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anonymous

  • 4 years ago

For the following function find the vertical, horizontal, and oblique asymptotes, if any f(x)= -23x^2/2x^2-3x^2-18x +27

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  1. dumbcow
    • 4 years ago
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    is that supposed to be 2x^3 in denominator ?

  2. anonymous
    • 4 years ago
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    yes

  3. dumbcow
    • 4 years ago
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    since the degree in denominator is greater, the horizontal asymptote is y=0 and there are no oblique to find the vertical asymptotes, set the denominator equal to zero and solve for x

  4. dumbcow
    • 4 years ago
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    you may be able to solve it using factor by grouping

  5. dumbcow
    • 4 years ago
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    were you able to factor the denominator?

  6. anonymous
    • 4 years ago
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    no

  7. anonymous
    • 4 years ago
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    im so confused

  8. dumbcow
    • 4 years ago
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    2x^3-3x^2-18x +27 group them (2x^3 -3x^2) +(-18x+27) factor each group 1st group has a x^2 in common 2nd group has a -9 in common --> x^2(2x -3) -9(2x-3) Notice the terms inside parenthesis are the same, so factor them out --> (2x-3)(x^2 -9)

  9. anonymous
    • 4 years ago
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    k

  10. anonymous
    • 4 years ago
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    so would this be my answer

  11. dumbcow
    • 4 years ago
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    no you are finding the vertical asymptotes right....so your answer will be x = ? take whats above and set it equal to 0

  12. anonymous
    • 4 years ago
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    how would i set it to equal 0

  13. dumbcow
    • 4 years ago
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    put a "=0" after it ??

  14. anonymous
    • 4 years ago
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    ok i did that

  15. dumbcow
    • 4 years ago
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    solve for x: 2x-3 = 0 x^2 -9 = 0

  16. anonymous
    • 4 years ago
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    ok

  17. anonymous
    • 4 years ago
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    ok what do i have to do now

  18. dumbcow
    • 4 years ago
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    solve the 2 equations...this is algebra, get x by itself if you're taking a class where you have to find asymptotes and evaluate functions, im assuming you know how to solve simple algebraic equations

  19. anonymous
    • 4 years ago
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    x=3/2 and x=-3 or x=3

  20. dumbcow
    • 4 years ago
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    yep :)

  21. anonymous
    • 4 years ago
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    so i'm done

  22. anonymous
    • 4 years ago
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    ty

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