Question 636236
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r(x)\ =\ 4x^2\ +\ \frac{9}{9x^2}\ +\ 1\ ]


has a vertical asymptote at *[tex \LARGE x\ =\ 0], in other words the *[tex \LARGE y]-axis, because that is where the denominator of the rational part of the function goes to zero.


However, I rather suspect that you really meant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r(x)\ =\ \frac{4x^2\ +\ 9}{9x^2\ +\ 1}]


which has no real number zeros for the denominator polynomial, and therefore has no vertical asymptotes.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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