Question 1027754
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In a rational function:


If the degree of the numerator polynomial is less than the degree of the denominator polynomial, then there is a horizontal asymptote with equation *[tex \Large y\ =\ 0].


If the degree of the numerator polynomial is equal to the degree of the denominator polynomial, then there is a horizontal asymptote with equation *[tex \Large y\ =\ \frac{p}{q}] where *[tex \Large p] is the lead coefficient of the numerator polynomial and *[tex \Large q] 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 no horizontal asymptote but there is an oblique asymptote.  Where *[tex \Large n(x)] is the numerator polynomial and *[tex \Large d(x)] is the denominator polynomial, then calculate the quotient polynomial, *[tex \Large q(x)] thus:  *[tex \Large \frac{n(x)}{d(x)}\ =\ q(x)\ +\ \frac{r(x)}{d(x)}].  And then the equation of the oblique asymptote is *[tex \Large y\ =\ q(x)]


You want a horizontal asymptote with equation *[tex \Large y\ =\ 2], therefore the degree of your numerator polynomial must be equal to the degree of your denominator polynomial and the lead coefficient of the numerator must be twice the lead coefficient of the denominator.


If the the denominator polynomial has a zero at *[tex \Large x\ =\ a] and the numerator does NOT have a factor *[tex \Large x\ -\ a], then the rational function will have a vertical asymptote at *[tex \Large x\ =\ a].


You need a denominator polynomial that is at least degree 2 because you want two vertical asymptotes and need at least two binomial factors that comprise the denominator.  Your asymptotes are *[tex \Large x\ =\ -3] and *[tex \Large x\ =\ 1], so the denominator polynomial must have factors of *[tex \Large (x\ +\ 3)(x\ -\ 1)].  Hence, *[tex \Large x^2\ +\ 2x\ -\ 3] will do nicely for a denominator.


The numerator must be something of the form *[tex \Large 2x^2\ +\ bx\ +\ c] so that neither *[tex \Large x\ -\ 1] nor *[tex \Large x\ +\ 3] is a factor.  Let's use *[tex \Large (2x\ +\ 1)(x\ -\ 3)\ =\ 2x^2\ -\ 5x\ -\ 3].


And the resulting rational function is:


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


*[illustration rationalwith2vertandahorizasym.jpg]


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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