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Axis of a Parabola

  1. Dec 4, 2006 #1
    If someone could tell me how to go about it, I'd really appreciate it. Just point me in the right direction please. Kinda stuck.

    The vertex of a parabola is (-3,1) and it touches the x axis at the origin. Find the equation of the axis of the parabola.
     
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  3. Dec 4, 2006 #2

    radou

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    Did you mean, the parabola intersects the x axis at the origin? Or am I missing something?
     
  4. Dec 4, 2006 #3
    No, the x-axis is the tangent to the parabola at the origin. I dont know what to do with it.
     
  5. Dec 4, 2006 #4
    I dont think that possible.
     
  6. Dec 4, 2006 #5
    Couldnt it be a tilted parabola of the type [tex]y^2=4ax[/tex] so that its lower arm touches the x-axis at the origin?
     
  7. Dec 4, 2006 #6
    yes it could. In that case, it would be opening to the right.
     
  8. Dec 4, 2006 #7
    So how would you find its axis?
     
  9. Dec 4, 2006 #8
    it would simply be y = 1.
     
  10. Dec 4, 2006 #9
    Why would it be y=1?
     
  11. Dec 4, 2006 #10

    arildno

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    Since the axis has to go through the parabola's vertex, and must be parallell to the x-axis.
     
  12. Dec 4, 2006 #11

    HallsofIvy

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    Why would the axis of the parabola have to parallel to the x-axis? We are told that the x-axis is tangent to the parabola at (0,0). No line parallel to the axis of a parabola is ever tangent to it.
     
  13. Dec 5, 2006 #12

    arildno

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    THe x-axis is tangent to the parabola at the origin.

    The symmetry axis of the parabola is parallell to the x-axis, because the the tangent to the vertex is parallell to the y-axis.
    The symmetry axis of a parabola is perpendicular to the tangent at the vertex.
     
  14. Dec 5, 2006 #13
    It doesnt say the tangent at the vertex is paralell to the y-axis. You dont know the equation so you cannot assume that. the parabola could be tiled so that the axis of the parabola makes some angle with the x-axis and one of the arms just touches the x-axis at the origin.
     
  15. Dec 5, 2006 #14

    HallsofIvy

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    Yes, that is given.

    Where was that said?

     
  16. Dec 5, 2006 #15
    Where do you begin on something like this?
     
  17. Dec 5, 2006 #16

    arildno

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    I'm sorry, I was wrong.
     
  18. Dec 5, 2006 #17
    Any ideas on how to go about this?
     
  19. Dec 5, 2006 #18
    Well, if it is a parabola, the equation is of the form y = f(x) = ax^2+bx+c

    Now, f(-3)=1 and

    f(0)=0

    Finally, f'(0)=0

    Also, since f(x)= ax^2+bx+c and f(0)=0 c must also equal zero.

    I'm not sure how any of this will help you though.

    EDIT: I think I'm wrong again (as usual). I guess the equation of a parabola does not have to be of the form ax^2+bx+c=f(x)
     
    Last edited: Dec 5, 2006
  20. Dec 6, 2006 #19
    Could it be that the question is incomplete?
     
  21. Dec 6, 2006 #20
    I think the equation should be y=a(x-h)^2+k
    Since (-3,1) is the vertex h=-3 and k=1
    we get y=a(x+3)^2+1
    If x=0 and y=0
    9a+1=0 a=-1/9 y=1-((x+3)^2)/9
     
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