Reply
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??
@daivinhtran,
2(x-3)? Look at that part again.
@JPB,
I'm talking about vertical asymtote.. THat part doesn't matter