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Math Help - Finding equation of asymptotes-Specialists maths

  1. #1
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    Finding equation of asymptotes-Specialists maths

    The equations of the asymptotes of the curve with the equation-

    y=4x+3\x^2-9
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    Quote Originally Posted by johnsy123 View Post
    The equations of the asymptotes of the curve with the equation-

    y=4x+3/x^2-9
    1. Re-arrange the given equation:

    y = \dfrac{4x+3}{x^2-9} = \dfrac{4x+3}{(x-3)(x+3)}

    2. The zeros of the denominator will yield the equations of the vertical asymptotes.

    3. Re-arrange the given equation by long division:

    (4x+3)\div (x^2-9)=\underbrace{0}_{\text{term of asymptote}}+\dfrac{4x+3}{x^2-9}

    That means you'll get a horizontal asymptote since the term of the asymptote is a constant (the slope is zero)
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    Sorry i made an error. The equation is y= 4x +3/x^2-9.
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    MHF Contributor FernandoRevilla's Avatar
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    These problems are just routine if you know the corresponding theory. For example:


    (a) Horizontal:

    Find a=\displaystyle\lim_{x \to \infty}{f(x)}

    (b) Vertical:

    Find finite values of x_0 such that y\rightarrow{\infty} as x\rightarrow x_0

    (iii) Oblique

    ...

    Try it.

    Fernando Revilla
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    So what your simplistically saying, is to find the horizontal asymptote i should sub in infinite x values and to find the vertical asymptote i should sub in infinite y values?
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  6. #6
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    Quote Originally Posted by earboth View Post
    1. Re-arrange the given equation:

    y = \dfrac{4x+3}{x^2-9} = \dfrac{4x+3}{(x-3)(x+3)}

    2. The zeros of the denominator will yield the equations of the vertical asymptotes.

    3. Re-arrange the given equation by long division:

    (4x+3)\div (x^2-9)=\underbrace{0}_{\text{term of asymptote}}+\dfrac{4x+3}{x^2-9}

    That means you'll get a horizontal asymptote since the term of the asymptote is a constant (the slope is zero)
    Better yet, long divide first Then your constant will be the horizontal asymptote and the zeros of the denominator will be your vertical (or other) asymptotes.

    And no johnsy, you can't substitute \displaystyle \infty because \displaystyle \infty is not a number. But you can think about what would happen as your function gets infinitely large in either direction...
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