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Math Help - Finding asymptotes to the curve.

  1. #1
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    Finding asymptotes to the curve.

    Find all the Asymptotes to the curve

    (2x+3)y = (x-1)^2
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    Quote Originally Posted by varunnayudu View Post
    Find all the Asymptotes to the curve

    (2x+3)y = (x-1)^2
    y = \frac{x^2 - 2x + 1}{2x + 3} = \frac{x}{2} - \frac{7}{4} + \frac{25}{4(2x + 3)}.

    The equations of the vertical asymptote and the oblique asymptote should now be clear .....
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    Quote Originally Posted by mr fantastic View Post
    y = \frac{x^2 - 2x + 1}{2x + 3} = \frac{x}{2} - \frac{7}{4} + \frac{25}{4(2x + 3)}.

    The equations of the vertical asymptote and the oblique asymptote should now be clear .....



    hw did u get this ........
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  4. #4
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    Quote Originally Posted by zorro View Post
    hw did u get this ........
    Divide 2x + 3 into x^2 - 2x + 1 using polynomial long division.
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    Quote Originally Posted by varunnayudu View Post
    Find all the Asymptotes to the curve

    (2x+3)y = (x-1)^2
    1. The equation of the vertical asymptote x = {\text{const}} of y = \frac{{\left( {x - 1} \right)^2 }}{{2x + 3}}:

    2x + 3 = 0{\text{ }} \Leftrightarrow {\text{ }}2x =  - 3{\text{ }} \Leftrightarrow {\text{ }}x =  - \frac{3}{2} is the vertical asymptote.

    2. The equation of the oblique asymptote y = ax + b of y = \frac{{\left( {x - 1} \right)^2 }}{{2x + 3}}:

    a = \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}<br />
{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {x - 1} \right)^2 }}<br />
{{\left( {2x + 3} \right)x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{x^2  - 2x + 1}}{{2x^2  + 3x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{1 - \frac{2}{x} + \frac{1}{{x^2 }}}}{{2 + \frac{3}{x}}} = \frac{1}{2}.

    b = \mathop {\lim }\limits_{x \to \infty } \left[ {f\left( x \right) - ax} \right] = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{\left( {x - 1} \right)^2 }}{{2x + 3}} - \frac{x}{2}} \right] = \mathop {\lim }\limits_{x \to \infty } \frac{{ - 7x + 2}}{{4x + 6}} = \mathop {\lim }\limits_{x \to \infty } \frac{{ - 7 + \frac{2}{x}}}{{4 + \frac{6}{x}}} = \frac{{ - 7}}{4}.

    The oblique asymptote is y = \frac{x}{2} - \frac{7}{4}.

    Last edited by DeMath; January 12th 2009 at 12:40 PM.
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    Quote Originally Posted by DeMath View Post
    1. The equations of the vertical asymptote x = {\text{const}} of y = \frac{{\left( {x - 1} \right)^2 }}{{2x + 3}}:

    2x + 3 = 0{\text{ }} \Leftrightarrow {\text{ }}2x = - 3{\text{ }} \Leftrightarrow {\text{ }}x = - \frac{3}{2} is the vertical asymptote.

    2. The equations of the oblique asymptote y = ax + b of y = \frac{{\left( {x - 1} \right)^2 }}{{2x + 3}}:

    a = \mathop {\lim }\limits_{x \to \infty } \frac{{f\left( x \right)}}<br />
{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {x - 1} \right)^2 }}<br />
{{\left( {2x + 3} \right)x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{x^2 - 2x + 1}}{{2x^2 + 3x}} = \mathop {\lim }\limits_{x \to \infty } \frac{{1 - \frac{2}{x} + \frac{1}{{x^2 }}}}{{2 + \frac{3}{x}}} = \frac{1}{2}.

    b = \mathop {\lim }\limits_{x \to \infty } \left[ {f\left( x \right) - ax} \right] = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{\left( {x - 1} \right)^2 }}{{2x + 3}} - \frac{x}{2}} \right] = \mathop {\lim }\limits_{x \to \infty } \frac{{ - 7x + 2}}{{4x + 6}} = \mathop {\lim }\limits_{x \to \infty } \frac{{ - 7 + \frac{2}{x}}}{{4 + \frac{6}{x}}} = \frac{{ - 7}}{4}.

    The oblique asymptote is y = \frac{x}{2} - \frac{7}{4}.

    I suspect that your method for #2 will confuse the OP. Dividing first is the best approach I think.
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  7. #7
    Senior Member DeMath's Avatar
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    Sorry, what is OP?
    Is it Originally Posted?
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  8. #8
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    Quote Originally Posted by DeMath View Post
    Sorry, what is OP?
    Is it Originally Posted?
    OP = Original Poster (the member who started the thread).
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  9. #9
    Senior Member DeMath's Avatar
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    Quote Originally Posted by mr fantastic View Post
    I suspect that your method for #2 will confuse the OP.
    Why?

    Dividing first is the best approach I think.
    The method which I offered is an analytical method, it is written in many textbooks, and we can prove elementaryly this method.
    Last edited by DeMath; January 5th 2009 at 02:01 PM.
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  10. #10
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    Thanks Demath and Mr fantastic

    Quote Originally Posted by DeMath View Post
    Why?



    The method which I offered is an analytical method, it is written in many textbooks, and we can prove elementaryly this method.

    Thanks DEMATH and also Mr fantastic
    cheers
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