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Question 77459: Ok stan h. you go this..
-(6x-24+3x^2)/(5-26x+ 5x^2)
=(3^2+6x-24)/(5x^2-26x+5)
=3(x^2+2x-12)/(5x^2-25x-x+5)
=3(x^2+2x-12)/(5x(x-5)-(x-5))
=3(x^2+2x-12)/(x-5)(5x-1)
But what would the x equal to??
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Graph the following rational functions, give any equations for vertical,horizontal, or oblique asymptotes.
-6x-24+3x^2
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5-26x+ 5x^2
1 solutions
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Answer 55556 by stanbon(7053) on 2007-04-08 11:12:48 (Show Source):
-(6x-24+3x^2)/(5-26x+ 5x^2)
=(3^2+6x-24)/(5x^2-26x+5)
=3(x^2+2x-12)/(5x^2-25x-x+5)
=3(x^2+2x-12)/(5x(x-5)-(x-5))
=3(x^2+2x-12)/(x-5)(5x-1)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! (6x-24+3x^2)/(5-26x+ 5x^2)
=(3^2+6x-24)/(5x^2-26x+5)
=3(x^2+2x-12)/(5x^2-25x-x+5)
=3(x^2+2x-12)/(5x(x-5)-(x-5))
=3(x^2+2x-12)/(x-5)(5x-1)
But what would the x equal to??
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You did not show any equal sign in your posting so
the problem had no "solution".
If you meant to past the following you could find
a solution:
(6x-24+3x^2)/(5-26x+ 5x^2)=0
then
3(x^2+2x-12)/[(x-5)(5x-1)] would have a solution.
The fraction can be zero only if the numerator is zero, so you get:
3(x^2+2x-12)=0
x^*2+2x-12=0
x=[-2+-sqrt(4-4*12]/2
x=[-2+-sqrt52]/2
x=[-2+-2sqrt13]/2
x=(-1+sqrt13) or x=(-1-sqrt13)
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Cheers,
Stan H.
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