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what is the vertical and horizontal asymptote of (-9x^2 + 1) / x^2 ...? are there any 'holes' in the graph? thanks

2006-11-12 16:49:42 · 2 answers · asked by OO 1 in Education & Reference Homework Help

2 answers

the horizontal asymptote is the limit as x->∞, the vertical as x->0. Since this equation can be written -9 + 1/x^2, when x gets large, 1/x^2 gets small with respect to -9, and can be ignored, leaving -9 as the symptote. As x->0, the 1/x^2 dominates, so the vertical asymptote is 1/x^2. There is a discontinuity at x=0 where the function is undefined.

2006-11-12 18:36:48 · answer #1 · answered by gp4rts 7 · 0 0

Here's a great "calculator" site to help solve these problem! It will give you the asmptotes and holes plus more...

http://www.webgraphing.com/

2006-11-13 00:59:30 · answer #2 · answered by mdetaos 3 · 0 0

fedest.com, questions and answers