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

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Question 1166001: I am a rational function having a vertical asymptote at
the lines x = 3 and x = -3, and a horizontal asymptote y = 1.
If my only x-intercept is 5, and y-intercept is -5/9, What
function am I?

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


As we will see, there are too many constraints; there is no rational function that meets all the requirements.

(a) Vertical asymptotes at x=3 and x=-3

This requires factors of (x-3) and (x+3) (and no others) in the denominator:



(b) Only x-intercept at x=5

This requires a factor of (x-5) in the numerator, and no other factors in the numerator except a constant:



(c) Horizontal asymptote at y=1

This requires that the degree of the numerator be equal to the degree of the denominator; it also requires that the constant be a=1.

The only factors in the denominator are (x+3) and (x-3), and the only factor in the numerator is (x-5). For the degrees of the numerator and denominator to be equal, the factor in the numerator needs to be there twice:



At this point, there are no unknowns in the function; it is completely determined by the horizontal and vertical asymptotes and the x-intercept.

A graph showing the vertical asymptotes and the single x-intercept; the horizontal asymptote doesn't show up well because the function value gets close to 1 only for very large positive or very large negative values of x....



(d) y-intercept -5/9

This is problematic. The function as it currently stands has a y-intercept of -25/9.

If we were to introduce a factor of 1/5 in the function to get a y-intercept of -5/9, then the horizontal asymptote would be y=1/5 instead of y=1.

So we can't find a rational function that satisfies all of the given conditions.


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