Question 198680
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<b><i>Horizontal Asymptotes</i></b>


A rational function has at most 1 horizontal or oblique asymptote.


Compare the degree of the numerator polynomial (*[tex \Large d_n]) to the degree of the denominator polynomial (*[tex \Large  d_d])


If *[tex \Large  d_n - d_d < 0] then there is a horizontal asymptote at the line *[tex \Large y = 0]


If *[tex \Large d_n - d_d = 0] then there is a horizontal asymptote at the line *[tex \Large y = \frac{a_1}{a_2}] where *[tex \Large a_1] is the lead coefficient on the numerator polynomial and *[tex \Large a_2] is the lead coefficient on the denominator polynomial.


If *[tex \Large d_n - d_d = 1] then there is an oblique asymptote defined by *[tex \Large y = \frac{a_1}{a_2}x] where *[tex \Large a_1] is the lead coefficient on the numerator polynomial and *[tex \Large a_2] is the lead coefficient on the denominator polynomial.


If *[tex \Large d_n - d_d > 1] then there is no horizontal or oblique asymptote, though there may be higher order curves to which the function is asymptotic.


<b><i>Vertical Asymptotes</i></b>


A rational function has a vertical asymptote at *[tex \Large x = a_n] for every factor *[tex \Large x - a_n] of the denominator polynomial that does not have a corresponding factor in the numerator.  For example:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x) = \frac{x^2 - 5x + 6}{x^3 - 3x^2 + 2x} = \frac{(x - 2)(x - 3)}{x(x - 2)(x - 1)}]


has vertical asymptotes at *[tex \Large x = 0] and *[tex \Large x = 1] but not at *[tex \Large x = 2]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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