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You are not logged in. #1 20121022 11:16:39
find domain and asymptotes and y interceptI have to find the domain, vertical and horizontal asymptotes, and the yintercept and graph it for. #2 20121022 15:20:54
Re: find domain and asymptotes and y interceptHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #3 20121022 15:34:44
Re: find domain and asymptotes and y interceptI believe my graph is correct. I will try it out. #4 20121022 15:52:45
Re: find domain and asymptotes and y interceptI'm also pretty sure both of the asymptotes are correct, not so sure about the domain and y intercept. The graph is also kind of sloppy. #5 20121022 19:15:43
Re: find domain and asymptotes and y intercepthi Skylor, when x = 0 The vertical asymptotes are at x = 2 and x = + 3. I'm never happy with these questions that ask for the domain. I think the setter should be declaring a domain as part of the function definition rather than expecting the student to guess what domain is expected. But, assuming the questioner means "The domain is the set of real numbers less any values that are inadmissible", then the inadmissible values are x = 2 and x = +3, because at these x values the f(x) value cannot be computed as it would involve division by zero. As x tends to + or  infinity the graph approaches f(x) = 3x^2 so there are no horizontal asymptotes. g(x) : the quadratic can be similarly factorised to give the asymptotes and inadmissible values. As x tends to + or  infinity the graph approaches g(x) = 2x. As this is a line, it counts as an asymptote, but it is not horizontal. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #6 20121023 03:29:18
Re: find domain and asymptotes and y interceptI'm sorry I must have written the problem wrong #7 20121023 05:01:28
Re: find domain and asymptotes and y interceptWell no I don't. I thought you meant and that's the question I tried to answer in post 5. If that isn't right then the only other interpretation I can think of is That has x = 0 as an asymptote and doesn't cross the y axis at all. This shows the importance of correct placing of brackets. Oh wait a minute. Did you mean: In which case, same asymptotes as post 5 and same crossing .... in fact all the answers are unchanged, except ..... As x tends to infinity the function tends towards So now we have a horizontal asymptote. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei 