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Quadratic Equations

  1. Jan 10, 2005 #1
    [tex]y=4x^2-8x-5[/tex] Find the coordinates of the turning point of the curve. Where do I even start?
     
  2. jcsd
  3. Jan 10, 2005 #2

    Hurkyl

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    With the definition of turning point.
     
  4. Jan 10, 2005 #3
    I don't know that.
     
  5. Jan 10, 2005 #4

    Hurkyl

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    Well, then, that's where you need to start. What does your book have to say, or your notes?

    (Incidentally, what level math is this?)
     
  6. Jan 10, 2005 #5
    My book doesn't say anything about the definition of a turning point. I'm in grade 9.
     
  7. Jan 10, 2005 #6
    If you mean the minimi/maximi-point of the curve, you can set the derivate equal to zero and solve. Wait a minute, it must be an another way... Try to completting the square on the function...
     
  8. Jan 10, 2005 #7

    Hurkyl

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    Where did you see the problem?


    If I had to guess what it meant, I would say the point where the graph of the polynomial stops going downwards and starts going upwards. Since it's a parabola, that would be its vertex.
     
  9. Jan 10, 2005 #8
    Oh Yes!! Now that you guys mention that, I know what to do now :rolleyes: Thanks!!
     
    Last edited: Jan 10, 2005
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