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Quadratics, i was good until here.

  1. Sep 16, 2008 #1
    im going to try to make this as short and easy to follow as possible haha

    and im just warning, this is probably so simple that i cant even grasp it because im reading too deeply into it.

    okay, im working with these three points:
    [tex]A (0, 1)
    C (2, 6)
    D (3, 10)[/tex]

    i worked them into three equations, the first one is killing me though!

    [tex] a(0)^2 + b (0) + c = 1[/tex] this is the one im having trouble with, i cant set it up properly because all i have is [tex]c = 1[/tex]

    i have for the other two:

    [tex]C. . ., 4a + 2b +c = 6[/tex]

    [tex]D. . ., 9a + 3b + c = 10[/tex]

    now, in order to get my model for [tex]y = ax^2 + bx + c[/tex]

    how would i set this up, other than that im doing alright..
  2. jcsd
  3. Sep 16, 2008 #2


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    Well from the first equation you have found that c=1, so just sub that into the following two equations and then you have a system of two equations in two variables (a and b). Why should the first one cause any trouble?
  4. Sep 16, 2008 #3


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    [tex] a(0)^2 + b (0) + c = 1[tex]
    this is the one im having trouble with, i cant set it up properly because all i have is
    [tex] c=1 [/tex]
    You SHOULD get a + 0 + c = 1

    Continue from there and you should have no trouble.
  5. Sep 16, 2008 #4
    like i said, so simple that i cant eve grasp it.

    thanks, ill see how it works out! :)


    i got:

    [tex]y = \frac{1}{2}x^2 + 3x + 1[/tex]
    Last edited: Sep 16, 2008
  6. Sep 17, 2008 #5


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    No, you shouldn't. 02= 0, not 1. c= 1 is correct.
  7. Sep 17, 2008 #6
    i used [tex]c = 1[/tex]. im quite confident that the answer i got[tex](
    y = \frac{1}{2}x^2 + 3x + 1)[/tex] is incorrect. when i plug in all the numbers, get the changes from the original y value and square it, i get all sorts of weird answers for this one. usually, its all 0's except for one value but thats not happening.
  8. Sep 17, 2008 #7


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    Perhaps you solved the system of equations incorrectly. Have a look at your working, and if you still cant figure it out, post your working here :smile:
  9. Sep 17, 2008 #8
    okay, heres what we did in class:

    four points;
    [tex]A (0, 1)[/tex]
    [tex]B (1, 2)[/tex]
    [tex]C (2, 6)[/tex]
    [tex]D (3, 10[/tex]

    we went over (what im sure everyone here, including my self, already know. its good to think about and keep fresh in my mind) how two points define a line and for a function such as a half of a parabola, (at least) three points are required.

    for our first model, we chose points [tex]ABC[/tex].

    i have three equations:

    [tex]c =1[/tex]
    [tex]1a + 1b + c = 2[/tex]
    [tex]4a + 2b + c = 6[/tex]

    with that info, i dont know how i got from there to here(our first model):

    [tex]\hat{y}_{1} = \frac{3}{2}x^2 - \frac{1}{2}x + 1[/tex]

    plug in all the numbers to get the residual (the problem i had last time, but now i can do it in my sleep).

    heres what im about to do, hopefully i can elaborate enough.

    im going to plug in the numbers from all four points, not just the three that i am using for this model.
    point [tex]A[/tex]

    [tex]\hat{y}_{1} = \frac{3}{2}(0)^2 - \frac{1}{2}(0) + 1 = 1[/tex]

    [tex](y - \hat{y}_{1})^2 = 0[/tex] (the residual)

    [tex](0 - 0)^2 = 0[/tex]

    point [tex]B[/tex]

    [tex]\hat{y}_{1} = \frac{3}{2}(1)^2 - \frac{1}{2}(1) + 1 = 2[/tex]

    [tex](y - \hat{y}_{1})^2 = 0[/tex] (the residual)

    [tex](2 - 2)^2 = 0[/tex]

    im posting this now and will continue immediately. this is just to be sure nothing happens and i have to redo the whole thing.
  10. Sep 17, 2008 #9
    point [tex]C[/tex]

    [tex]\hat{y}_{1} = \frac{3}{2}(2)^2 - \frac{1}{2}(2) + 1 = 6[/tex]

    [tex](y - \hat{y}_{1})^2 = 0[/tex]

    [tex](0 - 0)^2 = 0 [/tex]

    heres where it finally changes!

    [tex]\hat{y}_{1} = \frac{3}{2}(3)^2 - \frac{1}{2}(3) + 1 = 13[/tex]

    [tex](y - \hat{y}_{1})^2 = 9[/tex]

    [tex](10 - 13)^2 = 0 [/tex]

    so for model [tex]ABC[/tex]

    [tex]\hat{y}_{1} = \frac{3}{2}x^2 - \frac{1}{2}x + 1[/tex]

    [tex]\sum(y - \hat{y}_{1})^2 = 9[/tex]

    the sum of our residual is nine
  11. Sep 17, 2008 #10
    the second model is with points [tex]ABD[/tex]

    heres how it looks: [tex]\hat{y}_{2} = x^2 + 1[/tex]

    (this was done is class as well)

    plug in the numbers

    everything is 0 except for point [tex]C[/tex] which will look like this:

    heres how it looks: [tex]\hat{y}_{2} = 2^2 + 1 = 5[/tex]

    [tex](6 - 5)^2 = 1[/tex]


    [tex]\sum(y_{1} - \hat{y}_{2})^2 = 1[/tex]
  12. Sep 17, 2008 #11
    our third model, well, fourth, the third one is the one i am trying to figure out:

    ill keep it short, i demonstrated how everything was done in the first long post. :)

    this is done with points [tex]BCD[/tex]

    the model looks like this: [tex]\hat{y}_{4} = 4x^2 - 2[/tex]

    all but the first point were 0's. ill plug in the number

    [tex]\hat{y}_{4} = 4(o)^2 - 2 = -2[/tex]

    [tex](1 - 2)^2 = 1[/tex]

    so for: [tex]\hat{y}_{4} = 4x^2 - 2; \sum(y - \hat{y}_{4})^2 = 1[/tex]
  13. Sep 17, 2008 #12


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    In your original post you said the points were A(0,1), B(2, 6), C(3, 10)
    Writing y= ax2+ bx+ c, those give a02[/sub]+ b0+ c= 1 or c= 1, 4a+ 2b+ c= 6, and 9a+ 3b+ c= 10. Since c= 1, the last two are 4a+ 2b+ 1= c or 4a+ 2b= 5 and 9a+ 3b+ 1= 10 or 9a+ 3b= 9. If you multiply the first of those equations by 3 you get 12a+ 6b= 15 and if you multiply the second by 2 you get 18a+ 6b= 18. Subtracting the first from the second, 6a= 3 so a= 1/2. Then 4a+ 2b= 5 becomes 2+ 2b= 5 so 2b= 3 and b= 3/2, not b= 3.

    The equation if y= (1/2)x2+ (3/2)x+ 1.

    But later you give A(0, 1), B(1, 2), C(2, 6), D(3, 10). The equation y= (1/2)x2+ (3/2)x+ 1 gives y(0)= (1/2)(0)+ (3/2)(0)+ 1= 1, y(1)= (1/2)(1)+ (3/2)(1)+ 1= 3, y(2)= (1/2)4+ (3/2)2+ 1= 6, and y(3)= (1/2)9+ (3/2)3+ 1= 10. That is correct for all except B(1,2) which was NOT in the original problem. You can, in general, fit a parabola through 3 given points but not 4. There is no quadratic function whose graph will pass through all 4 points.

    If you have 4 points, you should try a cubic formula: y= ax3+ bx2+ cx+ d so that the 4 points will give 4 equations for the 4 coefficients.
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