<PRE><B>Question 583086</B>
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A rational function has a vertical asymptote wherever the function is undefined, that is wherever the denominator is zero.  If the denominator is zero only when *[tex \Large x\ =\ -2], then a possible expression for your denominator is *[tex \Large x\ +\ 2] since *[tex \Large x\ +\ 2\ =\ 0] iff *[tex \Large x\ =\ -2].  A more general expression that provides the same result is *[tex \Large (x\ +\ 2)^n] where *[tex \Large n\ \in\ \mathbb{N}].


A rational function has a horizontal asymptote of 0 only when the degree of the numerator is strictly less than the degree of the denominator.  For your specific case denominator above which has a degree of 1 you must have a numerator of degree zero, which is to say some constant. For the general case, your numerator must have a degree no greater than *[tex \Large n\ -\ 1].


&lt;i&gt;&lt;b&gt;Specific case:&lt;/i&gt;&lt;/b&gt;


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{C}{x\ +\ 2}]


&lt;i&gt;&lt;b&gt;General case:&lt;/i&gt;&lt;/b&gt;


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sum_{i=0}^{n-1}\alpha_ix^{(n-1)-i}}{(x\ +\ 2)^n}]


where at least one *[tex \Large \alpha_k\ \neq\ 0]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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