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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.**

Consider the following table of values.

$x$ | $f(x)$ |
---|---|

$1$ | $2$ |

$2$ | $3$ |

$3$ | $7$ |

$4$ | $16$ |

$5$ | $32$ |

$6$ | $57$ |

$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.

- 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.

Then the teacher asked the students to write the cubic function graphed. Help Zosia write the expression for the cubic function.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":false,"useShortLog":false,"variables":["x"],"constants":[]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:1em;vertical-align:-0.25em;\"><\/span><span class=\"mord mathdefault\" style=\"margin-right:0.10764em;\">f<\/span><span class=\"mopen\">(<\/span><span class=\"mord mathdefault\">x<\/span><span class=\"mclose\">)<\/span><span class=\"mspace\" style=\"margin-right:0.2777777777777778em;\"><\/span><span class=\"mrel\">=<\/span><\/span><\/span><\/span>","formTextAfter":null,"answer":{"text":["\\dfrac{1}{6}(x-7)(x+3)(x-2)","\\dfrac{1}{6}(x^3-6x^2-13x+42)","\\dfrac{1}{6}x^3-x^2-\\dfrac{13}{6}x+7"]}}

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

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.
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.

$(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

Multiply

$7=42a$

DivEqn

$LHS/42=RHS/42$

$427 =a$

SimpQuot

Simplify quotient

$61 =a$

RearrangeEqn

Rearrange equation

$a=61 $

$f(x)=61 (x−7)(x+3)(x−2) $

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.

- 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.

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.

- For a polynomial function of degree $n,$ the $nth$ differences of the function values corresponding to equally spaced values of the independent variable are constant and different from zero.
- 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 $n.$

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

$f_{1}(x)=ax+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.
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. $f_{2}(x)=ax_{2}+bx+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

$a(x+h)_{2}+b(x+h)+c−(ax_{2}+bx+c)$

Simplify

Distr

Distribute $-1$

$a(x+h)_{2}+b(x+h)+c−ax_{2}−bx−c$

ExpandPosPerfectSquare

$(a+b)_{2}=a_{2}+2ab+b_{2}$

$a(x_{2}+2xh+h_{2})+b(x+h)+c−ax_{2}−bx−c$

Distr

Distribute $a$

$ax_{2}+2axh+ah_{2}+b(x+h)+c−ax_{2}−bx−c$

Distr

Distribute $b$

$ax_{2}+2axh+ah_{2}+bx+bh+c−ax_{2}−bx−c$

SubTerms

Subtract terms

$2axh+ah_{2}+bh$

$f_{2}(x+h)−f_{2}(x)f_{2}(x+2h)−f_{2}(x+h) $

A second difference is obtained by subtracting the first expression from the second one.
$f_{2}(x+2h)−f_{2}(x+h)−(f_{2}(x+h)−f_{2}(x))$

Distr

Distribute $-1$

$f_{2}(x+2h)−f_{2}(x+h)−f_{2}(x+h)+f_{2}(x)$

SubTerms

Subtract terms

$f_{2}(x+2h)−2f_{2}(x+h)+f_{2}(x)$

SubstituteExpressions

Substitute expressions

$a(x+2h)_{2}+b(x+2h)+c−2(a(x+h)_{2}+b(x+h)+c)+ax_{2}+bx+c$

Simplify

ExpandPosPerfectSquare

$(a+b)_{2}=a_{2}+2ab+b_{2}$

$a(x_{2}+4xh+4h_{2})+b(x+2h)+c−2(a(x_{2}+2xh+h_{2})+b(x+h)+c)+ax_{2}+bx+c$

Distr

Distribute $a$

$ax_{2}+4axh+4ah_{2}+b(x+2h)+c−2(ax_{2}+2axh+ah_{2}+b(x+h)+c)+ax_{2}+bx+c$

Distr

Distribute $b$

$ax_{2}+4axh+4ah_{2}+bx+2bh+c−2(ax_{2}+2axh+ah_{2}+bx+bh+c)+ax_{2}+bx+c$

Distr

Distribute $-2$

$ax_{2}+4axh+4ah_{2}+bx+2bh+c−2ax_{2}−4axh−2ah_{2}−2bx−2bh−2c+ax_{2}+bx+c$

SubTerms

Subtract terms

$2ah_{2}$

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$ | $$14000$ | $$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.

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b Let $x$ be the year. Write the polynomial function.

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a Recall the Properties of Finite Differences.

b Use the corresponding regression function on a graphing calculator.

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.

$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.

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

$(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}=ax_{1}+by_{2}=ax_{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

$⎩⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎧ y_{1}=a_{1}x_{1}+a_{2}x_{1}+⋯+a_{n−1}x_{1}+a_{n}y_{2}=a_{1}x_{2}+a_{2}x_{2}+⋯+a_{n−1}x_{2}+a_{n}y_{2}=a_{1}x_{2}+a_{2}⋮y_{n}=a_{1}x_{n}+a_{2}x_{n}+⋯+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.

- 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.

$(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.

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b Use the polynomial found in Part A to estimate the distance between the plane and the airport $5$ minutes after takeoff.

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a Use the $(n+1)$ Point Principle.

b Substitute $5$ for $x$ in the function obtained in Part A.

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=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)+d9=a(2)_{3}+b(2)_{2}+c(2)+d15=a(3)_{3}+b(3)_{2}+c(3)+d24=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=58a+4b+2c+1d=927a+9b+3c+1d=1564a+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 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 $[A].$

$⎣⎢⎢⎢⎡ 1000 0100 0010 0001 0.166…02.833…2 ⎦⎥⎥⎥⎤ ⇕⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧ 1a+0b+0c+0d=0.166…0a+1b+0c+0d=00a+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 $

b Substituting $5$ for $x$ in the equation found in Part A results in an estimate for the requested distance.

$y=0.17x_{3}+2.83x+2$

Substitute

$x=5$

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

CalcPow

Calculate power

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

Multiply

Multiply

$y=21.25+14.15+2$

AddTerms

Add terms

$y=37.4$

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.

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.

a $C(t)=6.667×10_{-5}t_{4}−0.003t_{3}$ $+0.063t_{2}$ $+1.538t$ $+360.0$

b $435.8$ parts per million.

a Use the function

QuartRegon a graphing calculator.

b What does the value of $R_{2}$ indicate?

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.

$C(t)=6.667×10_{-5}t_{4}−0.003t_{3}+0.063t_{2}−1.538t+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 $1995$ as $0,$ the value that corresponds to $2025$ is $2025−1995=30.$

$C(t)=6.667×10_{-5}t_{4}−0.003t_{3}+0.063t_{2}+1.538t+360.0$

Substitute

$t=30$

$C(30)=6.667×10_{-5}(30)_{4}−0.003(30)_{3}+0.063(30)_{2}+1.538(30)+360$

$C(30)=435.8427$

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.

$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.