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Math Help - Finding asympotes, turning point and domain!

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    Finding asympotes, turning point and domain!

    Given the function f(x)=\frac{A}{x^2-(\alpha +\beta)x+\alpha\beta}, state the equations of all the asymptotes and find the turning points, stating the domain.

    Please show complete working out. I don't know how to find out the above.
    Last edited by Joker37; January 26th 2011 at 03:26 PM. Reason: Forgot to put the middle "x" in the denominator
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    domain of function is region for which function is defined, so solve
     x^2 - (\alpha + \beta) + \alpha \cdot \beta = 0

    from there you will go to asymptotes, vertical, horizontal .....
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    Quote Originally Posted by Joker37 View Post
    Given the function f(x)=\frac{A}{x^2-(\alpha +\beta)+\alpha\beta}, state the equations of all the asymptotes and find the turning points, stating the domain.

    Please show complete working out. I don't know how to find out the above.
    Is your denominator correct ?
    Should the sum of the roots be multiplied by x ?
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    Quote Originally Posted by Archie Meade View Post
    Is your denominator correct ?
    Should the sum of the roots be multiplied by x ?
    Sorry, yeah you're right. I made a mistake. Below is the correct equation:
    f(x)=\frac{A}{x^2-(\alpha +\beta)x+\alpha\beta},
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  5. #5
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    Ok,

    notice that

    x^2-(\alpha+\beta)x+\alpha\;\beta=(x-\alpha)(x-\beta)

    This shows you the vertical asymptotes, as the values of x that cause the denominator to be zero
    are evident.

    The horizontal asymptote is discovered by letting x approach infinity.
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    Quote Originally Posted by Archie Meade View Post
    Ok,

    notice that

    x^2-(\alpha+\beta)x+\alpha\;\beta=(x-\alpha)(x-\beta)

    This shows you the vertical asymptotes, as the values of x that cause the denominator to be zero
    are evident.

    The horizontal asymptote is discovered by letting x approach infinity.
    Makes sense. What about the domain?
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    Quote Originally Posted by Joker37 View Post
    Makes sense. What about the domain?
    The function is defined for all real x except for the values of x that cause the denominator to be zero.
    (The vertical asymptotes).
    You see these from the factored version of the denominator.

    Then you need the turning points, discovered from f'(x)=0.
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    Archie Meade, I'm having trouble finding the turning point. I'm wondering if you could post up a complete step-by-step working out to it. Any more help will be appreciated!
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    Your previous method of partial fractions is useful for calculating the derivative of quotients.

    Alternatively, use the Quotient Rule.

    However, in this case, it's convenient to use a "negative exponent" instead.

    f(x)=\displaystyle\frac{A}{x^2-(\alpha+\beta)x+\alpha\;\beta}=A\left[x^2-(\alpha+\beta)x+\alpha\;\beta\right]^{-1}

    Using the Chain Rule

    u=x^2-(\alpha+\beta)x+\alpha\;\beta

    \displaystyle\frac{d}{dx}Au^{-1}=A\frac{du}{dx}\;\frac{d}{du}u^{-1}

    =\displaystyle\ A\left[2x-(\alpha+\beta)\right]\left(-\left[u^{-2}\right]\right)=\frac{-A\left[2x-(\alpha+\beta)\right]}{\left[x^2-(\alpha+\beta)x+\alpha\;\beta\right]^2}

    This is zero when the numerator is zero, if we do not have an (x-\alpha) or (x-\beta) factor there.

    (all the work calculating the denominator is often a waste of time, but not always).

    Since A\ne\ 0

    \displaystyle\ 2x-(\alpha+\beta)=0\Rightarrow\ x=\frac{\alpha+\beta}{2}

    There is a turning point at that x, provided \alpha\ne\ \beta,

    otherwise, we'd have calculated a turning point at the asymptote, which is not possible.



    Then place that x into f(x) to find the vertical co-ordinate of the turning point.
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    Okay, thanks. I'm copying and pasting this on word to print for future reference!
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    Quote Originally Posted by Archie Meade View Post
    Since A\ne\ 0

    \displaystyle\ 2x-(\alpha+\beta)=0\Rightarrow\ x=\frac{\alpha+\beta}{2}

    There is a turning point at that x, provided \alpha\ne\ \beta,

    otherwise, we'd have calculated a turning point at the asymptote, which is not possible.
    How do you know that A does not equal 0?
    And why aren't we solving:

     <br />
-A[2x-(\alpha+\beta)]=0<br />
?

    Where does the A go?
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    Quote Originally Posted by Joker37 View Post
    How do you know that A does not equal 0?
    And why aren't we solving:

     <br />
-A[2x-(\alpha+\beta)]=0<br />
?

    Where does the A go?
    look at the original function ... what would f(x) equal if A = 0 ?

    since A cannot equal 0, divide both sides of the equation by -A ...

    \frac{-A[2x-(\alpha + \beta)]}{-A} = \frac{0}{-A}

    2x-(\alpha + \beta) = 0
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    Quote Originally Posted by skeeter View Post
    look at the original function ... what would f(x) equal if A = 0 ?

    since A cannot equal 0, divide both sides of the equation by -A ...

    \frac{-A[2x-(\alpha + \beta)]}{-A} = \frac{0}{-A}

    2x-(\alpha + \beta) = 0
    Alright, alright, makes sense. I don't really pretend to be good at all this maths stuff.

    Quote Originally Posted by Archie Meade View Post
    Then place that x into f(x) to find the vertical co-ordinate of the turning point.
    I'm also not getting the correct corresponding y value when I algebraically substitute the x value of the turning point in f(x). Would anyone be able to supply the rest of the working out to see where I went wrong?
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    f(\frac{\alpha+\beta}{2}) = A[(\frac{\alpha+\beta}{2})^2 -(\alpha+\beta)+\alpha\beta]^{-1}<br />
=A[\frac{\alpha^2 +\beta^2}{4} - \frac{4(\alpha+\beta)}{4} +\frac{4\alpha\beta}{4}]^{-1}
    =A[\frac{\alpha^2+\beta^2-4\alpha+4\beta+4\alpha\beta}{4}]^{-1}
    Where have I gone wrong?
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    Quote Originally Posted by Joker37 View Post
    f(\frac{\alpha+\beta}{2}) = A[\left(\frac{\alpha+\beta}{2}\right)^2 -(\alpha+\beta)+\alpha\beta]^{-1}<br />
=A[\frac{\alpha^2 +\beta^2}{4} - \frac4{\alpha+\beta}{4} +4{\alpha\beta}{4}]^{-1}

    =A[\frac{\alpha^2+\beta^2-4\alpha+4\beta+4\alpha\beta}{4}]^{-1}

    Where have I gone wrong?
    you dropped the x again!

    \displaystyle\ f\left(\frac{\alpha+\beta}{2}\right)=A\left[\left(\frac{\alpha+\beta}{2}\right)^2-(\alpha+\beta)\left[\frac{\alpha+\beta}{2}\right]+\alpha\;\beta\right]^{-1}
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