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3psilon Group TitleBest ResponseYou've already chosen the best response.0
of \[y = \frac{ x^2  3x + 2 }{ x^{2}1 }\]
 one year ago

josiahh Group TitleBest ResponseYou've already chosen the best response.1
\[((x1)(x2))/((x1)(x+1))\] \[(x2)/(x+1)\] you have vertical asymptote at x=1 horizontal asymptotes at negative infinity and positive infinity y=1 is your horizontal asymptote for both infinities x=1 is your only vertical one because you canceled out the (x1) factor
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
I'm not really understanding the horizontal asymtote part :(
 one year ago

josiahh Group TitleBest ResponseYou've already chosen the best response.1
you take limit as x approaches infinity and negative infinity. when you do that, you get 1. limit x approaches negative infinity = limit x approaches infinity =1
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
Do you always do that to find the horizontal asymtote ?
 one year ago

josiahh Group TitleBest ResponseYou've already chosen the best response.1
yea. horizontal retricemptotes describe the behavior of the graph as x increases
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
@josiahh Doing it algebraically I'm not so good with limits So can you show me algebraically how the limit = 1 please?
 one year ago

josiahh Group TitleBest ResponseYou've already chosen the best response.1
when you are taking limits at infinity, you look at the degree of the polynomials in the denominator and the numerator
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
Ohhhh and if they are equal its Coeffcient over coeffcient?
 one year ago

josiahh Group TitleBest ResponseYou've already chosen the best response.1
yea that rule...
 one year ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
Thank YOU!
 one year ago
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