Question 1143053
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I don't think this is possible.


If the function has vertical asymptotes of 3 and -3, then the denominator of the function must have factors of *[tex \Large x\ -\ 3] and *[tex \Large x\ +\ 3], hence the denominator must be *[tex \Large x^2\ -\ 9].


If the function has a non-zero constant as a horizontal asymptote, then the degree of the numerator function must be the same as the degree of the denominator function.  Consequently, the degree of the numerator function in this case must be 2.


And if the only *[tex \Large x]-intercept is 5, then the numerator function has to be *[tex \Large (x\ -\ 5)^2] or *[tex \Large x^2\ -\ 10x\ +\ 25].


However, if our function is *[tex \Large y\ =\ \frac{x^2\,-\,10x\,+\,25}{x^2\ -\ 9}], then the *[tex \Large y]-intercept must be *[tex \Large y\ =\ -\frac{25}{9}], not *[tex \Large y\ =\ -\frac{5}{9}] as specified.


On the other hand, if the function is *[tex \Large y\ =\ \frac{x\ -\ 5}{x^2\ -\ 9}] so that the vertical asymptotes, the *[tex \Large x]-intercept, and the *[tex \Large y]-intercept are as specified, then the horizontal asymptote would be *[tex \Large y\ =\ 0] rather than *[tex \Large y\ =\ 1] as required.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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