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Question

Calculate the second difference for data in the table. Use a graphing calculator to find the quadratic regression for each data set. Make a conjecture about the relationship between the a values in the quadratic models and the second difference of the data.

hintHint:

1. When the difference between 2 consecutive differences for output values (y values) for a given constant change in the input values (x values) is constant. i.e. dy(n)- dy(n-1) is constant for any value of n, the function is known as a quadratic function.
2. Regression is a statistical tool used to find a model that can represent the relation between a given change in dependant variable (output values/ y values) for a given change in independent variable (input values/ x values).
Quadratic Equation using regression can be represented as-
Y = aX2 + bX + c, where-
Σy = nc + b(Σx) + a(Σx2)
Σxy = c(Σx) + b(Σx2) + a(Σx3)
Σx2y = c(Σx2) + b(Σx3) + a(Σx4)

The correct answer is: Second difference for data in the given table is 52. Quadratic regression for each data set can be represented using the function Y = 6.5X2. Also, the second difference is 8 times the a value.


    Step-by-step solution:-

    From the given information, we get-
    x coordinates in the given table pertains to length of bubble wrap (in inches) and y coordinates pertain to the cost of such bubble wrap.
    Now, from the given table, we observe the following readings-

    x1 = 3, y1 = 58.5;
    x2 = 5, y2 = 162.5;
    x3 = 7, y3 = 318.5;
    x4 = 9, y4 = 526.5;
    x5 = 11, y5 = 786.5.
    a). Difference between 2 consecutive x values-
                                                                                         dx1 = x2 - x1 = 5 - 3 = 2
                                                                                         dx2 = x3 - x2 = 7 - 5 = 2
                                                                                         dx3 = x4 - x3 = 9 - 7 = 2
                                                                                        dx4 = x5 - x4 = 11 - 9 = 2
    Difference between 2 consecutive y values-
                                                                                 dy1 = y2 - y1 = 162.5 - 58.5 = 104
                                                                                dy2 = y3 - y2 = 318.5 - 162.5 = 156
                                                                                dy3 = y4 - y3 = 526.5 - 318.5 = 208
                                                                                dy4 = y5 - y4 = 786.5 - 526.5 = 260
    We observe that the difference for every consecutive x values is constant i.e. 1 but for y values the difference is not constant.
    Hence, the given function is not a linear function.
    b). Now, difference between 2 consecutive differences for y values-
                                                                                       dy2 - dy1 = 156 - 104 = 52
                                                                                       dy3 - dy2 = 208 - 156 = 52
                                                                                        dy4 - dy3 = 21 - 15 = 52
    We observe that the difference of differences of 2 consecutive y values are constant i.e. 52.
    Hence, the given function is a quadratic function.
    Using Quadratic Regression formula and values from the adjacent table-
                                                                                        Y = aX2 + bX + c, where-
                                                                                        Σy = nc + b(Σx) + a(Σx2)
                                                                                  ∴ 1,852.5 = 5c + b(35) + a(285)
                                                                                    ∴ 1,852.5 = 5c + 35b + 285a .................................................. (Equation i)
    Σxy = c(Σx) + b(Σx2) + a(Σx3)
                                                                          ∴ 16,607.5 = c(35) + b(285) + a(2,555)
                                                                             ∴ 16,607.5 = 35c + 285b + 2,555a ....................................... (Equation ii)
    Σx2y = c(Σx2) + b(Σx3) + a(Σx4)
                                                                     ∴ 1,58,008.5 = c(285) + b(2,555) + a(24,309)
                                                                       ∴ 1,58,008.5 = 285c + 2,555b + 24,309a ....................... (Equation iii)
    Multiplying Equation i by 7, we get-
                                                                              1,955a + 245b + 35c = 12,967.5 …............................................... (Equation iv)
    Subtracting Equation iv from Equation ii, we get-
                                                                               2555a + 285b + 35c = 16,607.5 …............................................... (Equation ii)
                                                                             - 1,955a + 245b + 35c = 12,967.5 …............................................... (Equation iv)
                                                                                        560a + 40b = 3,640 .................................................. (Equation v)
    Multiplying Equation i with 57, we get-
                                                                         16,245a + 1,995b + 285c = 1,05,592.5 ......................... (Equation vi)
    Subtracting Equation vi from Equation iii, we get-
                                                                          24,309a + 2,555b + 285c = 1,58,008.5 ......................... (Equation iii)
                                                                        - 16,245a + 1,995b + 285c = 1,05,592.5 ......................... (Equation vi)
                                                                                     8,064a + 560b = 52,416 ......................... (Equation vii)
    Dividing Equation vii with 14, we get-
                                                                                        576a + 40b = 3,744 ............................................... (Equation viii)
    Subtracting Equation v from Equation viii, we get-
                                                                                        576a + 40b = 3,744 ............................................... (Equation viii)
                                                                                      - 560a + 40b = 3,640 ............................................... (Equation v)
                                                                                               16a = 104
    i.e. 16a = 104
                                                                                             ∴ a = 104/ 16 ................................... (Dividing both sides by 16)
                                                                                                ∴ a = 6.5
    Substituting a = 6.5 in Equation viii, we get-
                                                                                            576a + 40b = 3,744 .................................................. (Equation viii)
                                                                                      ∴ 576(6.5) + 40b = 3,744
                                                                                        ∴ 3,744 + 40b = 3,744
                                                                                        ∴ 40b = 3,744 - 3,744 ........................................ (Taking all constants together)
                                                                                                 ∴ 40b = 0
                                                                                                ∴ b = 0/40 ............................................ (Dividing both sides by 40)
                                                                                                   ∴ b = 0
    Substituting a = 6.5 and b = 0 in Equation i, we get-
                                                                                    285a + 35b + 5c = 1,852.5 .............................. (Equation i)
                                                                              ∴ 285(6.5) + 35 (0) + 5c = 1,852.5
                                                                                   ∴ 1,852.5 + 0 + 5c = 1,852.5
                                                                                      ∴ 1,852.5 + 5c = 1,852.5
                                                                                       ∴ 5c = 1,852.5 - 1,852.5 ..................... (Taking all constants together)
                                                                                                   ∴ 5c = 0
                                                                                                  ∴ c = 0/5 ........................... (Dividing both sides by 5)
                                                                                                   ∴ c = 0
    ∴ The Quadratic Equation is-
                                                                                              Y = aX2 + bX + c
                                                                                          ∴ Y = 6.5X2 + 0X + 0
                                                                                                ∴ Y = 6.5X2
    From the above calculations, we can find the relation between a value in the quadratic model i.e. 6.5 and the second difference (d) of the data i.e. 6.
    We observe that-
                                                                                                   52 = 8 × 6.5
                                                                                                   ∴ d = 8 × a
    ∴ Second difference = 8 × a
    Final Answer:-
    ∴ Second difference for data in the given table is 52. Quadratic regression for each data set can be represented using the function Y = 6.5X2. Also, the second difference is 8 times the a value.

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