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Math Help - [SOLVED] asymptotes!

  1. #1
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    [SOLVED] asymptotes!

    Hi,
    This is on the Barron SAT Math II page 17 # 28

    Which of the following lines are asymptotes of the graph of y= frac{x}{x+1}?

    I. x=1
    II. x=-1
    III. y=1

    (For the answer you can choose two or more, or even none if its all incorrect.)

    I know that x=1 is WRONG, and that x=-1 is CORRECT. But what about y=1? Why is it also an asymptote?

    Thanks in advance
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by fabxx View Post
    Hi,
    This is on the Barron SAT Math II page 17 # 28

    Which of the following lines are asymptotes of the graph of y= frac{x}{x+1}?

    I. x=1
    II. x=-1
    III. y=1

    (For the answer you can choose two or more, or even none if its all incorrect.)

    I know that x=1 is WRONG, Yup

    and that x=-1 is CORRECT. Yup

    But what about y=1? Why is it also an asymptote?

    Thanks in advance
    Examine the behavior of the function as x\to{\pm\infty}

    So evaluate \lim_{x\to\pm\infty}\frac{x}{x+1} and see what you get. This will be the horizontal asymptote with an equation of the form y=k.

    --Chris
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  3. #3
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by fabxx View Post
    Hi,
    This is on the Barron SAT Math II page 17 # 28

    Which of the following lines are asymptotes of the graph of y= frac{x}{x+1}?

    I. x=1
    II. x=-1
    III. y=1

    (For the answer you can choose two or more, or even none if its all incorrect.)

    I know that x=1 is WRONG, and that x=-1 is CORRECT. But what about y=1? Why is it also an asymptote?

    Thanks in advance

    \lim_{x \to \pm \infty}~ \frac{x}{x+1}

    You can solve this using L'hopital rule or divide by the largest degree of x

    \lim_{x \to \pm \infty}~ \frac{\frac{x}{x}}{\frac{x}{x} +\frac{1}{x}}

    which simplfies to

    \lim_{x \to \pm \infty}~ \frac{1}{1+\frac{1}{x}} = 1
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    \lim_{x \to \pm \infty}~ \frac{x}{x+1}

    You can solve this using L'hopital rule or divide by the largest degree of x

    \lim_{x \to \pm \infty}~ \frac{\frac{x}{x}}{\frac{x}{x} +\frac{1}{x}}

    which simplfies to

    \lim_{x \to \pm \infty}~ \frac{1}{1+\frac{1}{x}} = 1
    Ha.

    Or you can cheat...

    \lim_{x\to\infty}\frac{x}{x+1}=\frac{1}{1}=1.

    In general,

    \lim_{x\to\infty}\frac{a_nx^n+a_{n-1}x^{n-1}+\dots+a_2x^2+a_1x+a_0}{b_mx^m+b_{m-1}x^{m-1}+\dots+b_2x^2+b_1x+b_0}=\left\{\begin{array}{rl}  0&\text{when }m>n\\ \displaystyle\frac{a_n}{b_m}&\text{when }m=n\\\infty&\text{when }m<n\end{array}\right.

    --Chris
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  5. #5
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Ha.

    Or you can cheat...

    \lim_{x\to\infty}\frac{x}{x+1}=\frac{1}{1}=1.

    In general,

    \lim_{x\to\infty}\frac{a_nx^n+a_{n-1}x^{n-1}+\dots+a_2x^2+a_1x+a_0}{b_mx^m+b_{m-1}x^{m-1}+\dots+b_2x^2+b_1x+b_0}=\left\{\begin{array}{rl}  0&\text{when }m>n\\ \displaystyle\frac{a_n}{b_m}&\text{when }m=n\\\infty&\text{when }m<n\end{array}\right.

    --Chris
    Lol
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Ha.

    Or you can cheat...

    \lim_{x\to\infty}\frac{x}{x+1}=\frac{1}{1}=1.

    In general,

    \lim_{x\to\infty}\frac{a_nx^n+a_{n-1}x^{n-1}+\dots+a_2x^2+a_1x+a_0}{b_mx^m+b_{m-1}x^{m-1}+\dots+b_2x^2+b_1x+b_0}=\left\{\begin{array}{rl}  0&\text{when }m>n\\ \displaystyle\frac{a_n}{b_m}&\text{when }m=n\\\infty&\text{when }m<n\end{array}\right.

    --Chris
    excellent summary. the OP should bare in mind that he should watch his signs for the m < n case. you can get - \infty depending on the coefficients of x^n and x^m.

    a similar outcome is of course generated by taking the limit as x \to - \infty, it this case, you also have to watch out how the values of n and m change the signs
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Jhevon View Post
    excellent summary. the OP should bare in mind that he should watch his signs for the m < n case. you can get - \infty depending on the coefficients of x^n and x^m.

    a similar outcome is of course generated by taking the limit as x \to - \infty
    Yup...I just focused on the case with x\to\infty

    I'm to tired to type out what happens when x\to-\infty

    --Chris
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