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Vertical distance between two parabola

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Write a function for d(x), the vertical distance between the two curves, and find the minimum value of d(x).

    2. Relevant equations

    The equation for parabola one is y = x^2 + 6, for parabola two, y = -(x-2)^2 + 6


    3. The attempt at a solution

    The answer in the back of the book is d(x) = 2x^2 - 4x + 4, with a minimum value of 2. This is from my old algebra 2 trig textbook and I have no teacher to ask for help, as I am doing self study. Any and all help would be very much appreciated.
     
  2. jcsd
  3. Feb 15, 2009 #2
    I figured it out, its just d(x) = y1 - y2, seems I was just over thinking things... but I have another question :)

    Find the Value(s) of k for which the graph y = kx intersects the graph of y = x^2 + 25 in only one point.

    If I set the equations equal, I get 0 = x^2 - kx + 25, and I know perfect squares have only one root, so it seems as though the answer is 10 or -10. Does this seem reasonable?
     
  4. Feb 15, 2009 #3

    Dick

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    Yes, I think it does.
     
  5. Oct 11, 2011 #4
    from 0 = x2 - kx + 25, use the discriminant method to find the value of k that will result to a single intercept only. that is

    b2 - 4ac = 0 for one root or one intercept
    b2 - 4ac > 0 for two roots or two intercepts
    b2 - 4ac < 0 if you wish them to have no intercepts
     
  6. Oct 11, 2011 #5
    oh i forgot...

    a = 1
    b = -k
    c = 25
    x2 = x squared

    tnx tnx ^__^__^
     
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