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Graphing a standard quadratic equation?

  1. Jun 19, 2012 #1
    How do we graph a quadratic equation in the form ax^2+bx+c =0?, or an equation like , f(x)(or y)= [(x+b/2a)^2 - (D/4a^2)], where D = b^2 - 4ac? or how do we get a parabolic curve intersecting at its roots(of the quadratic equation) in the x axis?
     
  2. jcsd
  3. Jun 19, 2012 #2

    danago

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    Do you mean an equation of the form ax^2+bx+c =y? Anyway, in any case, you can make graphs quite easily on any graphics calculator, or you could also download some freeware graphing software. I used to use this program:

    http://download.cnet.com/Graph/3000-2053_4-10063417.html

    It is quite simple, but does a very good job for a freeware program. The other alternative would be to use Microsoft Excel which requires a series of points to which it can draw a smooth line.
     
  4. Jun 19, 2012 #3

    danago

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    Hmm re-reading your question, I'm thinking I may have misinterpreted it. Did you want to know how to sketch it by hand? If so, then you could do it quite easily by finding the roots (by factorizing or quadratic equation) and the y-intercept, and then just draw a smooth curve through the points. In the second form you have mentioned, you could recognize it as "turning point form" (from which the turning point can be read without any calculations) and then finding the y-intercept. This will give you enough for a basic rough sketch.
     
  5. Jun 20, 2012 #4
    I think you can graph it yourself by determining the minima/maxima and the zeros of the function. The general form a quadratic takes is parabolic; so it should not be hard to draw it.
     
  6. Jun 20, 2012 #5

    HallsofIvy

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    Assuming you want to graph the function [itex]y= ax^2+ bx+ c[/itex] (you don't graph a function- to graph you have to have at least two variables) I would start by completing the square:
    [itex]y= a(x^2+ (b/a)x)+ c= a(x^2+ (b/a)x+ (b^2/4a^2)- (b^2/4a^2))+ c[/itex]
    [itex]y= a(x^2+ (b/a)x+ (b^2/4a^2))+ c- (b^2/4a)= a(x+ b/2a)^2+ c- (b^2/4a)[/itex]

    Now I know that the vertex is at (-b/2a, c- b^2/4a). If a> 0, the parabola opens upward, if a< 0, downward.
     
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