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Math - asymptotes

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Identify all asymptotes.

1. y= 1 / 2-√x^2 - 3x

the asymptotes I got were:
vertical: 1 & 4
horizontal: 0
Is this it or are there more?

2. y= x^3-2x / x^2 + 1

I believe there are no asymptotes? Is this correct?

  • Math - asymptotes -

    You have vertical asymptotes when the denominator is zero
    your values of x=1 and x=4 do not result in a zero denominator.

    I have a feeling that your function is
    y = 1/(2 - √(x^2-3x))
    in that case the denominator is zero when x = 4 or x = -1
    so your vertical asymptotes are
    x = 4 and x = -1
    (your horizontal is y = 0)

    2. Again, I think your meant your function to b
    y = (x^3 - 2x)/(x^2 + 1)
    Please confirm

  • Math - asymptotes -

    yes, those are what i meant..sorry about the notation

  • Math - asymptotes -

    PLease don't switch names.

    for y = (x^3 - 2x)/(x^2 + 1)
    there is no vertical or horizontal asymptote.

    but when you divide it you get

    y = x - 3x/(x^2+1)

    so there is a "slanted" asymptote of y = x

  • math -

    The area of a circular trampoline is 112.07 square feet. What is the radius
    of the trampoline? Round to the nearest hundredth

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