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Math Help - Radical function with no vertical asymptote

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    Radical function with no vertical asymptote

    Hi,
    I need help with a problem. I don't understand it.

    Describe the condition that will produce a rational function with a graph that has no vertical asymtote.

    Thanks for any help!
    Michele
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    Quote Originally Posted by mgannon93 View Post
    Hi,
    I need help with a problem. I don't understand it.

    Describe the condition that will produce a rational function with a graph that has no vertical asymtote.
    Where do the vertical asymptotes of a rational function occur? So then what can you do to prevent a vertical asymptote?
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    Quote Originally Posted by Reckoner View Post
    Where do the vertical asymptotes of a rational function occur? So then what can you do to prevent a vertical asymptote?
    The vertical asymptotes occur in the denominator, right? When they equal zero?

    So if there's no x's in the denominator?

    I'm sorry I'm difficult.. it sounds like such an easy problem but I don't get it.
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    Quote Originally Posted by mgannon93 View Post
    The vertical asymptotes occur in the denominator, right? When they equal zero?
    Right. And what happens if there are no zeroes?
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    Quote Originally Posted by Reckoner View Post
    Right. And what happens if there are no zeroes?
    Nothing crosses the x-axis?
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    Quote Originally Posted by mgannon93 View Post
    Nothing crosses the x-axis?
    I meant in the denominator. If the vertical asymptotes of a rational function only occur when the denominator is zero, then what happens if the denominator has no zeroes?

    Although technically, if the numerator and denominator are simultaneously zero at a point, it is possible that there is no vertical asymptote there; it depends on the multiplicity of the factors.
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    Quote Originally Posted by Reckoner View Post
    I meant in the denominator. If the vertical asymptotes of a rational function only occur when the denominator is zero, then what happens if the denominator has no zeroes?

    Although technically, if the numerator and denominator are simultaneously zero at a point, it is possible that there is no vertical asymptote there; it depends on the multiplicity of the factors.
    Then the denominator has to be just a number, without a variable?

    So as long as the denominator doesn't have an x in it, there won't be a vertical asymtote. So an equation like,
    (x-2)(x+5)=0 would have no vertical asymtote.

    ?
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    Quote Originally Posted by mgannon93 View Post
    Then the denominator has to be just a number, without a variable?

    So as long as the denominator doesn't have an x in it, there won't be a vertical asymtote. So an equation like,
    (x-2)(x+5)=0 would have no vertical asymtote.

    ?
    Tell me then, what are the (real) zeroes of x^2+1?
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    Quote Originally Posted by Reckoner View Post
    Tell me then, what are the (real) zeroes of x^2+1?
    OH! So equations that give you imaginary numbers!

    Wow I feel like an idiot. Thanks. Is this explained alright?

    --

    Describe the condition that will produce a rational function with a graph that has no vertical asymptote.

    Rational functions with no rational zeroes will have no vertical asymptote, so an equation like x^2+1 would be a rational function with no vertical asymptotes.

    Is that right?
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    Quote Originally Posted by mgannon93 View Post
    Rational functions with no rational zeroes will have no vertical asymptote, so an equation like x^2+1 would be a rational function with no vertical asymptotes.

    Is that right?
    No, you aren't quite getting it. Yes, the function f(x)=x^2+1 has no vertical asymptotes, and we may consider it a rational function. But what I was trying to get you to see was that you can put it in the denominator, and the denominator will always be nonzero (that is, the denominator doesn't have to be a constant).

    g(x)=\frac1{x^2+1}

    has no vertical asymptotes.
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    Quote Originally Posted by Reckoner View Post
    No, you aren't quite getting it. Yes, the function f(x)=x^2+1 has no vertical asymptotes, and we may consider it a rational function. But what I was trying to get you to see was that you can put it in the denominator, and the denominator will always be nonzero (that is, the denominator doesn't have to be a constant).

    g(x)=\frac1{x^2+1}

    has no vertical asymptotes.
    Oh! Thank you thank you thank you!
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