Question 454009: I need to create a rational function with the following:
Crosses x-axis at 2
Touches the x-axis at -1
One verticle asymptote at x=-5 and another at x=6
One horizontal asymptote at y=3
help! i need to know the steps
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
If your function crosses the  -axis at 2, then the numerator polynomial must have a zero at 2. Hence, one factor of the numerator is 
If your function just touches the -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 ^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  .
Likewise, for a vertical asymptote at 6, the denominator must have a factor of  .
In order for there to be a horizontal asymptote other than the  -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 -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 , hence the first factor fo the denominator becomes ^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  , insert a factor of 3 into the numerator.
Putting it all together:
\ =\ \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
My calculator said it, I believe it, that settles it
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