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let f(x)=(x^2+2x-15)/(3x^2+12x-15)

Find the horizontal and vertical asymptotes of f(x).

  • Calculus -

    To have a vertical asymptote , the denominator must be zero
    so, 3x^2 + 12x - 15 = 0
    x^2 + 4x - 5 = 0
    (x+5)(x-1) = 0
    x = -5 or x = 1

    so we have two vertical asymptotes,
    x = -5 and x = 1

    for the horizontal asymptote, let x become large
    then we are left with x^2/3x^2 or 1/3

    V.A. : y = 1/3

  • Calculus -

    I did the exact same thing as you did for the verticle asymptotes, but my online homework says that -5, or 1 are wrong. I even tried putting both of them in list form.

  • Calculus -

    sorry, but I am 100% confident that what I did was correct. Since I can't see what type of input your online website expects, we are at an impasse.

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