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Math Help - graph equations

  1. #1
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    graph equations

    the curve, C has the equation:
    4(x+1)^2 - (y-1)^2 =1

    Find the greatest value of k, where k is a positive integer, for which the curve y=ln(x+k) cuts C at only one point.


    what i tried to do is:
    4(x+1)^2 - (y-1)^2 =1
    4x^2 + 8x - y^2 +2y +2=0
    By substitution,
    4x^2+8x-[ln(x+k)]^2+2[ln(x+k)]+2=0<br />
    and then i'm stuck because of [ln(x+k)]^2.
    is the approach wrong or did i make a mistake somewhere?
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  2. #2
    Super Member Bacterius's Avatar
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    Hello !

    You have two curves, 4(x+1)^2 - (y-1)^2 =1 and y=\ln{(x+k)}. You wish to find the greatest number k for which the curves intersect only once.

    Let us substitute the second curve into the first one (since the y values are presumed equal as there is intersection) :

    4(x+1)^2 - (\ln{(x+k)}-1)^2 =1

    Now let us reformulate the problem : you wish to find the greatest number k for which this equation has only one solution for x (equivalent to the statement "the curves only intersect once").

    Expand this :

    4(x+1)^2 - (\ln{(x+k)}-1)^2 =1

    4(x^2 + 2x + 1) - (\ln{(x+k)}^2 - 2 \ln{(x+k)} + 1) = 1

    Ok, keep going :

    4x^2 + 8x + 4 - \ln{(x+k)}^2 + 2 \ln{(x+k)} - 1 = 1

    Let's try to set this as a quadratic equation to use the discriminant :

    4x^2 + 8x - \ln{(x+k)}^2 + 2 \ln{(x+k)} + 2 = 0

    Consider everything after the 8x as the constant coefficient (yeah, who said we couldn't put x's in it ?)

    You know that if the discriminant of a quadratic equation is equal to zero, then there is only one real solution. So, let us express the discriminant :

    \Delta = b^2 - 4ac = 64 - 16 \times (\ln{(x+k)}^2 + 2 \ln{(x+k)} + 2)

    You want it to be equal to zero, so attempt to solve the following for k (yes, now we care about k because we are looking for the value of k that makes the discriminant equal to zero) :

    64 - 16 \times (\ln{(x+k)}^2 + 2 \ln{(x+k)} + 2) = 0

    Simplify :

    64 = 16 \times (\ln{(x+k)}^2 + 2 \ln{(x+k)} + 2)

    Again :

    4 = \ln{(x+k)}^2 + 2 \ln{(x+k)} + 2

    0 = \ln{(x+k)}^2 + 2 \ln{(x+k)} - 2

    \ln{(x+k)}^2 + 2 \ln{(x+k)} - 2 = 0

    Cool, a straightforward quadratic equation ! Here we are using an advanced method : we are going to solve for \ln{(x + k)}, and this will serve as an intermediate result to determine which values of k make the discriminant null. Get the discriminant first, for ease :

    \Delta = 2^2 - 4 \times (-2) = 4 + 4 \times 2 = 12

    \ln{(x+k)} = \frac{-4 \pm \sqrt{12}}{2}

    So if some k makes the two curves intersect, then we have \ln{(x+k)} = \frac{-4 \pm \sqrt{12}}{2} for the x where the curves intersect (discard the negative value since a logarithm is always positive)

    Now you can apply the same method but solving for x instead, and then merge the results together to find a condition on k allowing you to find the greatest k that satisfies your problem.



    PS : I think I got way too complicated here, I don't even know if this actually works, there surely is a simpler way to do this, what have you been doing last with your class ? (or elsewhere)
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  3. #3
    Super Member Bacterius's Avatar
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    Never mind what I wrote before, it's just wrong (fail). However, in your post I see a mistake, it should be -2 and not +2 in the two last lines.

    I don't have time now, so here is a little hint : try to use the discriminant to avoid messing with the logarithms, and remember the graphic interpretation of the quadratic discriminant.

    Although 3 unknowns make this really harsh.
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  4. #4
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    thanks for your guidance.
    it looks quite complicated. ="=

    the teacher is teaching the chapter, Graph of Standard Functions.
    so, i'm learning graphic calculator,ellipses,hyperbolas, various transformations, and ....well....graphs.
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  5. #5
    Super Member Bacterius's Avatar
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    I see, I guess one must use the transformation formulas to find the result. Here is an idea : "since the standard graphs (ln(x) for instance), with a much simpler expression, intersect in x when k = [something easy to find]", shifting the two curves by three units to the left maintains the intersection point which is also shifted three units to the left".

    That's just an idea, I'm gonna think about it.

    PS : what really scares me in the problem is that there are three unknowns, and that one unknown must be a whole number which may very well make the problem very complicated, but I guess your teacher gave this problem for a reason so there must be a somewhat trivial way to solve it
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