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| 10 Theory slides |
| 8 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Consider the following table of values.
x | f(x) |
---|---|
1 | 2 |
2 | 3 |
3 | 7 |
4 | 16 |
5 | 32 |
6 | 57 |
Looking at the x-values, it can be noted that the values are equally spaced. It is possible to find the first finite differences by subtracting two consecutive values. Consider the following example. f(2) - f(1) = 3 - 2 ⇓ f(2) - f(1) = 1
The first differences for each pair of consecutive terms can be found by doing the same process. Then, the second differences are found by subtracting consecutive first differences, third differences by subtracting consecutive second differences, and so on. Find every first, second, and third difference. Is there something that can be noted about the differences?As part of an entrepreneur unit in school, Zosia is learning some interesting math. During her lesson, Zosia's teacher explained that modeling data is useful to understanding business. He explained that there are various methods to write a polynomial function that models data.
As an example, the teacher told the students about a certain company that has experienced ups and downs in earnings due to economic downturns.
The teacher showed the students the following graph, where y represents the profit, in millions of dollars, and x represents the years before and after 2000. Here, negative values of y represent losses.
Notice that three of the points are x-intercepts.
x= 0, f(x)= 7
Identity Property of Addition
Multiply
.LHS /42.=.RHS /42.
Simplify quotient
Rearrange equation
Sometimes it can be useful to examine the differences between function values when the input values are separated by a known quantity. These differences are known as finite differences.
A finite difference is a mathematical expression that can be written in the following form.
f(x+c_1)-f(x+c_2)
In this expression, c_1 and c_2 are constants. There are three basic finite differences that are especially useful.
These differences are shown below.
Finite differences can be used to determine the degree of a polynomial function that models a data set. This is because finite differences have some really useful properties.
The finite differences of polynomial functions satisfy the following two properties.
To justify these properties, the minor cases will be examined first.
Substitute expressions
Distribute -1
(a+b)^2=a^2+2ab+b^2
Distribute a
Distribute b
Subtract terms
Distribute -1
Subtract terms
Substitute expressions
(a+b)^2=a^2+2ab+b^2
Distribute a
Distribute b
Distribute -2
Subtract terms
It can be concluded from the previous cases that, for a polynomial of degree n, its nth differences are the last constant non-zero differences. Therefore, the nth differences of a polynomial of degree n are constant and nonzero. Note that this is an informal justification and should not be taken as a proof.
Interested about Zosia's recent entrepreneur lessons, her uncle told her about his own experience. Some years ago, Zosia's uncle started a local business selling art supplies.
Her uncle showed Zosia how his sales went the first few years of business. At the end of the first year, he was $1000 behind because of a starting loan. Then he started seeing returns.
Year | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|
Income | -$1000 | $1000 | $6000 | $14 000 | $25 000 |
Recalling her math lessons, Zosia decided to help her uncle by modeling the income with a polynomial function. Help Zosia do the following.
x | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|
y | -1 | 1 | 6 | 14 | 25 |
By the Properties of Finite Differences, the constant differences can help determine the degree of the polynomial that best fits the data. The first differences are shown below.
The first differences are not constant. This means that the polynomial is not of degree 1 and that other differences need to be examined. Now it is time to look at the second differences.
Looking at the differences, it can be noted that the second differences are constant and non-zero. By the Properties of Finite Differences, the degree of the polynomial is 2.
To find a polynomial function of degree 2 to fit the data, the quadratic regression function of the calculator will be used. To view this function, press STAT, scroll to the right to view the CALC options, and choose the fifth option in the list, QuadReg.
This function is similar to a line of best fit function, but instead of fitting the data with a line, the quadratic regressions fits the data using a parabola. Using the information given, it is possible to write the function. y = 1.5x^2 - 2.5x Remember that at the beginning the numbers were divided by 1000. Therefore, to obtain the real income, each coefficient must be multiplied by 1000. y = 1500x^2 - 2500x It should be noted that the function can also be found by setting and solving a system of equations, where the coefficients are the variables. This method is explained later in the lesson.
Indicate the degree of the polynomial function that best fits the data in each table of values.
When given a set of points, it is possible to determine a polynomial that perfectly fits these points. To do so, it is important to consider a principle called the (n+1) Point Principle.
Given a set of n + 1 points in the coordinate plane, where no pair of points lies on a vertical line, there is a unique polynomial of degree at most n that fits the points perfectly.
To justify this principle, the smaller cases will be examined to come to a conclusion.
In general, n+1 points generate a system of n+1 equations with n+1 variables, where each variable is a coefficient of a polynomial of degree n. y_1 = a_1 x_1^n + a_2 x_1^(n-1) + ⋯ + a_(n-1)x_1 + a_n y_2 = a_1 x_2^n + a_2 x_2^(n-1) + ⋯ + a_(n-1)x_2 + a_n ... y_n = a_1 x_n^n + a_2 x_n^(n-1) + ⋯ + a_(n-1)x_n + a_n A solution can result in a polynomial with a degree less than n if the leading coefficient a_1 is zero. It should be noted that this is an informal justification and should not be taken as a proof.
After seeing her enthusiasm about his experience, Zosia's uncle invited her out to visit and get some first hand experience with how a business works. Nervous about her first flight, Zosia watched a documentary about airplanes and their speeds at takeoff.
In the documentary, the horizontal distance traveled by the plane during the first few minutes after takeoff was analyzed.
Zosia thought she could model the distances by finding a polynomial that passes through the following four points. (1,5), (2,9) (3,15), (4,24) Here, the x-coordinates represent the minutes after takeoff and the y-coordinates represent the horizontal distance between the plane and the airport.
y = ax^3 + bx^2 + cx + d To find the values of the coefficients, the points are substituted into the equation above to generate a system of four equations. 5 = a(1)^3 + b(1)^2 + c(1) + d 9 = a(2)^3 + b(2)^2 + c(2) + d 15 = a(3)^3 + b(3)^2 + c(3) + d 24 = a(4)^3 + b(4)^2 + c(4) + d These equations can be rewritten by calculating the powers and rearranging the terms. 1a + 1b + 1c + 1d = 5 8a + 4b + 2c + 1d = 9 27a + 9b + 3c + 1d=15 64a + 16b + 4c + 1d = 24 Since the system is big, solving it by elimination or substitution can get messy. Because of this, it is preferred to use a graphing calculator to solve the system. To input a system of equations as a matrix, press 2ND, then x^(-1). Then, scroll to the right to reach the EDIT menu.
Selecting one of the matrices in the menu opens a window to modify its elements. To input a system of four equations with four variables, a 4*5 matrix will be needed. The last column stores the results at the right-hand side of each equation.
To solve the system of equations, press 2ND and x^(-1) once more. However, this time, select the MATH menu and scroll down until the rref(
option is selected.
Finally, input matrix [A] to the rref(
function. To do so, push 2ND and x^(-1) to bring up all available matrices. In the NAMES menu, choose matrix [A].
The solutions can be read vertically in the last column, using the remaining coefficients as reference points for the variables. [ cccc 1&0&0&0&0.166... 0&1&0&0&0 0&0&1&0&2.833... 0&0&0&1&2 ] ⇕ 1a + 0b + 0c + 0d = 0.166... 0a + 1b + 0c + 0d = 0 0a + 0b + 1c + 0d= 2.833... 0a + 0b + 0c + 1d = 2 Now that the coefficients have been calculated, it is possible to write the polynomial function. Remember to round each coefficient to two decimal places. y = 0.166... x^3 + 0x^2 + 2.833... x + 2 ⇓ y = 0.17x^3 + 2.83x + 2
x= 5
Calculate power
Multiply
Add terms
Zosia is really into environmentalism. After watching the documentary about planes, she became increasingly worried about the atmospheric carbon dioxide levels on Earth and decided to do some research.
In her research, Zosia found data about the average levels, in parts per million, of carbon dioxide in the atmosphere per year. Some results are listed below. The years are counted from the year 1995, so 0 is the year 1995, 5 is 2000, and so on.
Year | Carbon Dioxide Levels (ppm) |
---|---|
0 | 360 |
5 | 369 |
10 | 379 |
15 | 390 |
20 | 401 |
25 | 414 |
Recalling her recent introduction to polynomial functions, Zosia decided to do the following.
QuartRegon a graphing calculator.
To find a polynomial of degree four that fits the points, press STAT again and move to the CALC menu. Then, scroll down to select the function QuartReg.
Now that the coefficients are known, it is possible to write an function for the carbon levels. It is important to remember that the symbol E^(-5) is equivalent to multiplying the number by 10^(-5). C(t) = 6.667*10^(-5)t^4 -0.003t^3 + 0.063t^2 - 1.538t + 360.0
t= 30
DiagnosticOnand press ENTER. Finally, reopen the window from Part A.
The closer the value of R^2 is to 1, the better the polynomial function fits the points. Since R^2 is very close to 1, the function fits the points really well. Therefore, it is possible that the prediction is reliable, but more points would clarify if that is the case.
The following table of values was presented earlier in the lesson.
x | f(x) |
---|---|
1 | 2 |
2 | 3 |
3 | 7 |
4 | 16 |
5 | 32 |
6 | 57 |
Below are the first, second, and third differences of these values.
Note that the third differences are constant and different from zero. By the Properties of Finite Differences, the function values are represented by a cubic polynomial function. f(x) = ax^3 + bx^2 + cx + d Since many points are given, the best way to find the coefficients is to use the cubic regression function on a graphing calculator. The first thing to do is to input the data into the calculator. Push STAT, choose Edit, and then enter the values in the first two columns.
To find a polynomial of degree three that fits the points, press STAT again and move to the CALC menu. Then, scroll down to select the function CubicReg.
Now it is possible to write a polynomial function of degree three that fits the given data. y = 0.33x^3 - 0.5x^2 + 0.17x + 2
Since the value of R^2 is exactly 1, the polynomial function fits the points perfectly. This confirms the conclusion made considering the Properties of Finite Differences. In general, when encountering a data set, it can be useful to find a polynomial function to model the data, as it might lead to more accurate predictions or conclusions.
Ignacio's teacher asked him to order the following four functions according to their degrees, from least to greatest.
After thinking for some minutes, Ignacio told his teacher that he needed some extra information to complete the task. The teacher agreed and gave him the following pair of statements.
To order the given functions according to their degrees, we first need to determine their degrees. Let's consider each function separately.
From the given graph, we can see that the function crosses the x-axis four times. In other words, the graph has four x-intercepts.
This suggests that the degree of the polynomial function f(x) is at least 4. Since Ignacio's teacher said that f(x) has the least possible degree that it can have, we can conclude that the degree of f(x) is indeed 4.
Given a function rule, the degree of the polynomial is equal to the greatest exponent. Let's consider the given function rule again. g(x)=x^3-2x^2+3x-8 The greatest exponent in the function is 3, so the degree of g(x) is 3.
The function h(x) is given as a table of values. Notice that the x-values are equally spaced, meaning that we can use the properties of finite differences to determine the degree of the polynomial function that best fits the data set.
x | -2 | 0 | 2 | 4 | 6 | 8 |
---|---|---|---|---|---|---|
h(x) | 6 | 8 | 6 | 12 | 26 | 48 |
Let's find the first differences by subtracting each pair of consecutive function values.
Since the first differences are not constant, let's proceed to finding the second differences.
The second differences are constant and different from zero. Therefore, the degree of the polynomial h(x) is 2.
The function k(x) is also given as a table of values, but notice that the x-values are not equally spaced. This means that we cannot proceed as we did with h(x). However, no pair of points lie on a vertical line — all the x-values are different. Therefore, we can use The (n+1) Point Principle.
The (n+1) Point Principle |- Given a set of n+1 points in the coordinate plane, where no pair of points lies on a vertical line, there is a unique polynomial of degree at most n that fits the points perfectly.
In our case, we were given 5+1=6 different points. Therefore, there is a unique polynomial of degree at most 5 that fits the points perfectly. Since the teacher said that k(x) has the greatest possible degree that it can have, we can conclude that the degree of k(x) is 5.
Now that we know the degree of each polynomial function, we can order them according to their degrees, from least to greatest.
Polynomial Function | Degree |
---|---|
h(x) | 2 |
g(x) | 3 |
f(x) | 4 |
k(x) | 5 |
Over time, a certain food chain has experienced ups and downs in profits due to economic downturns. Consider the following notes about the finance history of the business.
The following graph models the finances of the food chain over the years, where y represents the profit in millions of dollars and x represents the years before and after 2010. Here, negative values of y represent losses.
The first thing we notice from the given information and graph is that it has three x-intercepts.
Each x-intercept represents a zero of the function. By the Factor Theorem, each zero gives us a factor the polynomial function.
x-intercept | Zero | Factor |
---|---|---|
(-4,0) | -4 | x+4 |
(1,0) | 1 | x-1 |
(5,0) | 5 | x-5 |
Since the function is cubic, it can be written in factored form as a constant a multiplied by the three factors we identified. f(x) = a(x+4)(x-1)(x-5) Let's use the y-intercept of the graph to determine the value of a. The y-intercept is (0,40),so we have that f(0)=40. If we evaluate f(x) at x=0, we can solve the resulting equation for a.
Now that we know the value of a, we can substitute it into the factored form. We can then perform the required multiplications to write the function rule in standard form.
The following table shows the study time and test scores of four students.
Study Time (min) ⏰ | 12 | 16 | 18 | 19 |
---|---|---|---|---|
Test Score 📝 | 92 | 32 | 26 | 50 |
Since we were given only 4 data values, we know that by the (n+1) Point Principle, a polynomial of degree at most 3 exists that fits the points perfectly. In other words, we are looking for a cubic polynomial. P(x) = ax^3 + bx^2 + cx + d To find the values of the coefficients, we can substitute the study times for x and the test scores for f(x). This will generate a system of four equations. 92 = a( 12)^3 + b( 12)^2 + c( 12) + d 32 = a( 16)^3 + b( 16)^2 + c( 16) + d 26 = a( 18)^3 + b( 18)^2 + c( 18) + d 50 = a( 19)^3 + b( 19)^2 + c( 19) + d Let's perform the required operations and rearrange the system so that the results are on the right-hand side. 1728a + 144b + 12c + d = 92 4096a + 256b + 16c + d = 32 5832a + 324b + 18c + d = 26 6859a + 361b + 19c + d = 50 Since the system is big, solving it by elimination or substitution can be messy. Instead, let's use a graphing calculator. First, we input the system of equations as a matrix. To do so, we press 2ND, then x^(-1), and scroll to the right to reach the EDIT menu.
In that menu, we can select any of the matrices. After that, the calculator shows a window where we can enter the elements of our system. To input a system of four equations with four variables, a 4* 5 matrix will be needed. The last column stores the results on the right-hand side of each equation.
To solve the system of equations, we press 2ND and x^(-1) again. However, this time, we will select the MATH menu and scroll down until we select the rref(
option.
Finally, we input the matrix [A] into the rref(
function. To do so, push 2ND and x^(-1) to bring up all available matrices. In the NAMES menu, choose matrix [A].
The solutions to the system of equation are the numbers in the last column. For example, the value of a is the first number, the value of b is the second one, and so on. a = 1 b=-44 c=625 d=-2800 Finally, let's substitute these values into the cubic polynomial. P(x) = x^3 - 44x^2 + 625x - 2800 The graph of this function looks as follows.