Linear algebra population question, fit second degree polynomial through 3 points?

A country s census lists the population of the country as 250 million in 1990, 284 million in 2000, and 310 million in 2010. Fit a second-degree polynomial passing through these three points. (Let the year 2000 be x = 0 and let p(x) represent the population in millions.)

P(x) =

2020

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  • david
    Lv 7
    4 weeks ago

    A country s census lists the population of the country as 250 million in 1990, 284 million in 2000, and 310 million in 2010. Fit a second-degree polynomial passing through these three points. (Let the year 2000 be x = 0 and let p(x) represent the population in millions.)

     (0, 284)  ...  (10,310)  ...  (-10, 250)

    y = ax^2 + bx + c

    284 = a(0)  +  b(0)  +c  .... c = 284

    310 = a(10^2) + b(10) + 284

    26 = 100a + 10b  

    =============================

    250 = a(-10)^2 + b(-10) + 284

     -34 = 100a - 10b

    -[26 = 100a + 10b]

    -8 = -20b  ... b= 0.4

    100a = 26 - 10(0.4)

      a = 0.22

    ============

    p(x) = 0.22x^2 + 0.4b + 284

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  • alex
    Lv 7
    4 weeks ago

    P(x) = ax^2+bx+c

    points (0,284) , (-10 , 250) , (10 , 310)

    set up 3 equations then solve for a , b , c

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  • 4 weeks ago

    If x is the number of years after 2000, then the year 1990 is when x = -10, etc.

    You are given three points:

    (-10, 250), (0, 284), and (10, 310)

    Where "y" is in millions.

    You know it's a quadratic, so we can start from that general form:

    y = ax² + bx + c

    We are given three points which is three sets of x's and y's.  Substitute each set into the general form to get a system of three equations and three unknowns:

    250 = a(-10)² + b(-10) + c and 284 = a(0)² + b(0) + c and 310 = a(10)² + b(10) + c

    250 = 100a - 10b + c and 284 = 0a + 0b + c and 310 = 100a + 10b + c

    250 = 100a - 10b + c and 284 = c and 310 = 100a + 10b + c

    we have a value for c.  Substitute into the other two to make a new system of two equations and two unknowns:

    250 = 100a - 10b + 284 and 310 = 100a + 10b + 284

    -34 = 100a - 10b and 26 = 100a + 10b

    Since b's coefficients are opposites, I'll solve them with elimination.  Add the two equations together to get:

    -8 = 200a

    a = -8/200

    a = -1/25

    Now that we have a, we can solve for b:

    -34 = 100a - 10b

    -34 = 100(-1/25) - 10b

    -34 = -4 - 10b

    -30 = 10b

    b = 3

    Your equation is:

    y = (-1/25)x² + 3x + 284

    Where x is the number of years after 2000 and y is the population in millions.

    • Puzzling
      Lv 7
      4 weeks agoReport

      I concur with the method and the solution. Here's a graph of the solution: https://www.desmos.com/calculator/stggad3jvo

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