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anonymous

  • 5 years ago

determine the vertical asymptotes and the end behavior of the following function. 2x^2+3x+1/((x^2-2)(5-x^2))

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  1. amistre64
    • 5 years ago
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    i dont see it canceling... got 2 linears up top and 2 irreduciles on the bottom

  2. amistre64
    • 5 years ago
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    x = +- sqrt(2) and +-sqrt(5) are your vas then

  3. amistre64
    • 5 years ago
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    ha is y=0

  4. amistre64
    • 5 years ago
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    its like i was never wrong ;)

  5. anonymous
    • 5 years ago
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    haha how did you get y=0? is it from the denominator ?

  6. amistre64
    • 5 years ago
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    it from the shortcut to finding has.... if the bottom degree is greater than the top degree; at all goes to 0 in the end

  7. anonymous
    • 5 years ago
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    so you expand the denomiator out first. right youll get 2X^2/x^4

  8. amistre64
    • 5 years ago
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    pretty much; the coeffs are meangingless in this since the x^4 overpowers everything else

  9. anonymous
    • 5 years ago
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    cool that was easy.. why does it look soo hard. LOL

  10. anonymous
    • 5 years ago
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    they make it seem that way to annoy you. big denominator means small number (by which i mean close to zero). think about \[\frac{(10^6)^2}{(10^6)^4}=\frac{10^{12}}{10^{24}}=\frac{1}{10^{12}}\] a number very close to zero

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spraguer (Moderator)
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