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Algebra 2 View details

4. Applying Polynomial Functions to Real-World Situations

Lesson

Exercises

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eCourses /

Algebra 2 /

Chapter 2

Applying Polynomial Functions to Real-World Situations

Polynomial functions have a remarkable presence in numerous areas of study. The degree of the polynomial, a defining attribute of these functions, dictates their behavior and the nature of their graphs. This lesson emphasizes the various ways these functions get employed in tangible situations. Whether it's in physics to describe certain motions, in economics to predict trends, or in engineering for design purposes, the role of polynomial functions is undeniably significant. By grasping the concept of the degree and its influence on the function's characteristics, individuals can make more informed decisions and interpretations in their respective fields.

Lesson Settings & Tools

	10 Theory slides
	8 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Image Credits

Slide of 10

Polynomials are used in a variety of fields to model diverse phenomena. This lesson aims to show how polynomials can model real-life situations.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Binomial
Polynomial
Zeros of a Function
Degree of a Polynomial
Table of Values

Explore

Differences of Function Values

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?

Example

Writing a Function From a Graph

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.

Before $1997,$ the company suffered losses due to the credits requested for the opening.
In the years $1997,$ $2002,$ and $2007,$ the company had neither profit nor losses.
From $1997$ to $2002,$ the company made a profit.
Between $2002$ and $2007,$ the company suffered losses.
From $2007$ onwards, the company's profits have been increasing.

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.

Graph of a cubic polynomial passing through (-3,0), (0,7), (2,0), and (7,0)

Then the teacher asked the students to write the cubic function graphed. Help Zosia write the expression for the cubic function.

Hint

Notice that three of the points are $x -$ intercepts.

Solution

The first thing to notice in the given graph is that three of the given points are zeros of the function. The Factor Theorem indicates that the zeros of a polynomial function are factors of the function. Since the function is cubic, the three zeros indicate the three factors.

(7, 0), (- 3, 0), (2, 0) ⇓ f (x) = a (x - 7) (x + 3) (x - 2)

In this expression, the factors are given by the zeros and

a

is a constant. To find the value of

a,

the fourth point is substituted into the expression.

f (x) = a (x - 7) (x + 3) (x - 2)

SubstituteII

$x = 0$ , $f (x) = 7$

7 = a (0 - 7) (0 + 3) (0 - 2)

▼

Solve for

a

IdPropAdd

Identity Property of Addition

7 = a (- 7) (3) (- 2)

Multiply

7 = 42 a

DivEqn

$LHS / 42 = RHS / 42$

\frac{7}{4 2} = a

SimpQuot

Simplify quotient

\frac{1}{6} = a

RearrangeEqn

Rearrange equation

a = \frac{1}{6}

Now that the value of

a

has been found, it is possible to finish writing the function shown the graph, which represents the profits of the company.

f (x) = \frac{1}{6} (x - 7) (x + 3) (x - 2)

Discussion

Finite Differences and Their Properties

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.

Rule

Finite Difference

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.

Forward difference: $f (x + h) - f (x)$
Backward difference: $f (x) - f (x - h)$
Central difference: $f (x + h) - f (x - h)$

These differences are shown below.

Backward, central, and forward difference shown with a polynomial curve

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.

Rule

Properties of Finite Differences

The finite differences of polynomial functions satisfy the following two properties.

For a polynomial function of degree $n,$ the $n th$ differences of the function values corresponding to equally spaced values of the independent variable are constant and different from zero.
If the $n th$ differences of equally spaced values are constant and different from zero, then the values can be represented by a polynomial function of degree $n .$

Proof

Informal Justification

To justify these properties, the minor cases will be examined first.

$n = 1$

Consider a linear function.

f_{1} (x) = a x + b

In this expression,

a

and

b

are constants different from zero. Suppose that the

x -

values are spaced by a constant

h .

The first differences between two arbitrary consecutive values of the function can be written as follows.

f_{1} (x + h) - f_{1} (x)

The expression for the linear function can be substituted here to see what happens.

f_{1} (x + h) - f_{1} (x)

SubstituteII

$f_{1} (x + h) = a (x + h) + b$ , $f_{1} (x) = a x + b$

a (x + h) + b - (a x + b)

▼

Simplify

Distr

Distribute $- 1$

a (x + h) + b - a x - b

Distr

Distribute $a$

a x + a h + b - a x - b

SubTerms

Subtract terms

a h

Since both

a

and

h

are non-zero constants, their product is different from zero. Therefore, the first differences are always a non-zero constant, which confirms the property. It should be noted that, since the first differences are constant, all the second differences are zero.

$n = 2$

Now consider a quadratic function.

f_{2} (x) = a x^{2} + b x + c

In this expression,

a,

b,

and

c

are non-zero constants. The values of

x

are separated by a constant

h .

A first difference can be obtained by subtracting two consecutive function values.

f_{2} (x + h) - f_{2} (x)

Again, it is possible to substitute the expression for

f_{2} (x)

to find an expression for the first difference.

f_{2} (x + h) - f_{2} (x)

SubstituteExpressions

Substitute expressions

a (x + h)^{2} + b (x + h) + c - (a x^{2} + b x + c)

▼

Simplify

Distr

Distribute $- 1$

a (x + h)^{2} + b (x + h) + c - a x^{2} - b x - c

ExpandPosPerfectSquare

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

a (x^{2} + 2 x h + h^{2}) + b (x + h) + c - a x^{2} - b x - c

Distr

Distribute $a$

a x^{2} + 2 a x h + a h^{2} + b (x + h) + c - a x^{2} - b x - c

Distr

Distribute $b$

a x^{2} + 2 a x h + a h^{2} + b x + b h + c - a x^{2} - b x - c

SubTerms

Subtract terms

2 a x h + a h^{2} + b h

Looking at the linear term

2 a x h,

it can be concluded that this difference is not constant. Now, try again with a second difference. Two consecutive first differences are needed to find a second difference.

f_{2} (x + h) - f_{2} (x) f_{2} (x + 2 h) - f_{2} (x + h)

A second difference is obtained by subtracting the first expression from the second one.

f_{2} (x + 2 h) - f_{2} (x + h) - (f_{2} (x + h) - f_{2} (x))

Distr

Distribute $- 1$

f_{2} (x + 2 h) - f_{2} (x + h) - f_{2} (x + h) + f_{2} (x)

SubTerms

Subtract terms

f_{2} (x + 2 h) - 2 f_{2} (x + h) + f_{2} (x)

SubstituteExpressions

Substitute expressions

a (x + 2 h)^{2} + b (x + 2 h) + c - 2 (a (x + h)^{2} + b (x + h) + c) + a x^{2} + b x + c

▼

Simplify

ExpandPosPerfectSquare

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

a (x^{2} + 4 x h + 4 h^{2}) + b (x + 2 h) + c - 2 (a (x^{2} + 2 x h + h^{2}) + b (x + h) + c) + a x^{2} + b x + c

Distr

Distribute $a$

a x^{2} + 4 a x h + 4 a h^{2} + b (x + 2 h) + c - 2 (a x^{2} + 2 a x h + a h^{2} + b (x + h) + c) + a x^{2} + b x + c

Distr

Distribute $b$

a x^{2} + 4 a x h + 4 a h^{2} + b x + 2 b h + c - 2 (a x^{2} + 2 a x h + a h^{2} + b x + b h + c) + a x^{2} + b x + c

Distr

Distribute $- 2$

a x^{2} + 4 a x h + 4 a h^{2} + b x + 2 b h + c - 2 a x^{2} - 4 a x h - 2 a h^{2} - 2 b x - 2 b h - 2 c + a x^{2} + b x + c

SubTerms

Subtract terms

2 a h^{2}

Since

a

and

h

are non-zero constants, the second difference is a constant different from zero, confirming the property again. Similar to the previous case, since the second differences are constant, the third differences are zero.

Conclusion

It can be concluded from the previous cases that, for a polynomial of degree $n,$ its $n th$ differences are the last constant non-zero differences. Therefore, the $n th$ 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.

Example

Writing a Function From a Table of Values

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$	$$ 14000$	$$ 25000$

Recalling her math lessons, Zosia decided to help her uncle by modeling the income with a polynomial function. Help Zosia do the following.

a Determine the degree of the polynomial function that better fits these values.

b Let

x

be the year. Write the polynomial function.

Hint

a Recall the Properties of Finite Differences.

b Use the corresponding regression function on a graphing calculator.

Solution

a To make the numbers smaller and easier to manage, the incomes can be divided by

1000 .

Doing so results in the following table.

$x$	$1$	$2$	$3$	$4$	$5$
$y$	$- 1$	$1$	$6$	$14$	$25$

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 $2 .$

b A graphing calculator can be used to find the polynomial function. To do so, first enter the given values into lists. Push

STAT,

choose Edit, and then enter the values in the first two columns. It should be noted that it is convenient to input the numbers divided by

1000 .

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 = 1.5 x^{2} - 2.5 x

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 = 1500 x^{2} - 2500 x

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.

Pop Quiz

Finding the Degree of a Polynomial Function

Indicate the degree of the polynomial function that best fits the data in each table of values.

Discussion

Polynomials That Perfectly Fit a Set of Points

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.

Rule

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.

Proof

Informal Justification

To justify this principle, the smaller cases will be examined to come to a conclusion.

$n = 0$

Given a single point, it is possible to fit it with a unique constant function, or a zero degree polynomial.

When considering polynomials of greater degree, starting from degree

1,

many different polynomials can include the point.

Therefore, the polynomials of greater degree that fit a single point are not unique.

$n = 1$

Now consider that

n = 1 .

In this case,

n + 1 = 2

points are given.

(x_{1}, y_{1}) (x_{2}, y_{2})

It is possible to write a system of equations by substituting these points into a linear function written in slope-intercept form.

{y_{1} = a x_{1} + b y_{2} = a x_{2} + b (I) (II)

In these functions,

a

is the slope of the line and

b

its

y -

intercept. Solving this system of equations gives the values of

a

and

b .

The value of

a

can be zero if the points lie on a horizontal line, which means that the data can be fit with a constant function.

Similarly to the previous case, there are infinitely many different polynomials of greater degree that fit these two points, making the polynomials not unique.

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} + \dots + a_{n - 1} x_{1} + a_{n} y_{2} = a_{1} x_{2}^{n} + a_{2} x_{2}^{n - 1} + \dots + a_{n - 1} x_{2} + a_{n} y_{2} = a_{1} x_{2}^{n} + a_{2} ⋮ y_{n} = a_{1} x_{n}^{n} + a_{2} x_{n}^{n - 1} + \dots + 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.

Example

Writing a Polynomial That Fits a Set of Points

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.

One minute after takeoff, the horizontal distance between the plane and the airport was $5$ kilometers.
Two minutes after takeoff, the plane was $9$ kilometers from the airport.
Three minutes after takeoff, the plane was $15$ kilometers from the airport.
Four minutes after takeoff, the plane was $24$ kilometers from the airport.

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.

a Write a polynomial function for the plane's distance from the airport after takeoff. Round the answer to two decimal places if needed.

b Use the polynomial found in Part A to estimate the distance between the plane and the airport

5

minutes after takeoff.

Hint

a Use the

(n + 1)

Point Principle.

b Substitute

5

for

x

in the function obtained in Part A.

Solution

a By the

(n + 1)

Point Principle, when given four points, there may exist a polynomial of degree at most

3

that fits the points.

y = a x^{3} + b x^{2} + c x + 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.

⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ 1 a + 1 b + 1 c + 1 d = 5 8 a + 4 b + 2 c + 1 d = 9 27 a + 9 b + 3 c + 1 d = 15 64 a + 16 b + 4 c + 1 d = 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

2 ND,

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 $4 \times 5$ 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 $2 ND$ 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 $2 ND$ 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.

⎣ ⎢ ⎢ ⎢ ⎡ 1000010000100001 0.166 \dots 0 2.833 \dots 2 ⎦ ⎥ ⎥ ⎥ ⎤ ⇕ ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ 1 a + 0 b + 0 c + 0 d = 0.166 \dots 0 a + 1 b + 0 c + 0 d = 0 0 a + 0 b + 1 c + 0 d = 2.833 \dots 0 a + 0 b + 0 c + 1 d = 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 \dots x^{3} + 0 x^{2} + 2.833 \dots x + 2 ⇓ y = 0.17 x^{3} + 2.83 x + 2

b Substituting

5

for

x

in the equation found in Part A results in an estimate for the requested distance.

y = 0.17 x^{3} + 2.83 x + 2

Substitute

$x = 5$

y = 0.17 (5)^{3} + 2.83 (5) + 2

CalcPow

Calculate power

y = 0.17 (125) + 2.83 (5) + 2

Multiply

y = 21.25 + 14.15 + 2

AddTerms

Add terms

y = 37.4

Therefore, the plane will be about

37.4

kilometers away from the airport

5

minutes after takeoff.

Example

Predicting Carbon Dioxide Levels

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.

Zosia looking through Carbon Dioxide Data

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.

a Use a calculator to find a polynomial of degree four that better fits the data in the table. Round every coefficient to four significant figures.

b Use the polynomial found in Part A to make a prediction for the year

2025 .

Is this prediction reliable? Round to four significant figures.

Answer

C (t) = 6.667 \times 1 0^{- 5} t^{4} - 0.003 t^{3}

+ 0.063 t^{2}

+ 1.538 t

+ 360.0

435.8

parts per million.

Hint

a Use the function QuartReg on a graphing calculator.

b What does the value of

R^{2}

indicate?

Solution

a The first thing that is needed is to input the data into the graphing calculator. To do so, press

STAT,

choose Edit, and enter the values in the first two columns.

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

1 0^{- 5} .

C (t) = 6.667 \times 1 0^{- 5} t^{4} - 0.003 t^{3} + 0.063 t^{2} - 1.538 t + 360.0

b To make the prediction for the carbon dioxide levels in

2025,

this value has to be substituted into the function found in Part A. However, since the values are considered from

1995

0,

the value that corresponds to

2025

2025 - 1995 = 30 .

C (t) = 6.667 \times 1 0^{- 5} t^{4} - 0.003 t^{3} + 0.063 t^{2} + 1.538 t + 360.0

Substitute

$t = 30$

C (30) = 6.667 \times 1 0^{- 5} (30)^{4} - 0.003 (30)^{3} + 0.063 (30)^{2} + 1.538 (30) + 360

▼

Simplify

CalcPow

Calculate power

C (30) = 6.667 \times 1 0^{- 5} (810000) - 0.003 (27000) + 0.063 (900) + 1.538 (30) + 360

Multiply

C (30) = 54.0027 - 81 + 56.7 + 46.14 + 360

AddTerms

Add terms

C (30) = 435.8427

Therefore, the carbon dioxide levels in

2025

will be about

435.8

parts per million. Scrolling down on the last screen in Part A, it is possible to find the value of

R^{2} .

R^{2}

does not appear, press

2ND,

press

0,

then select the DiagnosticOn and 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.

Closure

Using Polynomials to Fit the Given Data

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) = a x^{3} + b x^{2} + c x + 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 = 0.33 x^{3} - 0.5 x^{2} + 0.17 x + 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.

Level 1

Level 2

Level 3

Applying Polynomial Functions to Real-World Situations

Exercises

Find the second differences of the following data set.

$x$	$2$	$4$	$6$	$8$	$10$
$f (x)$	$8$	$76$	$256$	$596$	$1144$

Consider the following set of points.

(3, 169), (11, 161305), (5, 3001), (1, - 15), (9, 59041), (7, 16689)

Find the third differences of the

y -

values.

Notice that the x-values in the table are equally spaced. To find the second differences, we begin by finding the first differences. We do this by subtracting each pair of consecutive function values.

Now that we know the first differences, we can proceed to finding the second differences. To do this, we subtract each pair of consecutive first differences.

As we can see, the second differences are 112, 160, and 208.

We will follow the same procedure as we did in Part A. However, notice that the given points are not in order. Let's first organize the points so that the x-values are in increasing order. (1,-15), (3,169), (5,3001), (7,16 689) (9,59 041), (11,161 305) With the points organized, we can see that the x-values are equally spaced. Now let's begin by finding the first differences. As before, we will subtract each pair of consecutive y-values.

We are ready to find the second differences by subtracting the values we found for the first differences.

Repeating the same process one more time will give us the third differences.

The third differences for the given set of points are 8208, 17 808, and 31 248.

Indicate the degree of the polynomial function that best fits each of the following data sets.

$x$	$3$	$4$	$5$	$6$	$7$	$8$
$g (x)$	$- 11$	$- 35$	$- 79$	$- 149$	$- 251$	$- 391$

{(5, 11), (4, 0), (3, - 9) (6, 24), (2, - 16)}

According to the Properties of Finite Differences, if the nth differences of equally spaced values are constant and different from zero, then the values can be represented by a polynomial function of degree n. Let's begin by finding the first differences of the given function values.

As we can see, the first differences are not constant — the values are not equal. This means that a linear polynomial does not fit the data well. Let's continue and find the second differences.

Once more, we see that the second differences are not constant either. This means that a quadratic polynomial is not the best fit. Let's find the third differences.

This time, all the third differences are equal and different from zero — they are all equal to -6. This implies that the degree of polynomial function that best fits the data set is 3.

As in Part A, we will find the finite differences of the data set to determine the degree of the required polynomial. However, notice that the given points are not ordered. Therefore, let's organize them so that the x-values are in increasing order. (2,-16),(3,-9),(4,0),(5,11),(6,24) We can see that the x-values are equally spaced. Let's continue our process by finding the first differences. To do this, we subtract each pair of consecutive y-values.

Since the first differences are not constant, let's continue to find the second differences.

As we can see, the second differences are constant and different from zero. Therefore, the degree of the polynomial function that best fits the given data set is 2.

Consider the following statement.

If the $n th$ differences of equally spaced values are and zero, then the values can be represented by a polynomial function of degree $n .$

Which of the following combinations completes the statement correctly?

When we want to determine the degree of the polynomial function that best fits a set of values, we study the finite differences of y-values. In fact, if the nth differences of equally space values are constant and different from 0, the values can be represented by a polynomial function of degree n.

If the nth differences of equally spaced values are constant and different from zero, then the values can be represented by a polynomial function of degree n.

Consequently, the missing parts of the given statement are constant and different from.

Davontay is interested in finding a cubic polynomial that passes through the points $(3, 0),$ $(- 2, 0),$ $(1, 0),$ and $(4, 72) .$ In the process, he first wrote a factored function and then performed a certain substitution to find the value of the leading coefficient.

f(x) = a(x+3)(x-2)(x-1), Substitution: 72 = a(4+3)(4-2)(4-1); solution: a=7/12; cubic function: f(x) = 7/12(x+3)(x-2)(x-1)

However, Davontay's teacher said that he made a mistake.

What mistake did Davontay make?

Write the function rule, in factored form, that Davontay should get once he corrects the error.

Let's check Davontay's computations carefully to determine where he made a mistake. As we can see, he first wrote the points through which the cubic function passes. (3,0), (-2,0), (1,0), (4,72) A general cubic polynomial has the form ax^3+bx^2+cx+d. Since three of the given points are x-intercepts, we can use the factored form of a cubic polynomial, which is what Davontay did. ( 3,0), ( -2,0), ( 1,0) [0.75em] f(x) = a(x + 3)(x - 2)(x + 1) However, the factor corresponding to an x-intercept of the form (k,0) is x-k. In other words, to form the factor, the sign of the x-intercept is changed. This is where Davontay made his mistake — he forgot to change the signs. ( 3,0), ( -2,0), ( 1,0) [0.75em] f(x) = a(x + 3)(x - 2)(x + 1) * [0.75em] f(x) = a(x- 3)(x+ 2)(x- 1) ✓ Therefore, Davontay's mistake is that the sign on each factor is wrong.

To write the correct function rule, we can follow Davontay's steps but correct his mistake. Let's write the points through which the polynomial passes. (3,0), (-2,0), (1,0), (4,72) Since three of the given points are x-intercepts, we can use the factored form of a cubic polynomial. f(x) = a(x-k_1)(x-k_2)(x-k_3) Here, k_1, k_2, and k_3 are the x-intercepts. Let's substitute the corresponding values.

Our next step is to find the value of a. To do so, we can use the fact that the polynomial also passes through ( 4, 72), which implies that f( 4)= 72. Let's evaluate f(x) at x= 4.

We got that the value of a is 4. We are ready to write the required cubic polynomial. f(x) = 4(x-3)(x+2)(x-1)