SOLUTION: Rational functions end behavior and multiplicity? I have the graph for: (4(x-5))/(x+2)(10-x) I understand how to plot the middle piece. But I am having trouble understanding

Algebra ->  Graphs -> SOLUTION: Rational functions end behavior and multiplicity? I have the graph for: (4(x-5))/(x+2)(10-x) I understand how to plot the middle piece. But I am having trouble understanding      Log On


   



Question 917130: Rational functions end behavior and multiplicity?
I have the graph for:
(4(x-5))/(x+2)(10-x)
I understand how to plot the middle piece. But I am having trouble understanding the end behaviors on the regions outside of the vertical asymptotes. I read that when a rational function has an odd multiplicity/exponent it therefore approaching from the first and third quadrants. But this rational function I have is doing the exact opposite of that despite it have multiplicity of 1's in the bottom. What am I misunderstanding here?
http://i.imgur.com/mjjOGcD.png
please explain
Thanks

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Rational functions end behavior and multiplicity?
I have the graph for:
(4(x-5))/(x+2)(10-x)

I understand how to plot the middle piece. But I am having trouble understanding the end behaviors on the regions outside of the vertical asymptotes. I read that when a rational function has an odd multiplicity/exponent it therefore approaching from the first and third quadrants. But this rational function I have is doing the exact opposite of that despite it have multiplicity of 1's in the bottom. What am I misunderstanding here?
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Note:: End behavior decribes the value of y when x becomes very negative
and when x becomes very large.
To think straight about end behavior concentrate only on the "x's"
in your problem--- the numbers are not that important when x is hugh.
Your Problem becomes x/(x*-x) = -1/x
----
So when x is very negative, y approaches 0 from the positive side.
And when x is very positive, y approaches 0 from the negative side
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Cheers,
Stan H.
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