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anonymous

  • 5 years ago

find the domain for y =x^2square root (x^2+5)

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  1. myininaya
    • 5 years ago
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    all real numbers.

  2. myininaya
    • 5 years ago
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    that is \[y=x^2*\sqrt{x^2+4}\]

  3. myininaya
    • 5 years ago
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    5 not 4

  4. anonymous
    • 5 years ago
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    so how would you find the vertical asymptote ?

  5. myininaya
    • 5 years ago
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    just ask yourself where is this function not defined when where there be a negative under the square root never since x^2 is always positive so x^2+5 is always positive

  6. myininaya
    • 5 years ago
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    there are no vertical asy. the function is continuous everywhere

  7. anonymous
    • 5 years ago
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    is this since the domain is real for all values of x?

  8. myininaya
    • 5 years ago
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    yes

  9. myininaya
    • 5 years ago
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    if we had 1/(x-1) the vertical asy would be at x=1

  10. anonymous
    • 5 years ago
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    ahhhh thank you very much myininaya!

  11. myininaya
    • 5 years ago
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    if we had (x-1)/[(x-1)(x-2)] then there would be a hole at x=1 and a vertical asy at x=2

  12. myininaya
    • 5 years ago
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    since the x-1 cancel

  13. anonymous
    • 5 years ago
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    thank you ^.^

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