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imranmeah91Best ResponseYou've already chosen the best response.1
if numerator is one degree larger than denominator you have oblique asymtote
 2 years ago

askme12345Best ResponseYou've already chosen the best response.0
how do you find the horizontal asymptote/oblique?
 2 years ago

imranmeah91Best ResponseYou've already chosen the best response.1
horizontal 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
 2 years ago

askme12345Best ResponseYou've already chosen the best response.0
what about vertical? thanks so much!!
 2 years ago

imranmeah91Best ResponseYou've already chosen the best response.1
vertical or oblique?
 2 years ago

askme12345Best ResponseYou've already chosen the best response.0
I need to know both haha 
 2 years ago

imranmeah91Best ResponseYou've already chosen the best response.1
Vertical 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
 2 years ago

imranmeah91Best ResponseYou've already chosen the best response.1
Oblique 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
 2 years ago
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