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# Rational Function's Graph

Mon 16 May, 2011 02:46 pm
I have a function : f(x)= (6x-4)/(3^2+ 10x -8)
I have to find vertical asymtote, horizon asymtote, and hole

then I did like this 2(x-3) / ((x+4)(3x-2))

so the verical asymtote should be x=-4 and x=2/3

but when I graph on calculator, x=2/3 doesn't appear on that.... It's a error

but It's not a hole, because it's not a factor of both demoniator and numerator...

SO. what is the vertical asymtote??
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JPB

1
Mon 16 May, 2011 03:42 pm
@daivinhtran,
2(x-3)? Look at that part again.
daivinhtran

1
Mon 16 May, 2011 06:22 pm
@JPB,
I'm talking about vertical asymtote.. THat part doesn't matter
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