multiplicity of the Vertical Asymptote factors in the denominator.  
The behavior around x=-5 goes to both +infinity and -infinity, it needs to have an odd multiplicity.  This ensures that this term can be both negative and positive, based on the number plugged in (eg: -4.9 vs. -5.1).   
The behavior around x=2 goes to both -infinity, so it needs to have an even multiplicity.  This ensures that this term is always the same sign, regardless of negative or positive, based on the number plugged in (eg: 1.9 vs. 2.1)  
Is this a reasonable explanation?  I feel it's a bit clumsy.  Is there a better way of explaining why the exponent is either even or odd in those Vert.Asy. factors in the denominator?  
 A: Your theorem seems to be the following:

Odd multiplicity at $x=r_1$ implies $\lim_{x \to r^{+}}R(x)\ne \lim_{x \to r^{-}}R(x)$. Even multiplicity at $x=r_1$ implies $\lim_{x \to r^{+}}R(x)= \lim_{x \to r^{-}}R(x)=\infty.$ ($R(x)$ is a rational function with a denominator containing a polynomial with at least $x=r_1$ as a root.)

If you'll allow for intense informality, this actually makes very much sense. The intuitive explanation is the following: If the root has even multiplicity, this ensures that the sign resulting in the denominator is the same since $(x-r_1)^a$ will be positive (if $x=|c|$) when $a$ is even. For example, in $f(x)=\dfrac{1}{(x-1)^2},$ we have the following table to illustrate this:
$$
\begin{array}{c|c|c}
x & (x-1)^2  & \text{sign} \\
\hline\\
0.5 & (-0.5)^2 & +\\
0.75 & (-0.25)^2 & +\\
0.99 & (-.01)^2 & +\\
\vdots & \vdots & \vdots \\
1.01 & (0.1)^2 & +\\
1.25 & (.25)^2 & +\\
1.5 & (0.5)^2 & + \\
\hline
\end{array}
$$
Like I said, this is highly informal, but you're seeing the picture, right? The fact that the exponent is even results in a removal of the $(-1)$ factor in the denominator. I am going to attempt a more formal proof, but this seems to at least shed some light on your situation.
