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osanseviero

  • 2 years ago

asymptotes of this function:

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  1. osanseviero
    • 2 years ago
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    \[\frac{ 2x-1 }{ x+2 }\]

  2. osanseviero
    • 2 years ago
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    Horizontal: x+2= 0, x=-2...how for the vertcial one?

  3. anonymous
    • 2 years ago
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    The vertical asymptote would be x=-2 (because adding -2 to 2 would give you a zero in the denominator), and the horizontal asymptote would be 2 (as calculated by dividing 2x/x)

  4. osanseviero
    • 2 years ago
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    oh...sorry, I did the vertical before...and what would the horizontal be?

  5. ranga
    • 2 years ago
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    Vertical asymptote is when y goes to infinity. Horizontal asymptote is when x goes to infinity. To find the horizontal asymptote, find the limit of (2x - 1) / (x + 2) as goes to infinity.

  6. osanseviero
    • 2 years ago
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    y=2 :)

  7. ranga
    • 2 years ago
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    yes.

  8. osanseviero
    • 2 years ago
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    So the range is (-inf,2)U(2,INF) ?

  9. ranga
    • 2 years ago
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    Yes the range is correct. Domain is (-inf, -2) U (-2, inf)

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