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Quadratic graphs

  1. May 2, 2004 #1
    How do I graph parabolas/quadratic equations?
  2. jcsd
  3. May 2, 2004 #2
    Either use a calculator, or just plot points.

  4. May 2, 2004 #3
    Ick, your making me remember how to graph stuff with out calc?

    Ill start off by assuming you mean a straight forward parabola with out a rotation or anything.

    First you need to put the parobala in the form:

    Y = a(x – h)^2 + k
    X = a(y – h)^2 + h

    Where a=1/4p

    The position of your vertex (the extreme point) is (h,k)

    The focus is (h, k + p) and the directrix line is y = k – p

    The other 2nd degree quadratics have similar geometric formulas. Maybe you can show us what you need to graph…
    Last edited: May 2, 2004
  5. May 2, 2004 #4
    Oops forgot to tell you how to find which way it opens.

    X = a(y – h)^2 + h
    If a < 0 then it opens to the left
    If a > 0 then it opens to the right

    Y = a(x – h)^2 + k
    If a < 0 then it opens down
    If a > 0 then it opens up
  6. May 3, 2004 #5
    I think it will be easier to just plug in points and go from there.
  7. May 3, 2004 #6


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    Gold Member

    to make a parabola graph you only need three points, if less then it's not kosher.
  8. May 3, 2004 #7
    Aside from the simple y=(x-a)^2+b formulas, there's also conics equations for parabolas:
    4py=x^2, where p is the distance to your focus or directrix. (the focus is an arbirtrary point, and the directrix is a line whos equation is y=-d, where d is the distance from y=0 to focus).
    As far as graphing, either get a simple table of values, or get three points: two X intercepts and one Y intercept. To get the X intercepts, simply substitute 0 into y so that your equation looks like 0 = (x-a)^2 + b, which shouldn't be too hard to solve. For the Y intercept, do the same, but substitute x as 0, so that you end up with y=(0-a)^2+b , which should be pretty easy to get as well.
    after you're done that, plot your points, and draw a curve through them.
  9. May 3, 2004 #8
    how about graphing something like y=3x+5b+3?
  10. May 3, 2004 #9
    :eek: You have got more variables in there than im am comfortable to deal with!
  11. May 3, 2004 #10
    did u mean y = 3x^2 + 5x + 3?????

    if u didnt:
    thats just a simple linear equation (im assuming that b is a constant - if its a variable u cant graph it) that has a y-intercept of 5b + 3 and a slope of 3..
  12. May 3, 2004 #11
    You're right, of course. For some reason I was thinking of a line when I said what I did.
  13. May 4, 2004 #12
    Yes. (my bad) :rolleyes:
  14. May 5, 2004 #13
    for that kind i just use a graphics calculator, i think if you want to look at that and draw it you have to play with the equation a bit till its easier. i 4got the format i used last year for this type of thing.
  15. May 6, 2004 #14


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    Staff Emeritus
    Science Advisor
    Gold Member

    If b is a third variable (call it z, for familiarity), then I believe you are looking at the equation of a plane.
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