Question 1117633
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Note that the function will have vertical asymptotes at *[tex \Large x\ =\ 1] and *[tex \Large x\ =\ -1].  As the independent variable approaches -1 from the left, the value of the function will be a large negative number, and from the right, a large positive.  Approaching 1 from the left is also a large negative, and from the right, a large positive.


Evaluating *[tex \Large f(0)] we get *[tex \Large \frac{2}{3}]


Applying the common denominator of *[tex \Large 3(x^2\ -\ 1)] and adding all of the fractions (verification of this step left as an exercise for the student), we simplify to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \frac{-3x^3\ +\ 2x^2\ +\ 3x\ -\ 4}{3x^2\ -\ 3}]


So we note that as *[tex \Large x] increases and decreases without bound, the function begins to behave very much like *[tex \Large y\ =\ -x].


 *[illustration Rational_Function.jpg]


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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