Question 1019976
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The domain of a function is the set of values of the independent variable for which the function is defined.  For a rational function such as your example, that means the set of all real numbers except for those values that cause the denominator to equal zero.


The equation of a vertical asymptote is of the form *[tex \Large x\ =\ a] where *[tex \Large a] is a value that causes the denominator function to equal zero, but does NOT cause the numerator function to be zero.  If both numerator and denominator go to zero at a value, then the rational function has a discontinuity or "hole" at that point, but there is no asymptote.


If the degree of the numerator polynomial in a rational function is LESS than the degree of the denominator function, the *[tex \Large x]-axis is the horizontal asymptote.  If the degree of the numerator polynomial is EQUAL to the degree of the denominator polynomial, then the equation of the horizontal asymptote is *[tex \Large y\ =\ \frac{a}{b}] where *[tex \Large a] is the lead coefficient of the numerator polynomial and *[tex \Large b] is the lead coefficient of the denominator polynomial.  If the degree of the numerator polynomial is GREATER than the degree of the denominator polynomial then there is a slant (aka oblique) asymptote with an equation *[tex \Large y\ =\ Q(x)] where *[tex \Large Q(x)] is the quotient excluding the remainder when the denominator polynomial is divided into the numerator polynomial using polynomial long division.


a.  Set the denominator equal to zero and solve.  The domain is the set of real numbers excluding this(these) value(s).


b.  Evaluate the numerator for the value(s) excluded from the domain in part a.  If the numerator is NOT zero for a tested value, then there is a vertical asymptote at *[tex \Large x] equal to that value.  Since your numerator has a degree larger than the degree of your denominator, there is no horizontal asymptote.  Use polynomial long division to determine the equation of the slant asymptote.  If you need a refresher, click <a href="http://www.purplemath.com/modules/polydiv2.htm">Purple Math Polynomial Long Division</a>.  Be sure to read all of the relevant pages.


c. 


*[illustration rationalfunctionNGTD.jpg]


d.  By inspection of the graph.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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