SOLUTION: I am rational function having a vertical asymptote at the lines x=3 x=-3, and a horizontal asymptote at y=1. If only my x-intercept is 5 and my y-interxepr is 5/9, who am I?

Algebra ->  Rational-functions -> SOLUTION: I am rational function having a vertical asymptote at the lines x=3 x=-3, and a horizontal asymptote at y=1. If only my x-intercept is 5 and my y-interxepr is 5/9, who am I?       Log On


   

Question 1169077: I am rational function having a vertical asymptote at the lines x=3 x=-3, and a horizontal asymptote at y=1. If only my x-intercept is 5 and my y-interxepr is 5/9, who am I?
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Vertical asymptotes at x=3 and x=-3 --> factors of (x+3) and (x-3) in the denominator:

f%28x%29+=+a%2F%28%28x%2B3%29%28x-3%29%29

Only x-intercept at x=5 --> only linear factors in the numerator are (x-5):

f%28x%29+=+%28a%28x-5%29%5En%29%2F%28%28x%2B3%29%28x-3%29%29

Horizontal asymptote at y=1 --> degrees of numerator and denominator are the same, and leading coefficients are the same. So a=1; and the number of factors of (x-5) must be 2:

f%28x%29+=+%28%28x-5%29%5E2%29%2F%28%28x%2B3%29%28x-3%29%29

y-intercept 5/9....

This is a problem; the given requirements are inconsistent.

With the function we have at this point, the y-intercept is -25/9.

If we add a constant factor to make the y-intercept 5/9, then we no longer have a horizontal asymptote at y=1.

ANSWER: The given conditions are inconsistent; there is no rational function with all the given features.