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.
Hint:
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 1. Quadratic regression for each data set can be represented using the function Y = 0.5X2. Also, the second difference is 2 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 = 1, y1 = 0.5;
x2 = 2, y2 = 2;
x3 = 3, y3 = 4.5;
x4 = 4, y4 = 8;
x5 = 5, y5 = 12.5.
a). Difference between 2 consecutive x values-
dx1 = x2 - x1 = 2 - 1 = 1
dx2 = x3 - x2 = 3 - 2 = 1
dx3 = x4 - x3 = 4 - 3 = 1
dx4 = x5 - x4 = 5 - 4 = 1
Difference between 2 consecutive y values-
dy1 = y2 - y1 = 2 - 0.5 = 1.5
dy2 = y3 - y2 = 4.5 - 2 = 2.5
dy3 = y4 - y3 = 8 - 4.5 = 3.5
dy4 = y5 - y4 = 12.5 - 8 = 4.5
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 = 2.5 - 1.5 = 1
dy3 - dy2 = 3.5 - 2.5 = 1
dy4 - dy3 = 4.5 - 3.5 = 1
We observe that the difference of differences of 2 consecutive y values are constant i.e. 1.
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)
∴ 27.5 = 5c + b(15) + a(55)
∴ 27.5 = 5c + 15b + 55a .................................................. (Equation i)
Σxy = c(Σx) + b(Σx2) + a(Σx3)
∴ 112.5 = c(15) + b(55) + a(225)
∴ 112.5 = 15c + 55b + 225a ....................................... (Equation ii)
Σx2y = c(Σx2) + b(Σx3) + a(Σx4)
∴ 489.5 = c(55) + b(225) + a(979)
∴ 489.5 = 55c + 225b + 979a ....................... (Equation iii)
Multiplying Equation i by 3, we get-
165a + 45b + 15c = 82.5 ............................... (Equation iv)
Subtracting Equation i from iv, we get-
165a + 45b + 15c = 82.5 …............................................... (Equation iv)
- 55a + 15b + 5c = 27.5 …............................................... (Equation i)
110a + 30b = 55 .................................................. (Equation v)
Dividing Equation iii by 11, we get-
89a + 20.45b + 5c = 44.5 ......................... (Equation vi)
Subtracting Equation vi from Equation i, we get-
89a + 20.45b + 5c = 44.5 ......................... (Equation vi)
- 55a + 15b + 5c = 27.5 ......................... (Equation i)
34a + 5.45b = 17 ......................... (Equation vii)
Multiplying Equation vii with 110, we get-
3,740a + 599.5b = 1870 ............................................... (Equation viii)
Multiplying Equation v with 34, we get-
3,740a + 1,020b = 1870 ............................................... (Equation ix)
Subtracting Equation viii from Equation ix, we get-
3,740a + 1,020b = 1870 ............................................... (Equation ix)
- 3,740a + 599.5b = 1870 ............................................... (Equation viii)
420.5b = 0
i.e. 420.5b = 0
∴ b = 0/ 420.5 ................................... (Dividing both sides by 420.5)
∴ b = 0
Substituting b = 0 in Equation vii, we get-
34a + 5.45b = 17 .................................................. (Equation v)
∴ 34a + 5.45(0) = 17
∴ 34a + 0 = 17
∴ 34a = 17 - 0 ........................................ (Taking all constants together)
∴ 34a = 17/34 ....................................... (Dividing both sides by 34)
∴ a = 0.5
Substituting a = 0.5 and b = 0 in Equation i, we get-
55a + 15b + 5c = 27.5 .............................. (Equation i)
∴ 55(0.5) + 15(0) + 5c = 27.5
∴ 27.5 + 0 + 5c = 27.5
∴ 27.5 + 5c = 27.5
∴ 5c = 27.5 - 27.5 ..................... (Taking all constants together)
∴ 5c = 0
∴ c = 0/27.5 ........................... (Dividing both sides by 27.5)
∴ c = 0
∴ The Quadratic Equation is-
Y = aX2 + bX + c
∴ Y = 0.5 X2 + 0 X + 0
∴ Y = 0.5X2
From the above calculations, we can find the relation between a value in the quadratic model i.e. 0.5 and the second difference (d) of the data i.e. 1.
We observe that-
1 = 2 × 0.5
∴ 1 = 2 × a
∴ Second difference = 2 × a
Final Answer:-
∴ Second difference for data in the given table is 1. Quadratic regression for each data set can be represented using the function Y = 0.5X2. Also, the second difference is 2 times the a value.
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-
x2 = 2, y2 = 2;
x3 = 3, y3 = 4.5;
x4 = 4, y4 = 8;
x5 = 5, y5 = 12.5.
a). Difference between 2 consecutive x values-
dx1 = x2 - x1 = 2 - 1 = 1
dx2 = x3 - x2 = 3 - 2 = 1
dx3 = x4 - x3 = 4 - 3 = 1
dx4 = x5 - x4 = 5 - 4 = 1
Difference between 2 consecutive y values-
dy1 = y2 - y1 = 2 - 0.5 = 1.5
dy2 = y3 - y2 = 4.5 - 2 = 2.5
dy3 = y4 - y3 = 8 - 4.5 = 3.5
dy4 = y5 - y4 = 12.5 - 8 = 4.5
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 = 2.5 - 1.5 = 1
dy3 - dy2 = 3.5 - 2.5 = 1
dy4 - dy3 = 4.5 - 3.5 = 1
We observe that the difference of differences of 2 consecutive y values are constant i.e. 1.
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)
∴ 27.5 = 5c + b(15) + a(55)
∴ 27.5 = 5c + 15b + 55a .................................................. (Equation i)
Σxy = c(Σx) + b(Σx2) + a(Σx3)
∴ 112.5 = c(15) + b(55) + a(225)
∴ 112.5 = 15c + 55b + 225a ....................................... (Equation ii)
Σx2y = c(Σx2) + b(Σx3) + a(Σx4)
∴ 489.5 = c(55) + b(225) + a(979)
∴ 489.5 = 55c + 225b + 979a ....................... (Equation iii)
Multiplying Equation i by 3, we get-
165a + 45b + 15c = 82.5 ............................... (Equation iv)
Subtracting Equation i from iv, we get-
165a + 45b + 15c = 82.5 …............................................... (Equation iv)
- 55a + 15b + 5c = 27.5 …............................................... (Equation i)
110a + 30b = 55 .................................................. (Equation v)
Dividing Equation iii by 11, we get-
89a + 20.45b + 5c = 44.5 ......................... (Equation vi)
Subtracting Equation vi from Equation i, we get-
89a + 20.45b + 5c = 44.5 ......................... (Equation vi)
- 55a + 15b + 5c = 27.5 ......................... (Equation i)
34a + 5.45b = 17 ......................... (Equation vii)
Multiplying Equation vii with 110, we get-
3,740a + 599.5b = 1870 ............................................... (Equation viii)
Multiplying Equation v with 34, we get-
3,740a + 1,020b = 1870 ............................................... (Equation ix)
Subtracting Equation viii from Equation ix, we get-
3,740a + 1,020b = 1870 ............................................... (Equation ix)
- 3,740a + 599.5b = 1870 ............................................... (Equation viii)
420.5b = 0
i.e. 420.5b = 0
∴ b = 0/ 420.5 ................................... (Dividing both sides by 420.5)
∴ b = 0
Substituting b = 0 in Equation vii, we get-
34a + 5.45b = 17 .................................................. (Equation v)
∴ 34a + 5.45(0) = 17
∴ 34a + 0 = 17
∴ 34a = 17 - 0 ........................................ (Taking all constants together)
∴ 34a = 17/34 ....................................... (Dividing both sides by 34)
∴ a = 0.5
Substituting a = 0.5 and b = 0 in Equation i, we get-
55a + 15b + 5c = 27.5 .............................. (Equation i)
∴ 55(0.5) + 15(0) + 5c = 27.5
∴ 27.5 + 0 + 5c = 27.5
∴ 27.5 + 5c = 27.5
∴ 5c = 27.5 - 27.5 ..................... (Taking all constants together)
∴ 5c = 0
∴ c = 0/27.5 ........................... (Dividing both sides by 27.5)
∴ c = 0
∴ The Quadratic Equation is-
Y = aX2 + bX + c
∴ Y = 0.5 X2 + 0 X + 0
∴ Y = 0.5X2
From the above calculations, we can find the relation between a value in the quadratic model i.e. 0.5 and the second difference (d) of the data i.e. 1.
We observe that-
1 = 2 × 0.5
∴ 1 = 2 × a
∴ Second difference = 2 × a
Final Answer:-
∴ Second difference for data in the given table is 1. Quadratic regression for each data set can be represented using the function Y = 0.5X2. Also, the second difference is 2 times the a value.
. Check for extraneous solutions
. Check for extraneous solutions
. Check for extraneous solutions
Check for extraneous solutions
