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: Final Answer:- ∴ Second difference for data in the given table is 8. Quadratic regression for each data set can be represented using the function Y = 4X2 + 8X + 4. 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 = 0, y1 = 4;
x2 = 1, y2 = 16;
x3 = 2, y3 = 36;
x4 = 3, y4 = 64;
x5 = 4, y5 = 100.
a). Difference between 2 consecutive x values-
dx1 = x2 - x1 = 1 - 0 = 1
dx2 = x3 - x2 = 2 - 1 = 1
dx3 = x4 - x3 = 3 - 2 = 1
dx4 = x5 - x4 = 4 - 3 = 1
Difference between 2 consecutive y values-
dy1 = y2 - y1 = 16 - 4 = 12
dy2 = y3 - y2 = 36 - 16 = 20
dy3 = y4 - y3 = 64 - 36 = 28
dy4 = y5 - y4 = 100 - 64 = 36
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 = 20 - 12 = 8
dy3 - dy2 = 28 - 20 = 8
dy4 - dy3 = 36 - 28 = 8
We observe that the difference of differences of 2 consecutive y values are constant i.e. 8.
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)
∴ 220 = 5c + b(10) + a(30)
∴ 220 = 5c + 10b + 30a .................................................. (Equation i)
Σxy = c(Σx) + b(Σx2) + a(Σx3)
∴ 680 = c(10) + b(30) + a(100)
∴ 680 = 10c + 30b + 100a ....................................... (Equation ii)
Σx2y = c(Σx2) + b(Σx3) + a(Σx4)
∴ 2,336 = c(30) + b(100) + a(354)
∴ 2,336 = 30c + 100b + 354a ....................... (Equation iii)
Dividing Equation 2 by 2, we get-
50a + 15b + 5c = 340 …............................................... (Equation iv)
Subtracting Equation I from Equation iv, we get-
50a + 15b + 5c = 340 …............................................... (Equation iv)
- 30a + 10b + 5c = 220 …............................................... (Equation i)
20a + 5b = 120 .................................................. (Equation v)
Multiplying Equation ii with 3, we get-
300a + 90b + 30c = 2,040 ......................... (Equation vi)
Subtracting Equation vi from Equation iii, we get-
354a + 100b + 30c = 2,336 ......................... (Equation iii)
- 300a + 90b + 30c = 2,040 ......................... (Equation vi)
54a + 10b = 296 ......................... (Equation vii)
Multiplying Equation v with 2, we get-
40a + 10b = 240 ............................................... (Equation viii)
Subtracting Equation viii from Equation vii, we get-
54a + 10b = 296 ............................................... (Equation vii)
- 40a + 10b = 240 ............................................... (Equation viii)
14a = 56
i.e. 14a = 56
∴ a = 56/ 14 ................................... (Dividing both sides by 14)
∴ a = 4
Substituting a = 4 in Equation v, we get-
20a + 5b = 120 .................................................. (Equation v)
∴ 20(4) + 5b = 120
∴ 80 + 5b = 120
∴ 5b = 120 - 80 ........................................ (Taking all constants together)
∴ 5b = 40
∴ b = 40/5 ............................................ (Dividing both sides by 5)
∴ b = 8
Substituting a = 3 and b = 0 in Equation i, we get-
30a + 10b + 5c = 220 .............................. (Equation i)
∴ 30 (4) + 10 (8) + 5c = 220
∴ 120 + 80 + 5c = 220
∴ 200 + 5c = 220
∴ 5c = 220 - 200 ..................... (Taking all constants together)
∴ 5c = 20
∴ c = 20/5 ........................... (Dividing both sides by 5)
∴ c = 4
∴ The Quadratic Equation is-
Y = aX2 + bX + c
∴ Y = 4X2 + 8X + 4
From the above calculations, we can find the relation between a value in the quadratic model i.e. 4 and the second difference (d) of the data i.e. 8.
We observe that-
8 = 2 × 4
∴ d = 2 × a
∴ Second difference = 2 × a
Final Answer:-
∴ Second difference for data in the given table is 8. Quadratic regression for each data set can be represented using the function Y = 4X2 + 8X + 4. 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 = 1, y2 = 16;
x3 = 2, y3 = 36;
x4 = 3, y4 = 64;
x5 = 4, y5 = 100.
a). Difference between 2 consecutive x values-
dx1 = x2 - x1 = 1 - 0 = 1
dx2 = x3 - x2 = 2 - 1 = 1
dx3 = x4 - x3 = 3 - 2 = 1
dx4 = x5 - x4 = 4 - 3 = 1
Difference between 2 consecutive y values-
dy1 = y2 - y1 = 16 - 4 = 12
dy2 = y3 - y2 = 36 - 16 = 20
dy3 = y4 - y3 = 64 - 36 = 28
dy4 = y5 - y4 = 100 - 64 = 36
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 = 20 - 12 = 8
dy3 - dy2 = 28 - 20 = 8
dy4 - dy3 = 36 - 28 = 8
We observe that the difference of differences of 2 consecutive y values are constant i.e. 8.
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)
∴ 220 = 5c + b(10) + a(30)
∴ 220 = 5c + 10b + 30a .................................................. (Equation i)
Σxy = c(Σx) + b(Σx2) + a(Σx3)
∴ 680 = c(10) + b(30) + a(100)
∴ 680 = 10c + 30b + 100a ....................................... (Equation ii)
Σx2y = c(Σx2) + b(Σx3) + a(Σx4)
∴ 2,336 = c(30) + b(100) + a(354)
∴ 2,336 = 30c + 100b + 354a ....................... (Equation iii)
Dividing Equation 2 by 2, we get-
50a + 15b + 5c = 340 …............................................... (Equation iv)
Subtracting Equation I from Equation iv, we get-
50a + 15b + 5c = 340 …............................................... (Equation iv)
- 30a + 10b + 5c = 220 …............................................... (Equation i)
20a + 5b = 120 .................................................. (Equation v)
Multiplying Equation ii with 3, we get-
300a + 90b + 30c = 2,040 ......................... (Equation vi)
Subtracting Equation vi from Equation iii, we get-
354a + 100b + 30c = 2,336 ......................... (Equation iii)
- 300a + 90b + 30c = 2,040 ......................... (Equation vi)
54a + 10b = 296 ......................... (Equation vii)
Multiplying Equation v with 2, we get-
40a + 10b = 240 ............................................... (Equation viii)
Subtracting Equation viii from Equation vii, we get-
54a + 10b = 296 ............................................... (Equation vii)
- 40a + 10b = 240 ............................................... (Equation viii)
14a = 56
i.e. 14a = 56
∴ a = 56/ 14 ................................... (Dividing both sides by 14)
∴ a = 4
Substituting a = 4 in Equation v, we get-
20a + 5b = 120 .................................................. (Equation v)
∴ 20(4) + 5b = 120
∴ 80 + 5b = 120
∴ 5b = 120 - 80 ........................................ (Taking all constants together)
∴ 5b = 40
∴ b = 40/5 ............................................ (Dividing both sides by 5)
∴ b = 8
Substituting a = 3 and b = 0 in Equation i, we get-
30a + 10b + 5c = 220 .............................. (Equation i)
∴ 30 (4) + 10 (8) + 5c = 220
∴ 120 + 80 + 5c = 220
∴ 200 + 5c = 220
∴ 5c = 220 - 200 ..................... (Taking all constants together)
∴ 5c = 20
∴ c = 20/5 ........................... (Dividing both sides by 5)
∴ c = 4
∴ The Quadratic Equation is-
Y = aX2 + bX + c
∴ Y = 4X2 + 8X + 4
From the above calculations, we can find the relation between a value in the quadratic model i.e. 4 and the second difference (d) of the data i.e. 8.
We observe that-
8 = 2 × 4
∴ d = 2 × a
∴ Second difference = 2 × a
Final Answer:-
∴ Second difference for data in the given table is 8. Quadratic regression for each data set can be represented using the function Y = 4X2 + 8X + 4. Also, the second difference is 2 times the a value.
Round to the nearest tenth if necessary.
tells the amount of drink left in your system after t hours. How much of the soft drink will be left in your system after 16 hours?
Check for extraneous solutions
. Identify any extraneous solutions
