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Math Help - Horizontal and vertical asymptotes

  1. #1
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    Horizontal and vertical asymptotes

    Find the vertical and horizonal asymtotes of the function:

    (x-9)/sqrt(3x^2+4x+3)
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Asuhuman18 View Post
    Find the vertical and horizonal asymtotes of the function:

    \frac{x-9}{\sqrt{3x^2+4x+3}}
    Vertical asymptotes occur when the denominator is zero. Horizontal asymptotes are found by taking the limits as f(x) approaches positive and negative infinity.

    Spoiler:
    3x^2+4x+3=0 has no real solutions, so there are no vertical asymptotes.

    Before I take the infinite limits, I'll do a little work on f(x).

    \frac{x-9}{\sqrt{3x^2+4x+3}} = \frac{x-9}{\sqrt{3x^2(1+\frac{4}{3x}+\frac{1}{x^2})}} = \frac{x(1-\frac{9}{x})}{\sqrt{3}\cdot|x|\cdot\sqrt{1+\frac{4  }{3x}+\frac{1}{x^2}}}=\frac{sign(x)}{\sqrt{3}}\cdo  t\frac{1-\frac{9}{x}}{\sqrt{1+\frac{4}{3x}+\frac{1}{x^2}}}, where sign(x) is 1 if x is positive and -1 if x is negative.

    \lim_{x\to\infty}\frac{sign(x)}{\sqrt{3}}\cdot\fra  c{1-\frac{9}{x}}{\sqrt{1+\frac{4}{3x}+\frac{1}{x^2}}}=  \frac{1}{\sqrt{3}}\cdot\frac{1-0}{\sqrt{1+0+0}} = \frac{1}{\sqrt{3}}


    \lim_{x\to-\infty}\frac{sign(x)}{\sqrt{3}}\cdot\frac{1-\frac{9}{x}}{\sqrt{1+\frac{4}{3x}+\frac{1}{x^2}}}=  \frac{-1}{\sqrt{3}}\cdot\frac{1-0}{\sqrt{1+0+0}} = \frac{-1}{\sqrt{3}}

    So horizontal asymptotes occur at \boxed{y=\pm\frac{1}{\sqrt{3}}}
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