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anonymous

  • one year ago

Identify the vertical asymptotes of f(x) = 10 over quantity x squared minus 7x minus 30. x = 10 and x = 3 x = 10 and x = -3 x = -10 and x = 3 x = -10 and x = -3

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  1. anonymous
    • one year ago
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    \[f(x) = \frac{ 10 }{ x^2 -7x -30 }\]

  2. misty1212
    • one year ago
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    set \[x^2-7x-30=0\] solve for \(x\)

  3. anonymous
    • one year ago
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    okay give me a sec

  4. anonymous
    • one year ago
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    fortunalely for you this factors as \[(x-10)(x+3)=0\] so the zeros are easy to find

  5. misty1212
    • one year ago
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    yeah @satellite73 gave you a huge hint there

  6. anonymous
    • one year ago
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    so its x = -10 and x = 3

  7. misty1212
    • one year ago
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    no dear

  8. misty1212
    • one year ago
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    \[x-10=0\\ x=10\]

  9. misty1212
    • one year ago
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    also \[x+3=0\\ x=-3\]

  10. anonymous
    • one year ago
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    oh so i have to bring it to the other side okay i see

  11. misty1212
    • one year ago
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    riiiiiigght

  12. anonymous
    • one year ago
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    thnx i have one more do you mind helping with?

  13. misty1212
    • one year ago
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    sure

  14. anonymous
    • one year ago
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    Identify the horizontal asymptote of f(x) = quantity 2 x minus 1 over quantity x squared minus 7 x plus 3. y = 0 y = 1 over 2 y = 2 No horizontal asymptote

  15. anonymous
    • one year ago
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    \[f(x) = \frac{ 2x - 1 }{ x^2 - 7x +3 }\]

  16. anonymous
    • one year ago
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    do i basically do the same thing as before?

  17. misty1212
    • one year ago
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    no!!

  18. misty1212
    • one year ago
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    horizontal not vertical

  19. anonymous
    • one year ago
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    oh so what do i need to do?

  20. misty1212
    • one year ago
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    the degree of the numerator is less than the degree of the denominator

  21. misty1212
    • one year ago
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    the degree of the numerator is 1, the degree of the denominator i s2

  22. misty1212
    • one year ago
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    that means the denominator grows faster than the numerator so the horizontal asymptote is \[y=0\]

  23. anonymous
    • one year ago
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    thank you so much!!!

  24. anonymous
    • one year ago
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    i gtg but i have lie 5 more questions to do later can you help with?

  25. misty1212
    • one year ago
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    sure dear if i am on

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