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Quadratic, standard and standard form of a curve

  1. Jan 21, 2009 #1
    the curve points are:
    a) (25, 43)
    b) (28, 46)
    c) (32, 47)
    d) (35, 45)

    would i use y=mx+b i think thats a straight line
     
  2. jcsd
  3. Jan 21, 2009 #2

    gabbagabbahey

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    You are given four points on the curve, so in general, the smallest degree polynomial that would accommodate them is a cubic polynomial:

    [tex]y(x)=Ax^3+Bx^2+Cx+D[/tex]

    Where you can determine the constants [itex]A[/itex], [itex]B[/itex], [itex]C[/itex] and [itex]D[/itex] by plugging in the given points and solving the resulting system of 4 equations.
     
  4. Jan 21, 2009 #3
    so how would i put the points of A into

    [tex]y(x)=Ax^3+Bx^2+Cx+D[/tex]


    cause there are 2 points and only 1 A
     
  5. Jan 21, 2009 #4
    The points are not a straight-line. Try graphing them to see this for yourself. They look like a curve to me- probably quadratic and opening downwards (ie [tex]-x^2[/tex]).

    If you are studying Quadratic, standard and standard form of a curve, my guess is that you only need [tex]y(x) = Ax^2 + Bx + C[/tex].
     
  6. Jan 21, 2009 #5

    gabbagabbahey

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    Unfortunately the points don't all lie on the same quadratic, so a cubic is necessary.
     
  7. Jan 21, 2009 #6

    gabbagabbahey

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    I thought a), b) c) and d) were each points on the curve in the form [itex](x_1,y_1)=(25,43)[/itex] etc.....is this assumption incorrect?
     
  8. Jan 21, 2009 #7
    ya thats what it is, there are 4 points on the curve
    so like

    [itex](x_1,y_1)=(25,43)[/itex]

    is what i was thinking

    but thats only one point
     
  9. Jan 21, 2009 #8

    gabbagabbahey

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    Okay, well each point gives you an equation. For example, the first point tells you:

    [tex]43=A(25)^3+B(25)^2+C(25)+D[/tex]

    And so using all 4 points you get 4 equations and 4 unknowns ([itex]A[/itex], [itex]B[/itex], [itex]C[/itex], and [itex]D[/itex]) allowing you to solve for them.
     
  10. Jan 21, 2009 #9
    ohhh ok that makes sense thank you for making it more clear

    so the almost final will look like this

    [tex]43=A(15625)+B(625)+C(25)+D[/tex]

    then you divide them ??
     
  11. Jan 21, 2009 #10
    You need to setup a system of four equations, then you can solve by substitution, elimination, or matrix methods to find A,B,C, and D. So it should like this like:

    Equation 1: [tex]y_1(x)=Ax^3_1+Bx^2_1+Cx_1+D[/tex]

    Equation 2: [tex]y_2(x)=Ax^3_2+Bx^2_2+Cx_2+D[/tex]

    Equation 3: [tex]y_3(x)=Ax^3_3+Bx^2_3+Cx_3+D[/tex]

    Equation 4: [tex]y_4(x)=Ax^3_4+Bx^2_4+Cx_4+D[/tex]

    Plug in your coordinates for your (x,y) values above and then solve the system.
     
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