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

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

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

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

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

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