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

josiahh
 2 years ago
Best 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

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

josiahh
 2 years ago
Best ResponseYou've already chosen the best response.1you 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

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

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

3psilon
 2 years ago
Best 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?

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

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