SOLUTION: Finding rational functions off of graphs:
Hi! I don't understand what to do with the horizontal asymptote when I am finding the graph function. For example:
Zero's of the fun
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Question 631227: Finding rational functions off of graphs:
Hi! I don't understand what to do with the horizontal asymptote when I am finding the graph function. For example:
Zero's of the function: -2 and 3
Vertical asymptote: -1
Horizontal Asymptote: y=2x-4
Thank you very much!
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
WARNING! Danger, Will Robinson. Terminology Error! 
is NOT a horizontal asymptote. Horizontal asymptotes are horizontal, hence the name. Linear functions with a non-zero (2 is not zero) slope are NOT horizontal. Asymptotes that are slanted or oblique are called Slant Asymptotes or Oblique Asymptotes.
The fact that the function has zeros at -2 and 3 tells us that the factors of the numerator polynomial are )
, )
, and some constant 
(because all polynomial equations \right)\ =\ 0)
where )
is a polynomial with degree 
and 
have identical solution sets)
Hence, the numerator is 
The fact that the function has a vertical asymptote of 
means that the denominator polynomial has a zero at , therefore the denominator polynomial must be 
.
If a rational function has a numerator that is one degree greater than the degree of the denominator, then the function will have a slant asymptote equal to the quotient of a polynomial long division of the numerator by the denominator.
Perform the polynomial long division of 
.
Your quotient will have a factor of in it, but if you set the quotient equal to the given slant asymptote 
, you will very quickly see the value of .
Then it is simply a matter of constructing your function from the derived numerator and denominator.
Go to Purple Math Polynomial Long Division if you need a refresher on polynomial long division.
John
My calculator said it, I believe it, that settles it
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