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 6. Quadratic regression for each data set can be represented using the function Y = 3X2. 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 = 0;
x2 = 1, y2 = 3;
x3 = 2, y3 = 12;
x4 = 3, y4 = 27;
x5 = 4, y5 = 48.
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 = 3 - 0 = 3
dy2 = y3 - y2 = 12 - 3 = 9
dy3 = y4 - y3 = 27 - 12 = 15
dy4 = y5 - y4 = 48 - 27 = 21
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 = 9 - 3 = 6
dy3 - dy2 = 15 - 9 = 6
dy4 - dy3 = 21 - 15 = 6
We observe that the difference of differences of 2 consecutive y values are constant i.e. 6.
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)
∴ 90 = 5c + b(10) + a(30)
∴ 90 = 5c + 10b + 30a .................................................. (Equation i)
Σxy = c(Σx) + b(Σx2) + a(Σx3)
∴ 300 = c(10) + b(30) + a(100)
∴ 300 = 10c + 30b + 100a ....................................... (Equation ii)
Σx2y = c(Σx2) + b(Σx3) + a(Σx4)
∴ 1,062 = c(30) + b(100) + a(354)
∴ 1,062 = 30c + 100b + 354a ....................... (Equation iii)
Dividing Equation 2 by 2, we get-
50a + 15b + 5c = 150 …............................................... (Equation iv)
Subtracting Equation I from Equation iv, we get-
50a + 15b + 5c = 150 …............................................... (Equation iv)
- 30a + 10b + 5c = 90 …............................................... (Equation i)
20a + 5b = 60 .................................................. (Equation v)
Multiplying Equation ii with 3, we get-
300a + 90b + 30c = 900 ......................... (Equation vi)
Subtracting Equation vi from Equation iii, we get-
354a + 100b + 30c = 1,062 ......................... (Equation iii)
- 300a + 90b + 30c = 900 ......................... (Equation vi)
54a + 10b = 162 ......................... (Equation vii)
Multiplying Equation v with 2, we get-
40a + 10b = 120 ............................................... (Equation viii)
Subtracting Equation viii from Equation vii, we get-
54a + 10b = 162 ............................................... (Equation vii)
- 40a + 10b = 120 ............................................... (Equation viii)
14a = 42
i.e. 14a = 42
∴ a = 42/ 14 ................................... (Dividing both sides by 14)
∴ a = 3
Substituting a = 3 in Equation v, we get-
20a + 5b = 60 .................................................. (Equation v)
∴ 20(3) + 5b = 60
∴ 60 + 5b = 60
∴ 5b = 60 - 60 ........................................ (Taking all constants together)
∴ 5b = 0
∴ b = 0/5 ............................................ (Dividing both sides by 5)
∴ b = 0
Substituting a = 3 and b = 0 in Equation i, we get-
30a + 10b + 5c = 90 .............................. (Equation i)
∴ 30 (3) + 10 (0) + 5c = 90
∴ 90 + 0 + 5c = 90
∴ 90 + 5c = 90
∴ 5c = 90 - 90 ..................... (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 = 3X2 + 0 X + 0
∴ Y = 3 X2
From the above calculations, we can find the relation between a value in the quadratic model i.e. 3 and the second difference (d) of the data i.e. 6.
We observe that-
6 = 2 × 3
∴ d = 2 × a
∴ Second difference = 2 × a
Final Answer:-
∴ Second difference for data in the given table is 6. Quadratic regression for each data set can be represented using the function Y = 3X2. 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 = 3;
x3 = 2, y3 = 12;
x4 = 3, y4 = 27;
x5 = 4, y5 = 48.
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 = 3 - 0 = 3
dy2 = y3 - y2 = 12 - 3 = 9
dy3 = y4 - y3 = 27 - 12 = 15
dy4 = y5 - y4 = 48 - 27 = 21
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 = 9 - 3 = 6
dy3 - dy2 = 15 - 9 = 6
dy4 - dy3 = 21 - 15 = 6
We observe that the difference of differences of 2 consecutive y values are constant i.e. 6.
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)
∴ 90 = 5c + b(10) + a(30)
∴ 90 = 5c + 10b + 30a .................................................. (Equation i)
Σxy = c(Σx) + b(Σx2) + a(Σx3)
∴ 300 = c(10) + b(30) + a(100)
∴ 300 = 10c + 30b + 100a ....................................... (Equation ii)
Σx2y = c(Σx2) + b(Σx3) + a(Σx4)
∴ 1,062 = c(30) + b(100) + a(354)
∴ 1,062 = 30c + 100b + 354a ....................... (Equation iii)
Dividing Equation 2 by 2, we get-
50a + 15b + 5c = 150 …............................................... (Equation iv)
Subtracting Equation I from Equation iv, we get-
50a + 15b + 5c = 150 …............................................... (Equation iv)
- 30a + 10b + 5c = 90 …............................................... (Equation i)
20a + 5b = 60 .................................................. (Equation v)
Multiplying Equation ii with 3, we get-
300a + 90b + 30c = 900 ......................... (Equation vi)
Subtracting Equation vi from Equation iii, we get-
354a + 100b + 30c = 1,062 ......................... (Equation iii)
- 300a + 90b + 30c = 900 ......................... (Equation vi)
54a + 10b = 162 ......................... (Equation vii)
Multiplying Equation v with 2, we get-
40a + 10b = 120 ............................................... (Equation viii)
Subtracting Equation viii from Equation vii, we get-
54a + 10b = 162 ............................................... (Equation vii)
- 40a + 10b = 120 ............................................... (Equation viii)
14a = 42
i.e. 14a = 42
∴ a = 42/ 14 ................................... (Dividing both sides by 14)
∴ a = 3
Substituting a = 3 in Equation v, we get-
20a + 5b = 60 .................................................. (Equation v)
∴ 20(3) + 5b = 60
∴ 60 + 5b = 60
∴ 5b = 60 - 60 ........................................ (Taking all constants together)
∴ 5b = 0
∴ b = 0/5 ............................................ (Dividing both sides by 5)
∴ b = 0
Substituting a = 3 and b = 0 in Equation i, we get-
30a + 10b + 5c = 90 .............................. (Equation i)
∴ 30 (3) + 10 (0) + 5c = 90
∴ 90 + 0 + 5c = 90
∴ 90 + 5c = 90
∴ 5c = 90 - 90 ..................... (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 = 3X2 + 0 X + 0
∴ Y = 3 X2
From the above calculations, we can find the relation between a value in the quadratic model i.e. 3 and the second difference (d) of the data i.e. 6.
We observe that-
6 = 2 × 3
∴ d = 2 × a
∴ Second difference = 2 × a
Final Answer:-
∴ Second difference for data in the given table is 6. Quadratic regression for each data set can be represented using the function Y = 3X2. 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
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