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Finding the vertex of a parabola from an equation

  1. May 15, 2005 #1
    I need help with the following question.

    The parabola y^2 - x + 4y + k = 0 passes through the point (12,1). Find the vertex of the parabola.
  2. jcsd
  3. May 15, 2005 #2

    by differentiating ? or the line of symmetry between the roots of the equation ?

    or completin the square ?
    Last edited: May 15, 2005
  4. May 15, 2005 #3
    Plug in the point values to solve for k.

    What kind of parabola is this? Sideways or vertical? (hint)
  5. May 15, 2005 #4
    would be a sideways since the y is squared, but how would i find the h in (h,k)?
  6. May 15, 2005 #5
    Show me the equation in the form of x =. Start from there. Did you solve for k?
  7. May 15, 2005 #6
    x = y^2 + 4y + k. So to solve for k do I then plug in 12 and 1 for x and y?
  8. May 15, 2005 #7
    Alright so that would give me the k value, but what about the h value?
  9. May 15, 2005 #8
    plug in values , then just complete the square to put it in the form

    f(y) = x = (y+a)^2+b

    and b,-a is the vertex
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