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?

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(13206) (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:

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

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:

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.

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