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 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.
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 = 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.
. Check for extraneous solutions
. Check for extraneous solutions
. Check for extraneous solutions
. Check for extraneous solutions
Check for extraneous solutions
