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Math Help - I Need Help With Parabolas!!!

  1. #1
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    Unhappy I Need Help With Parabolas!!!

    Okay. I need help finding the vertex, focus, axis of symmetry, and directrix of the parabola:

    y^2 - 2y + 4x + 9 = 0

    This is my only hope!! I dont understand how to even start looking for it.

    My math book only has one example of this kind of problem... and its not even that close to looking like this. T_T!

    HELPPPPP!
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  2. #2
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    parabola

    One option is imply to draw it and anaylize.

    But here are some notes:

    "vertex, focus, axis of symmetry, and directrix of the parabola"

    But it may not show it in this form:
    y^2 - 2y + 4x + 9 = 0

    ay^2 + by + c = 0
    Axis of symmetry: y = -b/2a
    Vertex: where the y axis intercepts the parabola

    Put in standard form (I bet this doesn't help):
    (y-k)^2 = 4p(x-h)
    p is distance from vertex to focus.
    axis of symmetry: y = k
    focus: (h+p), k)
    directrix: x = h-p
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  3. #3
    Jen
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    Quote Originally Posted by madison_leavelle View Post
    Okay. I need help finding the vertex, focus, axis of symmetry, and directrix of the parabola:

    y^2 - 2y + 4x + 9 = 0

    This is my only hope!! I dont understand how to even start looking for it.

    My math book only has one example of this kind of problem... and its not even that close to looking like this. T_T!

    HELPPPPP!

    First we need to put the function in a standard form that will allow us to peel off the information we need. This standard form will look similar to

    (x-h)^2=4p(y-k)

    With vertex (h,k) and the focus will be P distance above the vertex with directrix the same distance below the vertex.

    Setting up our equation into the form we would like.

    Start by getting the y's on one side.

    y^2-2y+4x+9  =>  y^2-2y=-4x-9

    Then we look at what we have and add some number to both sides such that we will be able to factor them. This one appears to need a 1, so add a 1 to both sides gives us...

    y^2-2y+1=-4x-9+1=>(y-1)^2=4(-1)(x+2)

    This is the standard form that we need. Notice that the vertex will be (-2,1) and our focus will be -1 distance on the x axis from the vertex (so (-3,1))and our directrix will be
    1 on the other side (so x=-1). Our axis of symmetry is the line that passes through the vertex.


    I couldn't get this to graph on my graph program because it isn't a function so I drew one by hand
    In this case, since our y is the one that is squared, our parabola will be centered around the y axis.

    I Need Help With Parabolas!!!-capture.jpg


    Sorry my picture is a little sloppy. The red is the axis of symmetry at y=1 the green is the focus at (-3,1) and the directrix at x=-1

    Hope this helps. Sorry for the funky pic.
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  4. #4
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    Oops...I mean analyze the graph.

    I think you need to find a text with standard form of conics, pick out parabolas, and try to fit your equation into it. This is the kind of thing that you probably will soon forget unless you use it from time to time.
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