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Math Help - family of functions- verticle asymptotes

  1. #1
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    family of functions- verticle asymptotes

    im stuck on a question!

    a family of functions is given by:
    r(x)=1/a+(x-b)^2

    a) for what vales of a and b does the graph of r have a verticle asymptote? where are the verticle asymptotes in this case?
    b) find values of a and b so that the function r has a local maximum at the point (3,5)

    any help would be greatly appreciated
    thanks
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  2. #2
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    Hello,

    I'm not sure what you mean:

    r(x)=\frac1a + (x-b)^2

    or

    r(x)=\frac{1}{a + (x-b)^2}
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  3. #3
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    the second one!
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  4. #4
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    Have you tried solving a + (x - b)^2 = 0? I think that would pin it down for you.
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  5. #5
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    Quote Originally Posted by Paige05 View Post
    im stuck on a question!

    a family of functions is given by:
    r(x)=1/(a+(x-b)^2)

    a) for what vales of a and b does the graph of r have a verticle asymptote? where are the verticle asymptotes in this case?
    b) find values of a and b so that the function r has a local maximum at the point (3,5)
    A function has vertiacal asymptotes if the denominator equals zero:

    a+(x-b)2=0~\iff~(x-b)^2=-a~ \iff~x=b+\sqrt{-a}~ \vee~x=b-\sqrt{-a}

    That means:

    you get no vertical asymptote if a > 0

    you get exactly one vertical asymptote if a = 0 then the equation of the asymptote is : x = b

    you get 2 vertical asymptotes if a < 0. The equations of the asymptotes are: x=b+\sqrt{-a}~ \vee~x=b-\sqrt{-a}


    to b):

    r(x)=(a+(x-b)^2)^{-1} Use chain rule to derivate:

    r'(x)=(-1)(a+(x-b)^2)^{-2} \cdot (2(x-b)) Rearrange:

    r'(x)=-\frac{2(x-b)}{(a+(x-b)^2)^{2}}

    r'(x) = 0 ~\implies~2(x-b)=0~\iff~x=b

    From your problem you know that now b = 3. Calculate

    r(3)=\frac{1}{a+(3-3)^2} = 5 ~\implies~a = \frac15
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  6. #6
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    ahhhhh.... i see!
    thanks so much
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