Question 454009
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If your function crosses the *[tex \Large x]-axis at 2, then the numerator polynomial must have a zero at 2.  Hence, one factor of the numerator is *[tex \Large x\ -\ 2]


If your function just touches the *[tex \Large x]-axis at -1, then the numerator polynomial must have a zero at -1 with a multiplicity of 2.  Hence, another factor of the numerator is *[tex \Large \left(x\ +\ 1\right)^2]


If the function must have a vertical asymptote at -5, then the denominator polynomial must have a zero at -5.  Hence a factor of the denominator must be *[tex \Large x\ +\ 5].


Likewise, for a vertical asymptote at 6, the denominator must have a factor of *[tex \Large x\ -\ 6].


In order for there to be a horizontal asymptote other than the *[tex \Large y]-axis, the degree of the numerator and denominator polynomials must be the same and the ratio of the lead coefficients must be equal to the value of the *[tex \Large y]-coordinate of the ordered pairs in the asymptote.


In order to accomplish equal degree, we need to double up on one of the factors in the denominator.  I choose the *[tex \Large x\ +\ 5], hence the first factor fo the denominator becomes *[tex \Large \left(x\ +\ 5\right)^2]


In order for there to be a 3 to 1 ratio between the lead coefficients on the numerator and denominator such that the horizontal asymptote occurs at *[tex \Large y\ =\ 3], insert a factor of 3 into the numerator.


Putting it all together:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \rho(x)\ =\ \frac{3(x\,+\,1)^2(x\,-\,2)}{(x\,+\,5)^2(x\,-\,6)}]


You can multiply the factors yourself if that is the form of the answer you are required to submit.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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