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imranmeah91
 2 years ago
Best ResponseYou've already chosen the best response.1if numerator is one degree larger than denominator you have oblique asymtote

askme12345
 2 years ago
Best ResponseYou've already chosen the best response.0how do you find the horizontal asymptote/oblique?

imranmeah91
 2 years ago
Best ResponseYou've already chosen the best response.1horizontal is very easy. if you have same degree numerator and denominator , just divide the coeficcient 4x^2  = > 4/2 = > horizontal retricemtote=> y=2 2 x^2

askme12345
 2 years ago
Best ResponseYou've already chosen the best response.0what about vertical? thanks so much!!

imranmeah91
 2 years ago
Best ResponseYou've already chosen the best response.1vertical or oblique?

askme12345
 2 years ago
Best ResponseYou've already chosen the best response.0I need to know both haha 

imranmeah91
 2 years ago
Best ResponseYou've already chosen the best response.1Vertical for vertical asymtote , look for where function is inderminant(0 in denominator) (x^23x)  (x5) in this case when x=5 , denominator is 0 which makes function inderminant so vertical asymtote x=5

imranmeah91
 2 years ago
Best ResponseYou've already chosen the best response.1Oblique when numerator is one degree bigger than denominator (x^3+x^2+x+1)/(x^2+x+1) use long division to divide it out http://www4b.wolframalpha.com/Calculate/MSP/MSP53619icgb4di5hgchfe00003fc2e54f7d06c4a5?MSPStoreType=image/gif&s=21&w=204&h=131 y=x is our oblique asymtote
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