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Parabola Help

  1. Apr 29, 2010 #1
    Hey guys what's up? I am new to this forum, so I am sorry if I posted in the wrong area..

    Anyways, I have a question on Parabolas. I know how to find the Vertex, Axis of Symmetry, X,Y Int. and so on, but I am having trouble graphing it. My teacher said I need to find three good points to graph it, but I have no idea what she means. I mean I know how to graph it, but the A variable throws me off.. I'll give an example:

    Vertex: (1,-9)
    Axis of Symmetry: x=1
    Width- Narrow
    Direction- Up
    X Int.= 1 plus/minus square root of 3
    Y Int.= -6
    Min: y=-9

    How would you get three good points from this? Thank you so much to those that take the time to read and help!
    Last edited: Apr 29, 2010
  2. jcsd
  3. Apr 30, 2010 #2


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    Science Advisor

    So you're asking how would you get three points from the four that you already have?

    In a case like the above I'd normally start with just the vertex and the two x-intercepts to make a rough sketch, you can use the y-intercept also as guide to keep it somewhat accurate.

    Remember however that you wont always have real x-intercepts, but you can always calculate as many additional points as you wish if you want to make the sketch more accurate.

    All your teacher really wants is that you be able to make a quick rough (approximate) sketch of a parabola based on the vertex and few points either side of it.
  4. May 3, 2010 #3
    Uart makes a good point, I hope you do notice that with the information you posted there are indeed 4 points.

    1) the vertex ----> (1, -9)

    2 and 3) the X intercepts -----> (1+sqrt(3), 0) and (1 - sqrt(3), 0)

    4) the y intercept (0, -6)

    You could certain use these 4.

    One thing to add though, is it is possible that your parabola won't have any x-intercepts, and/or your vertex will also be one or both of your intercepts.

    So one thing to do is to pick two points to the left and right of your vertex.

    For simplification imagine your parabolas equation is y = 4x^2

    The Vertex and all the intercepts will be the origin (0, 0). So what you could do is just pick an x value less then 0 and greater then 0 to find two new points. say x = +/-2. Plug these two x values into the original equation and solve for y. In this case you will get the points (2, 16) and (-2, 16).

    Good luck with your conic sections.
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