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Using 3 Points to Find Quadratic Equation in General Form

  1. Jul 28, 2013 #1
    Help! Using 3 Points to Find Quadratic Equation in General Form

    I have a quadratic equation question that I have to solve, but I can't seem to understand it.
    The question is: "Using points (1,0);(3,0);(0,-6), find the quadratic formula in General Form
    (ax^2 + bx + c).

    My teacher says that I have use the method of y= a(x-p) + q to solve it, no other method is allowed for me to use.

    I had a try at it:

    { a(1-p)^2 + q= 0 ...(1)
    { a(3-p)^2 + q= 0 ...(2)
    { a(p)^2 + q= -6 ...(3)

    { a - 2ap + ap^2 + q= 0 ...(4)
    { 9a - 6ap + ap^2 + q= 0...(5)

    { -8a + 4ap= 0 ...(6)
    { -9a + 6ap= -6 ...(7)

    { -48a + 24ap= 0 ...(6)
    { -36a + 24ap= -6 ...(7)

    { -12a= 24 ...(8)
    a= -2


    I have the constant "a" down, but I can't get "p" or "q"
    Any help is appreciated. Thanks a bunch!
     
  2. jcsd
  3. Jul 28, 2013 #2

    ehild

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    Subtract equation 1 from eq 2.

    ehild
     
  4. Jul 28, 2013 #3
    I did that to get equation 6
     
  5. Jul 28, 2013 #4

    ehild

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    You know a already, substitute it into eq,.6.


    ehild
     
  6. Jul 28, 2013 #5
    Would you get P = -2?
     
  7. Jul 28, 2013 #6

    LCKurtz

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    You mean ##y = a(x-p)^2+q##

    So put ##a=-2## in your equations 1 and 2 and see if you can solve them. It's easy.
     
  8. Jul 28, 2013 #7
    ok, I'll try that
     
  9. Jul 28, 2013 #8

    ehild

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    Why minus????

    ehild
     
  10. Jul 28, 2013 #9
    (1,0);(3,0);(0,-6)

    You must have made a typo; those points can't form a parabola.
     
  11. Jul 28, 2013 #10

    D H

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    Yes, they do.
     
  12. Jul 29, 2013 #11

    symbolipoint

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    What do you mean, "no other method is allowed"? Starting and proceeding through standard form is much more complicated. Why not use general form, create your three equations, perform row operations to find a, b, and c (as for general form), and then you can write the general form equation; from that, you can complete the square to find the standard form version of the parabola.

    You would start with this matrix:
    [ 1 1 1 0 ]
    [ 9 3 1 0 ]
    [ 0 0 1 -6 ]
    For columns representing a, b, c, and y; the variables now being a, b, and c.

    You would find the equation, y=-2x2+8x-6, and the three given points each satisfy this equation. (That is still in general form...)
     
    Last edited: Jul 29, 2013
  13. Jul 29, 2013 #12

    D H

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    He means he can't use the Vandermonde matrix approach that you outlined.
     
  14. Jul 29, 2013 #13

    Mentallic

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    Given any 3 points, as long as they're not collinear and they all have different x-values to each other, you can form a parabola passing through them.
     
  15. Jul 29, 2013 #14

    Ray Vickson

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    A much, much easier way is to note that the two zeros of the quadratic form are given as part of the input data, so y = A*(x-1)*(x-3). Get the constant A from the requirement that y = -6 when x = 0.

    OK, that is not in the teacher's standard form, but it can easily be re-written to conform.
     
  16. Jul 29, 2013 #15
    Not strictly true. The curve [itex]x=y^2[/itex] is also a parabola, isn't it?
     
  17. Jul 29, 2013 #16

    D H

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    Strictly true.

    If you generate three (x,y) pairs from x=y2 such that all of the x values are different you won't recreate x=y2 when trying to fit to y=ax^2+bx+c. You will however find a unique set of values for a, b, and c.
     
  18. Jul 29, 2013 #17

    Mentallic

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    Yes :smile: Although I left it out to avoid going on and on about the situations of when it's a function of x versus a function of y, and we could have even gone further and mentioned rotated parabolas etc.
     
  19. Jul 29, 2013 #18

    symbolipoint

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    Yes, I almost had that but become confused during that analysis. Although I solved the problem in the way as I indicated earlier, not the way xpetey wanted, at least converting into standard form made sense.
     
  20. Jul 29, 2013 #19
    Ok, strictly true. The 3 x-values being different is a sufficient condition but it isn't a necessary condition, which is what I thought Mentallic was suggesting (and his/her subsequent response supports that).

    But I've no idea why you're trying to fit any general parabola to [itex]y=ax^2+bx+c[/itex]

    That may be relevant to this thread's OP, but not what I was saying to Mentallic.
     
  21. Jul 30, 2013 #20

    PeterO

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    Was your teacher requiring you to use that form in solution, or suggesting that would be the best way.

    Those 2 points (1,0) and (3,0) make that method pretty useful,
    but then the form y = a(x-m)(x-n) is also pretty handy.
     
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