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mamma17

  • one year ago

Describe the vertical asymptote(s) and hole(s) for the graph of y=(x-1)(x+3)/(x+3)(x-5) [1:06:17 PM] Akora: asymptote: x = –1 and hole: x = 5 asymptote: x = 5 and hole: x = –3 asymptote: x = –5 and hole: x = 3 asymptotes: x = 5 and x = –3

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  1. rajathsbhat
    • one year ago
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    why did you close the last question? It looks like you're pasting the questions from a test.

  2. mamma17
    • one year ago
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    no it aint from a testitis jux i did not neeit any more ani figured it out and copy them from skype cause i am asng some oer fiends as well

  3. rajathsbhat
    • one year ago
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    oh? so what was the answer to the last question?

  4. mamma17
    • one year ago
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    i got A

  5. rajathsbhat
    • one year ago
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    OK that's right. If you're really not cheating, I'm sorry I doubted you.

  6. mamma17
    • one year ago
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    iaint it is a quick checkand idid the lesson jux d not undrsand it

  7. satellite73
    • one year ago
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    \[ y=\frac{(x-1)(x+3)}{(x+3)(x-5)}\]

  8. rajathsbhat
    • one year ago
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    ok then. This question is exactly like the previous questions. Do the exact same thing.

  9. satellite73
    • one year ago
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    you can cancel the \(x+3\) factor, which means it has a "hole" at \(x=-3\)

  10. satellite73
    • one year ago
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    however you cannot cancel the factor of \(x-5\) so it has a vertical asymptote at \(x=5\)

  11. mamma17
    • one year ago
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    figured it out

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