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?