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3psilon

  • 2 years ago

Help finding ALL asymtotes

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  1. 3psilon
    • 2 years ago
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    of \[y = \frac{ x^2 - 3x + 2 }{ x^{2}-1 }\]

  2. josiahh
    • 2 years ago
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    \[((x-1)(x-2))/((x-1)(x+1))\] \[(x-2)/(x+1)\] 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 (x-1) factor

  3. 3psilon
    • 2 years ago
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    I'm not really understanding the horizontal asymtote part :(

  4. josiahh
    • 2 years ago
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    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

  5. 3psilon
    • 2 years ago
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    Do you always do that to find the horizontal asymtote ?

  6. josiahh
    • 2 years ago
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    yea. horizontal retricemptotes describe the behavior of the graph as x increases

  7. 3psilon
    • 2 years ago
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    @josiahh Doing it algebraically I'm not so good with limits So can you show me algebraically how the limit = 1 please?

  8. josiahh
    • 2 years ago
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    when you are taking limits at infinity, you look at the degree of the polynomials in the denominator and the numerator

  9. 3psilon
    • 2 years ago
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    Ohhhh and if they are equal its Coeffcient over coeffcient?

  10. josiahh
    • 2 years ago
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    yea that rule...

  11. 3psilon
    • 2 years ago
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    Thank YOU!

  12. josiahh
    • 2 years ago
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    welcome

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