Question 1039544
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The numerator polynomial must have a smaller degree than the denominator polynomial.  The denominator polynomial must have factors of *[tex \Large x\ -\ 1] and *[tex \Large x\ -\ 2], but neither of these factors can appear in the numerator polynomial.


Other than that, the sky is the limit. Note that no zeros for the function are specified, so it may have one, or many, or none.  Also, the problem doesn't say that *[tex \Large x\ =\ 1] and *[tex \Large x\ =\ 2] are the only vertical asymptotes so there may be others, and it doesn't prohibit removable discontinuities, so those might exist as well.


A minimalist answer to this question would have a zero degree polynomial in the numerator and the product of *[tex \Large (x\ -\ 1)(x\ -\ 2)] in the denominator.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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