1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Gold Member

    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


    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    [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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Quadratics, i was good until here.