Post by: tram on October 12, 2010, 06:48:25 pm
(x^3 - 1)/ x
are?
(soz for the lack of latex)
Post by: Souljette_93 on October 12, 2010, 06:58:04 pm
Post by: Martoman on October 12, 2010, 08:20:40 pm
as x gets very big or very big negatively the 1/x fraction becomes stupidly close to 0 as the 1/x graph has an asymptote there. The other reasoning if you don't like looking at it graphically is that
Post by: tram on October 12, 2010, 09:28:00 pm
Post by: brightsky on October 12, 2010, 09:58:49 pm
yeah, see my think was EXACTLY what both of you guys said, only i had a srsly lengthy discission with my friends about whether or not y= -1/x was also an asymptote. I said that the asymptote x=0 covered this but they insisted that it was necessary. Opinions????
Hmm..I think it's clear enough from the graph that y = -1/x isn't an asymptote. Also, it doesn't correlate to any of the usual techniques used to find asymptotes. Are you able to provide the reasoning behind their answer?
Post by: tram on October 13, 2010, 05:51:51 pm
Post by: brightsky on October 13, 2010, 07:47:02 pm
well, it had to do with the fact that you were effectively adding ordinates and thus as x approaches 0, then obvsiously x^2 approaches zero, and that means that the graph is apporaching whatever -1/x is => it is an asymptote.
Why let x tend to zero?
Post by: supernatural on October 13, 2010, 07:48:00 pm
the same question was in the mav 2010 trial. thats what the answers say as well.
because when x -> ∞ f(x) ->
and x cannot be zero too, so that's an asymptote as well
I said that the asymptote x=0 covered this
i think so as well. when x-> 0 [from the left or right] f(x)->
and what are the asymptotes for
so x=0 covers this i guess