According to my study guide this should require calculating six limits???
I am lost. I think 2 and -2 are asymptotes by finding zeros in the bottom but dont know how to calculate six limits to find asymptotes.
Thank You in Advance!
2007-06-20 20:25:02 · 4 answers · asked by Lanksta 1 in Science & Mathematics ➔ Mathematics
Yes, there are vertical asymptotes at x = ± 2.
There are also horizontal asymptotes:
lim (x->∞) 3x^2 / (x^2 - 4) = 3
lim (x->-∞) 3x^2 / (x^2 - 4) = 3
So there is a horizontal asymptote at y = 3 (asymptotic in both directions).
There are no other asymptotes.
2007-06-20 20:36:38 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
there is a vertical asymptote at y=+3 . Just check by tending x to infinity and using LH rule.
there is also a horizontal asymptote at x=2,x=-2.
There are no inclined asymptotes
2007-06-21 06:11:04 · answer #2 · answered by Sal 2 · 0⤊ 0⤋
Vertical asymptotes: x = ±2
Horizontal asymptotes: y = 3
2007-06-21 03:34:02 · answer #3 · answered by sahsjing 7 · 0⤊ 0⤋
im not too sure on this but maybe you can figure something from it
so you have 2 and -2 as asymptotes
so you look at the lim as x->2 from the neg. side and one from the pos. side.
you do that again as x-> -2
then you look at as x-> infinity
and again as x-> - infinity
the infinities should give you an asymptote at y=3 for both pos and neg infinity
2007-06-21 03:41:25 · answer #4 · answered by blahman 2 · 0⤊ 0⤋