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Math Help - Asymptote-Mid Terms

  1. #1
    Member OnMyWayToBeAMathProffesor's Avatar
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    Asymptote-Mid Terms

    Hello,

    I have mid terms tomorrow, and i completely blanked on how to do this:

    \frac{x^2-3}{x+2}

    I have to find the domain, range, vertical/horizontal asymptote and discontinuities, if any, what kind; removable to or continuous.

    I know the vertical asymptote is -2 but how do i find the horizontal ones. i graphed it and the graph did have multiple horizontal asymptotes.

    Thank you.


    (new addition)

    \frac{7x}{\sqrt{x^2-10}}

    same things
    Last edited by OnMyWayToBeAMathProffesor; January 16th 2008 at 03:54 PM. Reason: new addition
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    Hello,

    I have mid terms tomorrow, and i completely blanked on how to do this:

    \frac{x^2-3}{x+2}

    I have to find the domain, range, vertical/horizontal asymptote and discontinuities, if any, what kind; removable to or continuous.

    I know the vertical asymptote is -2 but how do i find the horizontal ones. i graphed it and the graph did have multiple horizontal asymptotes.

    Thank you.


    (new addition)

    \frac{7x}{\sqrt{x^2-10}}

    same things
    Ask yourself this question... is there a horizontal asymptote? Take infinite limits of this function, and you will find that it goes to negative and positive infinity on its ends.
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  3. #3
    Math Engineering Student
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    We don't even need to take limits, algebra just tells us the answer.

    f(x) = \frac{{x^2  - 3}}<br />
{{x + 2}} = \frac{{(x + 2)(x - 2) + 1}}<br />
{{x + 2}} = x - 2 + \frac{1}<br />
{{x + 2}}.

    We have a slant asymptote, so it doesn't exist horizontal asymptote.
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  4. #4
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    Hello,

    I have mid terms tomorrow, and i completely blanked on how to do this:

    \frac{x^2-3}{x+2}

    I have to find the domain, range, vertical/horizontal asymptote and discontinuities, if any, what kind; removable to or continuous.

    I know the vertical asymptote is -2 but how do i find the horizontal ones. i graphed it and the graph did have multiple horizontal asymptotes.

    Thank you.


    (new addition)

    \frac{7x}{\sqrt{x^2-10}}

    same things
    y = \frac{7x}{\sqrt{x^2-10}} = \frac{7}{\sqrt{1 - \frac{10}{x^2}}} over the domain. What does y do in the limit x \rightarrow \pm \infty ?
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  5. #5
    Member OnMyWayToBeAMathProffesor's Avatar
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    reply to mr. F

    as x approaches infinity, y approaches 7. and as x approaches -infinity, y approaches 7. but how? also, how did you get, <br />
y = \frac{7x}{\sqrt{x^2-10}} = \frac{7}{\sqrt{1 - \frac{10}{x^2}}}<br />

    Thank you
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  6. #6
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    Quote Originally Posted by Krizalid View Post
    We don't even need to take limits, algebra just tells us the answer.

    f(x) = \frac{{x^2  - 3}}<br />
{{x + 2}} = \frac{{(x + 2)(x - 2) + 1}}<br />
{{x + 2}} = x - 2 + \frac{1}<br />
{{x + 2}}.

    We have a slant asymptote [snip]
    *Ahem* Krizalid ... how does one know that there's a slant asymptote without considering some sort of limit .....

    I know what you mean , but I think it encourages the superficial rote learning of a fact, rather than an understanding of the theory behind that fact .... If a student asked does y = \frac{1}{x} + 1, say, have a horizontal asymptote, I would very much hope that the answer they got used the concept of a limit at some stage .....
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  7. #7
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    as x approaches infinity, y approaches 7. and as x approaches -infinity, y approaches 7. but how? also, how did you get, <br />
y = \frac{7x}{\sqrt{x^2-10}} = \frac{7}{\sqrt{1 - \frac{10}{x^2}}}<br />

    Thank you
    y = \frac{7x}{\sqrt{x^2-10}} = \frac{7x}{\sqrt{x^2 \left( 1 - \frac{10}{x^2} \right)}} = \frac{7x}{x \sqrt{1 - \frac{10}{x^2}}} = ....
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  8. #8
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    as x approaches infinity, y approaches 7. and as x approaches -infinity, y approaches 7. but how? also, how did you get, <br />
y = \frac{7x}{\sqrt{x^2-10}} = \frac{7}{\sqrt{1 - \frac{10}{x^2}}}<br />

    Thank you
    \frac{10}{x^2} \rightarrow 0 as x \rightarrow \pm \infty. So y \rightarrow \frac{7}{\sqrt{1 - 0}} = ....
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  9. #9
    Member OnMyWayToBeAMathProffesor's Avatar
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    so the numerator basically defines the end behavior? so:

    <br />
y \rightarrow \frac{7}{\sqrt{1 - 0}} = ....<br />
as <br />
x \rightarrow \pm \infty<br />
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  10. #10
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    so the numerator basically defines the end behavior? so:

    <br />
y \rightarrow \frac{7}{\sqrt{1 - 0}} =<br />
7 as <br />
x \rightarrow \pm \infty<br />
Mr F edit: added the 7.
    Not quite. It only seems that way since the denominator --> 1 .....

    To understand what I'm saying, try finding the horizontal asymptote of y = \frac{7x}{\sqrt{3x^2-10}}, say ......
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post

    \frac{7x}{\sqrt{x^2-10}}

    same things
    this question is not as simple as you are all making it out to be

    first note that the 10 in the denominator is immaterial as x goes to infinity or negative infinity.

    also note, that an alternative definition for |x| is \sqrt{x^2}

    thus we have: \lim_{x \to \pm \infty} \frac {7x}{\sqrt{ x^2 - 10}} = \lim_{x \to \pm \infty} \frac {7x}{\sqrt{x^2}} = \lim_{x \to \pm \infty} \frac {7x}{|x|} = \left \{ \begin{array}{lr} 7 & \mbox {as } x \to \infty \\ & \\ -7 & \mbox{ as } x \to - \infty \end{array} \right.

    so there are two horizontal asymptotes here


    for the vertical asymptotes, find where the function is undefined, that is, where the denominator is zero. such x's are vertical asymptotoes. these are also the x's that are NOT in the domain (among other x's perhaps, here for instance, we have to find out what x's causes the square root to mess up)

    are you good on everything else, MathProfessor?
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  12. #12
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    Thumbs up Good get.

    Quote Originally Posted by Jhevon View Post
    this question is not as simple as you are all making it out to be
    [snip]
    also note, that an alternative definition for |x| is \sqrt{x^2}

    thus we have: \lim_{x \to \pm \infty} \frac {7x}{\sqrt{ x^2 - 10}} = \lim_{x \to \pm \infty} \frac {7x}{\sqrt{x^2}} = \lim_{x \to \pm \infty} \frac {7x}{|x|} = \left \{ \begin{array}{lr} 7 & \mbox {as } x \to \infty \\ & \\ -7 & \mbox{ as } x \to - \infty \end{array} \right.

    so there are two horizontal asymptotes here
    Quite right - good pick up.
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  13. #13
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    Quote Originally Posted by mr fantastic View Post
    y = \frac{7x}{\sqrt{x^2-10}} = \frac{7x}{\sqrt{x^2 \left( 1 - \frac{10}{x^2} \right)}} = \frac{7x}{x \sqrt{1 - \frac{10}{x^2}}} = ....
    Correction:


    y = \frac{7x}{\sqrt{x^2-10}} = \frac{7x}{\sqrt{x^2 \left( 1 - \frac{10}{x^2} \right)}} = \frac{7x}{|x| \sqrt{1 - \frac{10}{x^2}}} = \frac{7 \, \text{sgn}(x)}{\sqrt{1 - \frac{10}{x^2}}}


    The appropriate corrections in my subsequent posts are easy to see and simple to make.
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