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Thread: Identify any vertical, horizontal, or slant asymptotes

  1. #1

    Identify any vertical, horizontal, or slant asymptotes

    Identify any vertical, horizontal, or slant asymptotes


    b) f(x)=x-4x-6 / 2x-2


    b) Vertical: Take the denominator and simplify
    2x^(2)-2 ---> 2x^(2)-2=0 ----> 2x^(2)=2 ---> x^(2)=1 ----> x=+/-1 (is this correct or should it be +/- square root of 1?)

    The exponent 3>2 so there are no horizontal asymptotes

    oblique is= x/2-2 ???? How do I get oblique?

  2. #2
    Elite Member
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    Hello, lovetolearn!

    Identify any vertical, horizontal, or slant asymptotes

    . . (b) .f(x) .= .(x3 - 4x2 - 6)/(2x2 - 2)

    Vertical: Take the denominator and simplify
    . . 2x2 - 2 .= .0 . . . . 2x2 = 2 . . . . x2 = 1 . . . . x = +1

    The exponent 3 > 2, so there are no horizontal asymptotes

    oblique is = x/2 - 2 ? . . . How do I get oblique?

    Long division.

    (x3 - 4x2 - 6 ) divided by (2x2 - 2) .= .x/2 - 2 . . . remainder x - 10


    We have:
    . . . . . . . x3 - 4x2 - 6 . . . . .x . . . . . .. x - 10
    f(x) . = . -------------- . = . -- .- .2 .+ .--------
    . . . . . . . . . 2x2 - 2 . . . . . .2 . . .. . . . 2x2 - 2


    As x → ∞, the second fraction → 0.

    And the graph approaches: .y .= .x/2 - 2

  3. #3
    Thank you so much! I understand the concept now!

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