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Graphing a parabola for a calculus problem

  1. Aug 25, 2005 #1
    I am graphing a parabola for a calculus problem. I understand how you factor and use the two x values as your x-intercepts, but i'm not sure how this one would be graphed.

    y=4x^2 - 25. I understand how the vertex of the parabola would be at negative 25, but I have no clue what to do woith the 4x^2.
    Help would be appreciated. thanks.
  2. jcsd
  3. Aug 25, 2005 #2


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    Well, like you said factoring and getting the two x-intercepts is a start. You also want to know the turning point. If you don't remember the formula for that, just remember that the derivative is zero at the turning point and solve.
  4. Aug 26, 2005 #3
    factoring you get (2x-5) (2x+5) set these equal to zero
  5. Aug 27, 2005 #4


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    By the way, the vertex is not "at negative 25"- that's just the y-value of the point. The vertex is at (0, -25).
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