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Sketching this curve

  1. Feb 2, 2010 #1
    1. The problem statement, all variables and given/known data

    I need to make a sketch of this function:

    4x = 4y - y^2

    2. Relevant equations

    3. The attempt at a solution

    So I see that it's a kind of parabola in terms of y.. so I try to make it into parabola form:

    x = y - (1/4)y^2
    = (-1/4)(y^2-4y+4-4+0)
    = (-1/4)((y-2)^2-4))
    = (-1/4)(y-2)^2+1

    Now I try to write it in terms of x:
    x-1 = (-1/4)(y-2)^2
    -4x+4 = (y-2)^2
    +/-sqrt(-4x+4) = y-2
    y = +/-sqrt(-4x+4)+2

    But these curves look like they are different?
  2. jcsd
  3. Feb 2, 2010 #2
    your functions should look like this:

    Attached Files:

    • 1.jpg
      File size:
      4.6 KB
  4. Feb 2, 2010 #3


    Staff: Mentor

    Leaving it in the form above is helpful, as you can tell that the graph is similar to the graph of x = -y^2. This is a parabola whose axis of symmetry is horizontal, and that opens to the left. Your parabola can be thought of as the translation to the right by 1 unit and up 2 units of the graph of x = -y^2. This puts the vertex at (1, 2).
    Yes. What you are getting by solving for y are equations for the upper and lower halves of the parabola. The pos. square root gives the upper half, and the neg. sq. root gives the lower half.
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