intothevoidx Posted October 30, 2007 Share Posted October 30, 2007 The function y= (3x^3-2x+1)/(2x^2) has me confused. I am trying to figure out why this function has a horizontal asymptote when the degree of the numerator is larger than the denominator, which I thought was supposed to mean it has no horizontal asymptote? Thanks Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted October 30, 2007 Share Posted October 30, 2007 It doesn't have a horizontal asymptote. Try graphing it. Link to comment Share on other sites More sharing options...
intothevoidx Posted October 30, 2007 Author Share Posted October 30, 2007 Well, then my wonderful textbook(sarcasm) has an error in it....It says the answer is y=3/2*x. Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted October 30, 2007 Share Posted October 30, 2007 That's roughly the slope of the line you get when you graph the original function. Check the directions again to see what it's asking for. Link to comment Share on other sites More sharing options...
intothevoidx Posted October 31, 2007 Author Share Posted October 31, 2007 It says determine any horizontal, vertical or slant asymptotes. I think it may be the slant asymptote. My bad.... I figure I should ask my other question on this post as well. This problem is from a Cliff's Book The function (2x-5)/(x^2-4x-5) has a horizontal asymptote of y=2. This function seems like it violates the rules........When I worked it out I got that y=0 because the the numerator has a smaller power than the denominator? Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted October 31, 2007 Share Posted October 31, 2007 You're right again. I graphed it and it approaches 0. Link to comment Share on other sites More sharing options...
intothevoidx Posted October 31, 2007 Author Share Posted October 31, 2007 Well then, I guess I did find an error this time. Thanks for the help. Link to comment Share on other sites More sharing options...
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