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Question 1027754: Please help me with these problem did any of this before.
Give an example of each of the following:
a. A function with vertical asymptotes at x=-3 and x=1, and a horizontal asymptote at y=2.
b. A word problem whose solution is 12!/9!3!
c. The equation of an ellipse with vertices at (0,0) and (-8,0).
d. A logarithmic expression equivalent to 5.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
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  .
If the degree of the numerator polynomial is equal to the degree of the denominator polynomial, then there is a horizontal asymptote with equation  where  is the lead coefficient of the numerator polynomial and  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 ) is the numerator polynomial and ) is the denominator polynomial, then calculate the quotient polynomial, ) thus: }{d(x)}\ =\ q(x)\ +\ \frac{r(x)}{d(x)}) . And then the equation of the oblique asymptote is )
You want a horizontal asymptote with equation  , 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  and the numerator does NOT have a factor  , then the rational function will have a vertical asymptote at .
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  and  , so the denominator polynomial must have factors of (x\ -\ 1)) . Hence,  will do nicely for a denominator.
The numerator must be something of the form  so that neither  nor  is a factor. Let's use (x\ -\ 3)\ =\ 2x^2\ -\ 5x\ -\ 3) .
And the resulting rational function is:
\ =\ \frac{2x^2\ -\ 5x\ -\ 3}{x^2\ +\ 2x\ -\ 3})
John
My calculator said it, I believe it, that settles it
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