assignment 3 graphs of algebraic sentences

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Unit 4: Graphing lines and slope

About this unit.

Use the power of algebra to understand and interpret points and lines (something we typically do in geometry). Topics you'll explore include the slope and the equation of a line.

Coordinate plane

• Quadrants of the coordinate plane (Opens a modal)
• Points on the coordinate plane (Opens a modal)
• Points and quadrants example (Opens a modal)
• Quadrants on the coordinate plane Get 5 of 7 questions to level up!

Solutions to two-variable linear equations

• Two-variable linear equations intro (Opens a modal)
• Solutions to 2-variable equations (Opens a modal)
• Worked example: solutions to 2-variable equations (Opens a modal)
• Completing solutions to 2-variable equations (Opens a modal)
• Solutions to 2-variable equations Get 3 of 4 questions to level up!
• Complete solutions to 2-variable equations Get 3 of 4 questions to level up!

x-intercepts and y-intercepts

• Intro to intercepts (Opens a modal)
• x-intercept of a line (Opens a modal)
• Intercepts from an equation (Opens a modal)
• Intercepts from a graph Get 3 of 4 questions to level up!
• Intercepts from an equation Get 3 of 4 questions to level up!
• Intro to slope (Opens a modal)
• Positive & negative slope (Opens a modal)
• Worked example: slope from graph (Opens a modal)
• Worked example: slope from two points (Opens a modal)
• Slope (more examples) (Opens a modal)
• Slope review (Opens a modal)
• Slope from graph Get 3 of 4 questions to level up!
• Slope from two points Get 3 of 4 questions to level up!

Horizontal & vertical lines

• Slope of a horizontal line (Opens a modal)
• Horizontal & vertical lines Get 5 of 7 questions to level up!

Slope-intercept form intro

• Intro to slope-intercept form (Opens a modal)
• Graph from slope-intercept equation (Opens a modal)
• Graphing slope-intercept form (Opens a modal)
• Graphing lines from slope-intercept form review (Opens a modal)
• Slope-intercept intro Get 3 of 4 questions to level up!
• Graph from slope-intercept form Get 3 of 4 questions to level up!

Writing slope-intercept equations

• Slope-intercept equation from graph (Opens a modal)
• Writing slope-intercept equations (Opens a modal)
• Slope-intercept equation from slope & point (Opens a modal)
• Slope-intercept equation from two points (Opens a modal)
• Slope-intercept form problems (Opens a modal)
• Slope-intercept form review (Opens a modal)
• Slope-intercept equation from graph Get 3 of 4 questions to level up!
• Slope-intercept from two points Get 3 of 4 questions to level up!

Graphing two-variable inequalities

• Intro to graphing two-variable inequalities (Opens a modal)
• Graphing two-variable inequalities (Opens a modal)
• Graphs of inequalities Get 3 of 4 questions to level up!

2.3 Models and Applications

Learning objectives.

In this section, you will:

• Set up a linear equation to solve a real-world application.
• Use a formula to solve a real-world application.

Josh is hoping to get an A in his college algebra class. He has scores of 75, 82, 95, 91, and 94 on his first five tests. Only the final exam remains, and the maximum of points that can be earned is 100. Is it possible for Josh to end the course with an A? A simple linear equation will give Josh his answer.

Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talk-time; it costs a widget manufacturer a certain amount to produce x widgets per month plus monthly operating charges; a car rental company charges a daily fee plus an amount per mile driven. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems.

Setting up a Linear Equation to Solve a Real-World Application

To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $0.10/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write 0.10 x . 0.10 x . This expression represents a variable cost because it changes according to the number of miles driven.

If a quantity is independent of a variable, we usually just add or subtract it, according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10/mi plus a daily fee of $50. We can use these quantities to model an equation that can be used to find the daily car rental cost C . C .

When dealing with real-world applications, there are certain expressions that we can translate directly into math. Table 1 lists some common verbal expressions and their equivalent mathematical expressions.

Given a real-world problem, model a linear equation to fit it.

• Identify known quantities.
• Assign a variable to represent the unknown quantity.
• If there is more than one unknown quantity, find a way to write the second unknown in terms of the first.
• Write an equation interpreting the words as mathematical operations.
• Solve the equation. Be sure the solution can be explained in words, including the units of measure.

Modeling a Linear Equation to Solve an Unknown Number Problem

Find a linear equation to solve for the following unknown quantities: One number exceeds another number by 17 17 and their sum is 31. 31. Find the two numbers.

Let x x equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as x + 17. x + 17. The sum of the two numbers is 31. We usually interpret the word is as an equal sign.

The two numbers are 7 7 and 24. 24.

Find a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is 36 , 36 , find the numbers.

Setting Up a Linear Equation to Solve a Real-World Application

There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus $.05/min talk-time. Company B charges a monthly service fee of $40 plus $.04/min talk-time.

• ⓐ Write a linear equation that models the packages offered by both companies.
• ⓑ If the average number of minutes used each month is 1,160, which company offers the better plan?
• ⓒ If the average number of minutes used each month is 420, which company offers the better plan?
• ⓓ How many minutes of talk-time would yield equal monthly statements from both companies?
• ⓐ The model for Company A can be written as A = 0.05 x + 34. A = 0.05 x + 34. This includes the variable cost of 0.05 x 0.05 x plus the monthly service charge of $34. Company B ’s package charges a higher monthly fee of $40, but a lower variable cost of 0.04 x . 0.04 x . Company B ’s model can be written as B = 0.04 x + $ 40. B = 0.04 x + $ 40.

If the average number of minutes used each month is 1,160, we have the following:

So, Company B offers the lower monthly cost of $86.40 as compared with the $92 monthly cost offered by Company A when the average number of minutes used each month is 1,160.

If the average number of minutes used each month is 420, we have the following:

If the average number of minutes used each month is 420, then Company A offers a lower monthly cost of $55 compared to Company B ’s monthly cost of $56.80.

To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of ( x , y ) ( x , y ) coordinates: At what point are both the x- value and the y- value equal? We can find this point by setting the equations equal to each other and solving for x.

Check the x- value in each equation.

Therefore, a monthly average of 600 talk-time minutes renders the plans equal. See Figure 2

Find a linear equation to model this real-world application: It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of $350 for utilities and $3,300 for salaries. What are the company’s monthly expenses?

Using a Formula to Solve a Real-World Application

Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem’s question is answered. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the area of a rectangular region, A = L W ; A = L W ; the perimeter of a rectangle, P = 2 L + 2 W ; P = 2 L + 2 W ; and the volume of a rectangular solid, V = L W H . V = L W H . When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time.

Solving an Application Using a Formula

It takes Andrew 30 min to drive to work in the morning. He drives home using the same route, but it takes 10 min longer, and he averages 10 mi/h less than in the morning. How far does Andrew drive to work?

This is a distance problem, so we can use the formula d = r t , d = r t , where distance equals rate multiplied by time. Note that when rate is given in mi/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution.

First, we identify the known and unknown quantities. Andrew’s morning drive to work takes 30 min, or 1 2 1 2 h at rate r . r . His drive home takes 40 min, or 2 3 2 3 h, and his speed averages 10 mi/h less than the morning drive. Both trips cover distance d . d . A table, such as Table 2 , is often helpful for keeping track of information in these types of problems.

Write two equations, one for each trip.

As both equations equal the same distance, we set them equal to each other and solve for r .

We have solved for the rate of speed to work, 40 mph. Substituting 40 into the rate on the return trip yields 30 mi/h. Now we can answer the question. Substitute the rate back into either equation and solve for d.

The distance between home and work is 20 mi.

Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for r . r .

On Saturday morning, it took Jennifer 3.6 h to drive to her mother’s house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer 4 h to return home. Her speed was 5 mi/h slower on Sunday than on Saturday. What was her speed on Sunday?

Solving a Perimeter Problem

The perimeter of a rectangular outdoor patio is 54 54 ft. The length is 3 3 ft greater than the width. What are the dimensions of the patio?

The perimeter formula is standard: P = 2 L + 2 W . P = 2 L + 2 W . We have two unknown quantities, length and width. However, we can write the length in terms of the width as L = W + 3. L = W + 3. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as in Figure 3 .

Now we can solve for the width and then calculate the length.

The dimensions are L = 15 L = 15 ft and W = 12 W = 12 ft.

Find the dimensions of a rectangle given that the perimeter is 110 110 cm and the length is 1 cm more than twice the width.

Solving an Area Problem

The perimeter of a tablet of graph paper is 48 in. The length is 6 6 in. more than the width. Find the area of the graph paper.

The standard formula for area is A = L W ; A = L W ; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one.

We know that the length is 6 in. more than the width, so we can write length as L = W + 6. L = W + 6. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length.

Now, we find the area given the dimensions of L = 15 L = 15 in. and W = 9 W = 9 in.

The area is 135 135 in. 2 .

A game room has a perimeter of 70 ft. The length is five more than twice the width. How many ft 2 of new carpeting should be ordered?

Solving a Volume Problem

Find the dimensions of a shipping box given that the length is twice the width, the height is 8 8 inches, and the volume is 1,600 in. 3 .

The formula for the volume of a box is given as V = L W H , V = L W H , the product of length, width, and height. We are given that L = 2 W , L = 2 W , and H = 8. H = 8. The volume is 1,600 1,600 cubic inches.

The dimensions are L = 20 L = 20 in., W = 10 W = 10 in., and H = 8 H = 8 in.

Note that the square root of W 2 W 2 would result in a positive and a negative value. However, because we are describing width, we can use only the positive result.

Access these online resources for additional instruction and practice with models and applications of linear equations.

• Problem solving using linear equations
• Problem solving using equations
• Finding the dimensions of area given the perimeter
• Find the distance between the cities using the distance = rate * time formula
• Linear equation application (Write a cost equation)

2.3 Section Exercises

To set up a model linear equation to fit real-world applications, what should always be the first step?

Use your own words to describe this equation where n is a number: 5 ( n + 3 ) = 2 n 5 ( n + 3 ) = 2 n

If the total amount of money you had to invest was $2,000 and you deposit x x amount in one investment, how can you represent the remaining amount?

If a man sawed a 10-ft board into two sections and one section was n n ft long, how long would the other section be in terms of n n ?

If Bill was traveling v v mi/h, how would you represent Daemon’s speed if he was traveling 10 mi/h faster?

Real-World Applications

For the following exercises, use the information to find a linear algebraic equation model to use to answer the question being asked.

Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?

Beth and Ann are joking that their combined ages equal Sam’s age. If Beth is twice Ann’s age and Sam is 69 yr old, what are Beth and Ann’s ages?

Ben originally filled out 8 more applications than Henry. Then each boy filled out 3 additional applications, bringing the total to 28. How many applications did each boy originally fill out?

For the following exercises, use this scenario: Two different telephone carriers offer the following plans that a person is considering. Company A has a monthly fee of $20 and charges of $.05/min for calls. Company B has a monthly fee of $5 and charges $.10/min for calls.

Find the model of the total cost of Company A’s plan, using m m for the minutes.

Find the model of the total cost of Company B’s plan, using m m for the minutes.

Find out how many minutes of calling would make the two plans equal.

If the person makes a monthly average of 200 min of calls, which plan should for the person choose?

For the following exercises, use this scenario: A wireless carrier offers the following plans that a person is considering. The Family Plan: $90 monthly fee, unlimited talk and text on up to 8 lines, and data charges of $40 for each device for up to 2 GB of data per device. The Mobile Share Plan: $120 monthly fee for up to 10 devices, unlimited talk and text for all the lines, and data charges of $35 for each device up to a shared total of 10 GB of data. Use P P for the number of devices that need data plans as part of their cost.

Find the model of the total cost of the Family Plan.

Find the model of the total cost of the Mobile Share Plan.

Assuming they stay under their data limit, find the number of devices that would make the two plans equal in cost.

If a family has 3 smart phones, which plan should they choose?

For exercises 17 and 18, use this scenario: A retired woman has $50,000 to invest but needs to make $6,000 a year from the interest to meet certain living expenses. One bond investment pays 15% annual interest. The rest of it she wants to put in a CD that pays 7%.

If we let x x be the amount the woman invests in the 15% bond, how much will she be able to invest in the CD?

Set up and solve the equation for how much the woman should invest in each option to sustain a $6,000 annual return.

Two planes fly in opposite directions. One travels 450 mi/h and the other 550 mi/h. How long will it take before they are 4,000 mi apart?

Ben starts walking along a path at 4 mi/h. One and a half hours after Ben leaves, his sister Amanda begins jogging along the same path at 6 mi/h. How long will it be before Amanda catches up to Ben?

Fiora starts riding her bike at 20 mi/h. After a while, she slows down to 12 mi/h, and maintains that speed for the rest of the trip. The whole trip of 70 mi takes her 4.5 h. For what distance did she travel at 20 mi/h?

A chemistry teacher needs to mix a 30% salt solution with a 70% salt solution to make 20 qt of a 40% salt solution. How many quarts of each solution should the teacher mix to get the desired result?

Paul has $20,000 to invest. His intent is to earn 11% interest on his investment. He can invest part of his money at 8% interest and part at 12% interest. How much does Paul need to invest in each option to make get a total 11% return on his $20,000?

For the following exercises, use this scenario: A truck rental agency offers two kinds of plans. Plan A charges $75/wk plus $.10/mi driven. Plan B charges $100/wk plus $.05/mi driven.

Write the model equation for the cost of renting a truck with plan A.

Write the model equation for the cost of renting a truck with plan B.

Find the number of miles that would generate the same cost for both plans.

If Tim knows he has to travel 300 mi, which plan should he choose?

For the following exercises, use the formula given to solve for the required value.

A = P ( 1 + r t ) A = P ( 1 + r t ) is used to find the principal amount P deposited, earning r % interest, for t years. Use this to find what principal amount P David invested at a 3% rate for 20 yr if A = $ 8,000. A = $ 8,000.

The formula F = m v 2 R F = m v 2 R relates force ( F ) ( F ) , velocity ( v ) ( v ) , mass , and resistance ( m ) ( m ) . Find R R when m = 45 , m = 45 , v = 7 , v = 7 , and F = 245. F = 245.

F = m a F = m a indicates that force ( F ) equals mass ( m ) times acceleration ( a ). Find the acceleration of a mass of 50 kg if a force of 12 N is exerted on it.

S u m = 1 1 − r S u m = 1 1 − r is the formula for an infinite series sum. If the sum is 5, find r . r .

For the following exercises, solve for the given variable in the formula. After obtaining a new version of the formula, you will use it to solve a question.

Solve for W : P = 2 L + 2 W P = 2 L + 2 W

Use the formula from the previous question to find the width, W , W , of a rectangle whose length is 15 and whose perimeter is 58.

Solve for f : 1 p + 1 q = 1 f f : 1 p + 1 q = 1 f

Use the formula from the previous question to find f f when p = 8 and  q = 13. p = 8 and  q = 13.

Solve for m m in the slope-intercept formula: y = m x + b y = m x + b

Use the formula from the previous question to find m m when the coordinates of the point are ( 4 , 7 ) ( 4 , 7 ) and b = 12. b = 12.

The area of a trapezoid is given by A = 1 2 h ( b 1 + b 2 ) . A = 1 2 h ( b 1 + b 2 ) . Use the formula to find the area of a trapezoid with h = 6 , b 1 = 14 , and  b 2 = 8. h = 6 , b 1 = 14 , and  b 2 = 8.

Solve for h: A = 1 2 h ( b 1 + b 2 ) A = 1 2 h ( b 1 + b 2 )

Use the formula from the previous question to find the height of a trapezoid with A = 150 , b 1 = 19 A = 150 , b 1 = 19 , and  b 2 = 11. b 2 = 11.

Find the dimensions of an American football field. The length is 200 ft more than the width, and the perimeter is 1,040 ft. Find the length and width. Use the perimeter formula P = 2 L + 2 W . P = 2 L + 2 W .

Distance equals rate times time, d = r t . d = r t . Find the distance Tom travels if he is moving at a rate of 55 mi/h for 3.5 h.

Using the formula in the previous exercise, find the distance that Susan travels if she is moving at a rate of 60 mi/h for 6.75 h.

What is the total distance that two people travel in 3 h if one of them is riding a bike at 15 mi/h and the other is walking at 3 mi/h?

If the area model for a triangle is A = 1 2 b h , A = 1 2 b h , find the area of a triangle with a height of 16 in. and a base of 11 in.

Solve for h: A = 1 2 b h A = 1 2 b h

Use the formula from the previous question to find the height to the nearest tenth of a triangle with a base of 15 and an area of 215.

The volume formula for a cylinder is V = π r 2 h . V = π r 2 h . Using the symbol π π in your answer, find the volume of a cylinder with a radius, r , r , of 4 cm and a height of 14 cm.

Solve for h: V = π r 2 h V = π r 2 h

Use the formula from the previous question to find the height of a cylinder with a radius of 8 and a volume of 16 π 16 π

Solve for r: V = π r 2 h V = π r 2 h

Use the formula from the previous question to find the radius of a cylinder with a height of 36 and a volume of 324 π . 324 π .

The formula for the circumference of a circle is C = 2 π r . C = 2 π r . Find the circumference of a circle with a diameter of 12 in. (diameter = 2 r ). Use the symbol π π in your final answer.

Solve the formula from the previous question for π . π . Notice why π π is sometimes defined as the ratio of the circumference to its diameter.

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• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: College Algebra
• Publication date: Feb 13, 2015
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• Book URL: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites
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Chapter 5 Polynomial and Rational Functions

5.3 Graphs of Polynomial Functions

Learning objectives.

In this section, you will:

• Recognize characteristics of graphs of polynomial functions.
• Use factoring to find zeros of polynomial functions.
• Identify zeros and their multiplicities.
• Determine end behavior.
• Understand the relationship between degree and turning points.
• Graph polynomial functions.
• Use the Intermediate Value Theorem.

The revenue in millions of dollars for a fictional cable company from 2006 through 2013 is shown in Table 1 .

The revenue can be modeled by the polynomial function

where[latex]\,R\,[/latex]represents the revenue in millions of dollars and[latex]\,t\,[/latex]represents the year, with[latex]\,t=6\,[/latex]corresponding to 2006. Over which intervals is the revenue for the company increasing? Over which intervals is the revenue for the company decreasing? These questions, along with many others, can be answered by examining the graph of the polynomial function. We have already explored the local behavior of quadratics, a special case of polynomials. In this section we will explore the local behavior of polynomials in general.

Recognizing Characteristics of Graphs of Polynomial Functions

Polynomial functions of degree 2 or more have graphs that do not have sharp corners; recall that these types of graphs are called smooth curves. Polynomial functions also display graphs that have no breaks. Curves with no breaks are called continuous. Figure 1 shows a graph that represents a polynomial function and a graph that represents a function that is not a polynomial.

Recognizing Polynomial Functions

Which of the graphs in Figure 2 represents a polynomial function?

The graphs of[latex]\,f\,[/latex]and[latex]\,h\,[/latex]are graphs of polynomial functions. They are smooth and continuous .

The graphs of[latex]\,g\,[/latex]and[latex]\,k\,[/latex]are graphs of functions that are not polynomials. The graph of function[latex]\,g\,[/latex]has a sharp corner. The graph of function[latex]\,k\,[/latex]is not continuous.

Do all polynomial functions have all real numbers as their domain?

Yes. Any real number is a valid input for a polynomial function.

Using Factoring to Find Zeros of Polynomial Functions

Recall that if[latex]\,f\,[/latex]is a polynomial function, the values of[latex]\,x\,[/latex]for which[latex]\,f\left(x\right)=0\,[/latex]are called zeros of[latex]\,f.\,[/latex]If the equation of the polynomial function can be factored, we can set each factor equal to zero and solve for the zeros .

We can use this method to find[latex]\,x\text{-}[/latex]intercepts because at the[latex]\,x\text{-}[/latex]intercepts we find the input values when the output value is zero. For general polynomials, this can be a challenging prospect. While quadratics can be solved using the relatively simple quadratic formula, the corresponding formulas for cubic and fourth-degree polynomials are not simple enough to remember, and formulas do not exist for general higher-degree polynomials. Consequently, we will limit ourselves to three cases:

• The polynomial can be factored using known methods: greatest common factor and trinomial factoring.
• The polynomial is given in factored form.
• Technology is used to determine the intercepts.

Given a polynomial function[latex]\,f,\,[/latex]find the x -intercepts by factoring.

• Set[latex]\,f\left(x\right)=0.\,[/latex]
• Factor out any common monomial factors.
• Factor any factorable binomials or trinomials.
• Set each factor equal to zero and solve to find the[latex]\,x\text{-}[/latex]intercepts.

Finding the x -Intercepts of a Polynomial Function by Factoring

Find the x -intercepts of[latex]\,f\left(x\right)={x}^{6}-3{x}^{4}+2{x}^{2}.[/latex]

We can attempt to factor this polynomial to find solutions for [latex]\,f\left(x\right)=0.[/latex]

This gives us five x -intercepts:[latex]\,\left(0,0\right),\left(1,0\right),\left(-1,0\right),\left(\sqrt{2},0\right),\,[/latex]and[latex]\,\left(-\sqrt{2},0\right).\,[/latex]See Figure 3. We can see that this is an even function because it is symmetric about the y -axis.

Find the x -intercepts of[latex]\,f\left(x\right)={x}^{3}-5{x}^{2}-x+5.[/latex]

Find solutions for[latex]\,f\left(x\right)=0\,[/latex] by factoring.

There are three x -intercepts:[latex]\,\left(-1,0\right),\left(1,0\right),\,[/latex]and[latex]\,\left(5,0\right).\,[/latex]See Figure 4.

Finding the y – and x -Intercepts of a Polynomial in Factored Form

Find the y – and x -intercepts of[latex]\,g\left(x\right)={\left(x-2\right)}^{2}\left(2x+3\right).[/latex]

The y -intercept can be found by evaluating[latex]\,g\left(0\right).[/latex]

So the y -intercept is[latex]\,\left(0,12\right).[/latex]

The x -intercepts can be found by solving[latex]\,g\left(x\right)=0.[/latex]

So the x -intercepts are[latex]\,\left(2,0\right)\,[/latex]and[latex]\,\left(-\frac{3}{2},0\right).[/latex]

We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial as shown in Figure 5.

Finding the x -Intercepts of a Polynomial Function Using a Graph

Find the x -intercepts of[latex]\,h\left(x\right)={x}^{3}+4{x}^{2}+x-6.[/latex]

This polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities.

Looking at the graph of this function, as shown in Figure 6, it appears that there are x -intercepts at[latex]\,x=-3,-2,\,[/latex]and[latex]\,1.[/latex]

We can check whether these are correct by substituting these values for[latex]\,x\,[/latex]and verifying that

Since[latex]\,h\left(x\right)={x}^{3}+4{x}^{2}+x-6,\,[/latex]we have:

Each x -intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now write the polynomial in factored form.

Find the y – and x -intercepts of the function[latex]\,f\left(x\right)={x}^{4}-19{x}^{2}+30x.[/latex]

y -intercept[latex]\,\left(0,0\right);\,[/latex] x -intercepts[latex]\,\left(0,0\right),\left(–5,0\right),\left(2,0\right),\,[/latex]and[latex]\,\left(3,0\right)[/latex]

Identifying Zeros and Their Multiplicities

Graphs behave differently at various x -intercepts. Sometimes, the graph will cross over the horizontal axis at an intercept. Other times, the graph will touch the horizontal axis and “bounce” off.

Suppose, for example, we graph the function shown.

Notice in Figure 7 that the behavior of the function at each of the x -intercepts is different.

The x -intercept[latex]\,x=-3\,[/latex] is the solution of equation[latex]\,\left(x+3\right)=0.\,[/latex]The graph passes directly through the x -intercept at[latex]\,x=-3.\,[/latex]The factor is linear (has a degree of 1), so the behavior near the intercept is like that of a line—it passes directly through the intercept. We call this a single zero because the zero corresponds to a single factor of the function.

The x -intercept[latex]\,x=2\,[/latex]is the repeated solution of equation[latex]\,{\left(x-2\right)}^{2}=0.\,[/latex]The graph touches the axis at the intercept and changes direction. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept.

The factor is repeated, that is, the factor[latex]\,\left(x-2\right)\,[/latex]appears twice. The number of times a given factor appears in the factored form of the equation of a polynomial is called the multiplicity. The zero associated with this factor,[latex]\,x=2,\,[/latex]has multiplicity 2 because the factor[latex]\,\left(x-2\right)\,[/latex]occurs twice.

The x -intercept[latex]\,x=-1\,[/latex]is the repeated solution of factor[latex]\,{\left(x+1\right)}^{3}=0.\,[/latex]The graph passes through the axis at the intercept, but flattens out a bit first. This factor is cubic (degree 3), so the behavior near the intercept is like that of a cubic—with the same S-shape near the intercept as the toolkit function[latex]\,f\left(x\right)={x}^{3}.\,[/latex]We call this a triple zero, or a zero with multiplicity 3.

For zeros with even multiplicities, the graphs touch or are tangent to the x -axis. For zeros with odd multiplicities, the graphs cross or intersect the x -axis. See Figure 8 for examples of graphs of polynomial functions with multiplicity 1, 2, and 3.

For higher even powers, such as 4, 6, and 8, the graph will still touch and bounce off of the horizontal axis but, for each increasing even power, the graph will appear flatter as it approaches and leaves the x -axis.

For higher odd powers, such as 5, 7, and 9, the graph will still cross through the horizontal axis, but for each increasing odd power, the graph will appear flatter as it approaches and leaves the x -axis.

Graphical Behavior of Polynomials at x -Intercepts

If a polynomial contains a factor of the form[latex]\,{\left(x-h\right)}^{p},\,[/latex]the behavior near the[latex]\,x\text{-}[/latex]intercept[latex]\,h\,[/latex]is determined by the power[latex]\,p.\,[/latex]We say that[latex]\,x=h\,[/latex]is a zero of multiplicity[latex]\,p.[/latex]

The graph of a polynomial function will touch the x -axis at zeros with even multiplicities. The graph will cross the x -axis at zeros with odd multiplicities.

The sum of the multiplicities is the degree of the polynomial function.

Given a graph of a polynomial function of degree [latex]\,n,\,[/latex] identify the zeros and their multiplicities.

• If the graph crosses the x -axis and appears almost linear at the intercept, it is a single zero.
• If the graph touches the x -axis and bounces off of the axis, it is a zero with even multiplicity.
• If the graph crosses the x -axis at a zero, it is a zero with odd multiplicity.
• The sum of the multiplicities is[latex]\,n.[/latex]

Use the graph of the function of degree 6 in Figure 9 to identify the zeros of the function and their possible multiplicities.

The polynomial function is of degree 6. The sum of the multiplicities must be 6.

Starting from the left, the first zero occurs at[latex]\,x=-3.\,[/latex]The graph touches the x -axis, so the multiplicity of the zero must be even. The zero of[latex]\,-3\,[/latex]most likely has multiplicity[latex]\,2.[/latex]

The next zero occurs at[latex]\,x=-1.\,[/latex]The graph looks almost linear at this point. This is a single zero of multiplicity 1.

The last zero occurs at[latex]\,x=4.\,[/latex]The graph crosses the x -axis, so the multiplicity of the zero must be odd. We know that the multiplicity is likely 3 and that the sum of the multiplicities is 6.

Use the graph of the function of degree 9 in Figure 10. to identify the zeros of the function and their multiplicities.

The graph has a zero of –5 with multiplicity 3, a zero of -1 with multiplicity 2, and a zero of 3 with multiplicity 4.

Determining End Behavior

As we have already learned, the behavior of a graph of a polynomial function of the form

will either ultimately rise or fall as[latex]\,x\,[/latex] increases without bound and will either rise or fall as[latex]\,x\,[/latex] decreases without bound. This is because for very large inputs, say 100 or 1,000, the leading term dominates the size of the output. The same is true for very small inputs, say –100 or –1,000.

Recall that we call this behavior the end behavior of a function. As we pointed out when discussing quadratic equations, when the leading term of a polynomial function,[latex]\,{a}_{n}{x}^{n},\,[/latex]is an even power function, as[latex]\,x\,[/latex]increases or decreases without bound,[latex]\,f\left(x\right)\,[/latex]increases without bound. When the leading term is an odd power function, as[latex]\,x\,[/latex]decreases without bound,[latex]\,f\left(x\right)\,[/latex]also decreases without bound; as[latex]\,x\,[/latex]increases without bound,[latex]\,f\left(x\right)\,[/latex]also increases without bound. If the leading term is negative, it will change the direction of the end behavior. Figure 11 summarizes all four cases.

Understanding the Relationship between Degree and Turning Points

In addition to the end behavior, recall that we can analyze a polynomial function’s local behavior. It may have a turning point where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). Look at the graph of the polynomial function[latex]\,f\left(x\right)={x}^{4}-{x}^{3}-4{x}^{2}+4x\,[/latex]in Figure 12. The graph has three turning points.

This function[latex]\,f\,[/latex] is a 4 th degree polynomial function and has 3 turning points. The maximum number of turning points of a polynomial function is always one less than the degree of the function.

Interpreting Turning Points

A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising).

A polynomial of degree[latex]\,n\,[/latex]will have at most[latex]\,n-1\,[/latex]turning points.

Finding the Maximum Number of Turning Points Using the Degree of a Polynomial Function

Find the maximum number of turning points of each polynomial function.

• [latex]\,f\left(x\right)=-{x}^{3}+4{x}^{5}-3{x}^{2}+1[/latex]
• [latex]\,f\left(x\right)=-{\left(x-1\right)}^{2}\left(1+2{x}^{2}\right)[/latex]

First, rewrite the polynomial function in descending order:[latex]\,f\left(x\right)=4{x}^{5}-{x}^{3}-3{x}^{2}+1[/latex]

Identify the degree of the polynomial function. This polynomial function is of degree 5.

The maximum number of turning points is[latex]\,5-1=4.[/latex]

First, identify the leading term of the polynomial function if the function were expanded.

Then, identify the degree of the polynomial function. This polynomial function is of degree 4.

The maximum number of turning points is[latex]\,4-1=3.[/latex]

Graphing Polynomial Functions

We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph polynomial functions.

Given a polynomial function, sketch the graph.

• Find the intercepts.
• Check for symmetry. If the function is an even function, its graph is symmetrical about the[latex]\,y\text{-}[/latex]axis, that is,[latex]\,f\left(-x\right)=f\left(x\right).\,[/latex] If a function is an odd function, its graph is symmetrical about the origin, that is,[latex]\,f\left(-x\right)=-f\left(x\right).[/latex]
• Use the multiplicities of the zeros to determine the behavior of the polynomial at the[latex]\,x\text{-}[/latex]intercepts.
• Determine the end behavior by examining the leading term.
• Use the end behavior and the behavior at the intercepts to sketch a graph.
• Ensure that the number of turning points does not exceed one less than the degree of the polynomial.
• Optionally, use technology to check the graph.

Sketching the Graph of a Polynomial Function

Sketch a graph of[latex]\,f\left(x\right)=-2{\left(x+3\right)}^{2}\left(x-5\right).[/latex]

This graph has two x -intercepts. At[latex]\,x=-3,\,[/latex]the factor is squared, indicating a multiplicity of 2. The graph will bounce at this x -intercept. At[latex]\,x=5,\,[/latex]the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept.

The y -intercept is found by evaluating[latex]\,f\left(0\right).[/latex]

The y -intercept is[latex]\,\left(0,90\right).[/latex]

Additionally, we can see the leading term, if this polynomial were multiplied out, would be[latex]\,-2{x}^{3},\,[/latex] so the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity. See Figure 13.

To sketch this, we consider that:

• As[latex]\,x\to -\infty \,[/latex]the function[latex]\,f\left(x\right)\to \infty ,\,[/latex]so we know the graph starts in the second quadrant and is decreasing toward the[latex]\,x\text{-}[/latex]axis.
• Since[latex]\,f\left(-x\right)=-2{\left(-x+3\right)}^{2}\left(-x–5\right)\,[/latex] is not equal to[latex]\,f\left(x\right),\,[/latex]the graph does not display symmetry.

At[latex]\,\left(0,90\right),\,[/latex]the graph crosses the y -axis at the y -intercept. See Figure 14.

Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at[latex]\,\left(5,0\right).\,[/latex]See Figure 15.

As[latex]\,x\to \infty \,[/latex]the function[latex]\,f\left(x\right)\to \mathrm{-\infty },\,[/latex]so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant.

Using technology, we can create the graph for the polynomial function, shown in Figure 16, and verify that the resulting graph looks like our sketch in Figure 15.

Using the Intermediate Value Theorem

In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the x -axis, we can confirm that there is a zero between them. Consider a polynomial function[latex]\,f\,[/latex]whose graph is smooth and continuous. The Intermediate Value Theorem states that for two numbers[latex]\,a\,[/latex]and[latex]\,b\,[/latex]in the domain of[latex]\,f,[/latex] if [latex]a \lt b[/latex] and [latex]f\left(a\right)\ne f\left(b\right),[/latex] then the function[latex]\,f\,[/latex]takes on every value between[latex]\,f\left(a\right)\,[/latex]and[latex]\,f\left(b\right).\,[/latex](While the theorem is intuitive, the proof is actually quite complicated and requires higher mathematics.) We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function[latex]\,f\,[/latex]at[latex]\,x=a\,[/latex]lies above the[latex]\,x\text{-}[/latex]axis and another point at[latex]\,x=b\,[/latex]lies below the[latex]\,x\text{-}[/latex]axis, there must exist a third point between[latex]\,x=a\,[/latex]and[latex]\,x=b\,[/latex]where the graph crosses the[latex]\,x\text{-}[/latex]axis. Call this point[latex]\,\left(c,\text{ }f\left(c\right)\right).\,[/latex]This means that we are assured there is a solution[latex]\,c\,[/latex]where [latex]f\left(c\right)=0.[/latex]

In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the[latex]\,x\text{-}[/latex]axis. Figure 17 shows that there is a zero between[latex]\,a\,[/latex] and[latex]\,b.\,[/latex]

Intermediate Value Theorem

Let[latex]\,f\,[/latex] be a polynomial function. The Intermediate Value Theorem states that if[latex]\,f\left(a\right)\,[/latex] and[latex]\,f\left(b\right)\,[/latex] have opposite signs, then there exists at least one value[latex]\,c\,[/latex] between[latex]\,a\,[/latex] and[latex]\,b\,[/latex] for which[latex]\,f\left(c\right)=0.[/latex]

Show that the function[latex]\,f\left(x\right)={x}^{3}-5{x}^{2}+3x+6\,[/latex] has at least two real zeros between[latex]\,x=1\,[/latex] and[latex]\,x=4.[/latex]

As a start, evaluate[latex]\,f\left(x\right)\,[/latex] at the integer values[latex]\,x=1,2,3,[/latex]and [latex]4.\,[/latex]See Table 2.

We see that one zero occurs at[latex]\,x=2.\,[/latex]Also, since[latex]\,f\left(3\right)\,[/latex]is negative and[latex]\,f\left(4\right)\,[/latex]is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4.

We have shown that there are at least two real zeros between[latex]\,x=1\,[/latex] and [latex]\,x=4.[/latex]

We can also see on the graph of the function in Figure 18 that there are two real zeros between[latex]\,x=1\,[/latex] and[latex]\,x=4.[/latex]

Show that the function[latex]\,f\left(x\right)=7{x}^{5}-9{x}^{4}-{x}^{2}\,[/latex]has at least one real zero between [latex]\,x=1\,[/latex]and[latex]\,x=2.[/latex]

Because[latex]\,f\,[/latex]is a polynomial function and since[latex]\,f\left(1\right)\,\,[/latex]is negative and[latex]\,f\left(2\right)\,[/latex]is positive, there is at least one real zero between[latex]\,x=1\,[/latex]and[latex]\,x=2.\,[/latex]

Writing Formulas for Polynomial Functions

Now that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. Because a polynomial function written in factored form will have an x -intercept where each factor is equal to zero, we can form a function that will pass through a set of x -intercepts by introducing a corresponding set of factors.

Factored Form of Polynomials

If a polynomial of lowest degree[latex]\,p\,[/latex] has horizontal intercepts at[latex]\,x={x}_{1},{x}_{2},\dots ,{x}_{n},\,[/latex] then the polynomial can be written in the factored form:[latex]\,f\left(x\right)=a{\left(x-{x}_{1}\right)}^{{p}_{1}}{\left(x-{x}_{2}\right)}^{{p}_{2}}\cdots {\left(x-{x}_{n}\right)}^{{p}_{n}}\,[/latex] where the powers[latex]\,{p}_{i}\,[/latex] on each factor can be determined by the behavior of the graph at the corresponding intercept, and the stretch factor[latex]\,a\,[/latex] can be determined given a value of the function other than the x -intercept.

Given a graph of a polynomial function, write a formula for the function.

• Identify the x -intercepts of the graph to find the factors of the polynomial.
• Examine the behavior of the graph at the x -intercepts to determine the multiplicity of each factor.
• Find the polynomial of least degree containing all the factors found in the previous step.
• Use any other point on the graph (the y -intercept may be easiest) to determine the stretch factor.

Writing a Formula for a Polynomial Function from the Graph

Write a formula for the polynomial function shown in Figure 19.

This graph has three x -intercepts:[latex]\,x=-3,2,\,[/latex]and[latex]\,5.\,[/latex]The y -intercept is located at[latex]\,\left(0,2\right).\,[/latex]At[latex]\,x=-3\,[/latex]and[latex]\,x=5,\,[/latex] the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. At[latex]\,x=2,\,[/latex]the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). Together, this gives us

To determine the stretch factor, we utilize another point on the graph. We will use the[latex]\,y\text{-}[/latex]intercept[latex]\,\left(0,–2\right),\,[/latex]to solve for[latex]\,a.[/latex]

The graphed polynomial appears to represent the function[latex]\,f\left(x\right)=\frac{1}{30}\left(x+3\right){\left(x-2\right)}^{2}\left(x-5\right).[/latex]

Given the graph shown in Figure 20, write a formula for the function shown.

[latex]f\left(x\right)=-\frac{1}{8}{\left(x-2\right)}^{3}{\left(x+1\right)}^{2}\left(x-4\right)[/latex]

Using Local and Global Extrema

With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph.

Each turning point represents a local minimum or maximum. Sometimes, a turning point is the highest or lowest point on the entire graph. In these cases, we say that the turning point is a global maximum or a global minimum. These are also referred to as the absolute maximum and absolute minimum values of the function.

Local and Global Extrema

A local maximum or local minimum at[latex]\,x=a\,[/latex](sometimes called the relative maximum or minimum, respectively) is the output at the highest or lowest point on the graph in an open interval around[latex]\,x=a.\,[/latex]If a function has a local maximum at[latex]\,a,\,[/latex]then[latex]\,f\left(a\right)\ge f\left(x\right)\,[/latex]for all[latex]\,x\,[/latex]in an open interval around[latex]\,x=a.\,[/latex]If a function has a local minimum at[latex]\,a,\,[/latex]then[latex]\,f\left(a\right)\le f\left(x\right)\,[/latex]for all[latex]\,x\,[/latex]in an open interval around[latex]\,x=a.[/latex]

A global maximum or global minimum is the output at the highest or lowest point of the function. If a function has a global maximum at [latex]\,a,\,[/latex]then[latex]\,f\left(a\right)\ge f\left(x\right)\,[/latex]for all[latex]\,x.\,[/latex]If a function has a global minimum at[latex]\,a,\,[/latex]then[latex]\,f\left(a\right)\le f\left(x\right)\,[/latex]for all[latex]\,x.[/latex]

Do all polynomial functions have a global minimum or maximum?

No. Only polynomial functions of even degree have a global minimum or maximum. For example,[latex]\,f\left(x\right)=x\,[/latex]has neither a global maximum nor a global minimum.

Using Local Extrema to Solve Applications

An open-top box is to be constructed by cutting out squares from each corner of a 14 cm by 20 cm sheet of plastic and then folding up the sides. Find the size of squares that should be cut out to maximize the volume enclosed by the box.

We will start this problem by drawing a picture like that in Figure 22, labeling the width of the cut-out squares with a variable,[latex]w.[/latex]

Notice that after a square is cut out from each end, it leaves a[latex]\,\left(14-2w\right)\,[/latex]cm by[latex]\,\left(20-2w\right)\,[/latex]cm rectangle for the base of the box, and the box will be[latex]\, w\,[/latex]cm tall. This gives the volume

Notice, since the factors are[latex]\,w,\,[/latex][latex]\,20–2w\,[/latex]and[latex]\,14–2w,\,[/latex]the three zeros are 10, 7, and 0, respectively. Because a height of 0 cm is not reasonable, we consider the only the zeros 10 and 7. The shortest side is 14 and we are cutting off two squares, so values[latex]\,w\,[/latex]may take on are greater than zero or less than 7. This means we will restrict the domain of this function to[latex]\,0 7.\,[/latex]Using technology to sketch the graph of[latex]\,V\left(w\right)\,[/latex]on this reasonable domain, we get a graph like that in Figure 23. We can use this graph to estimate the maximum value for the volume, restricted to values for[latex]\,w\,[/latex]that are reasonable for this problem—values from 0 to 7.

From this graph, we turn our focus to only the portion on the reasonable domain,[latex]\,\left[0,\text{ }7\right].\,[/latex]We can estimate the maximum value to be around 340 cubic cm, which occurs when the squares are about 2.75 cm on each side. To improve this estimate, we could use advanced features of our technology, if available, or simply change our window to zoom in on our graph to produce Figure 24.

From this zoomed-in view, we can refine our estimate for the maximum volume to about 339 cubic cm, when the squares measure approximately 2.7 cm on each side.

Use technology to find the maximum and minimum values on the interval[latex]\,\left[-1,4\right]\,[/latex]of the function[latex]\,f\left(x\right)=-0.2{\left(x-2\right)}^{3}{\left(x+1\right)}^{2}\left(x-4\right).[/latex]

The minimum occurs at approximately the point[latex]\,\left(0,-6.5\right),\,[/latex]and the maximum occurs at approximately the point[latex]\,\left(3.5,7\right).[/latex]

Key Concepts

• Polynomial functions of degree 2 or more are smooth, continuous functions.
• To find the zeros of a polynomial function, if it can be factored, factor the function and set each factor equal to zero.
• Another way to find the[latex]\,x\text{-}[/latex]intercepts of a polynomial function is to graph the function and identify the points at which the graph crosses the[latex]\,x\text{-}[/latex]axis.
• The multiplicity of a zero determines how the graph behaves at the[latex]\,x\text{-}[/latex]intercepts.
• The graph of a polynomial will cross the horizontal axis at a zero with odd multiplicity.
• The graph of a polynomial will touch the horizontal axis at a zero with even multiplicity.
• The end behavior of a polynomial function depends on the leading term.
• The graph of a polynomial function changes direction at its turning points.
• A polynomial function of degree[latex]\,n\,[/latex] has at most[latex]\,n-1\,[/latex] turning points.
• To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at most[latex]\,n-1\,[/latex] turning points.
• Graphing a polynomial function helps to estimate local and global extremas.
• The Intermediate Value Theorem tells us that if[latex]\,f\left(a\right) \text{and} f\left(b\right)\,[/latex] have opposite signs, then there exists at least one value[latex]\,c\,[/latex] between[latex]\,a\,[/latex] and[latex]\,b\,[/latex] for which[latex]\,f\left(c\right)=0.\,[/latex]

Section Exercises

• What is the difference between an[latex]\,x\text{-}[/latex]intercept and a zero of a polynomial function[latex]\,f?\,[/latex]

The[latex]\,x\text{-}[/latex]intercept is where the graph of the function crosses the[latex]\,x\text{-}[/latex]axis, and the zero of the function is the input value for which[latex]\,f\left(x\right)=0.[/latex]

2. If a polynomial function of degree[latex]\,n\,[/latex] has[latex]\,n\,[/latex] distinct zeros, what do you know about the graph of the function?

3. Explain how the Intermediate Value Theorem can assist us in finding a zero of a function.

If we evaluate the function at[latex]\,a\,[/latex] and at[latex]\,b\,[/latex] and the sign of the function value changes, then we know a zero exists between[latex]\,a\,[/latex] and[latex]\,b.[/latex]

4. Explain how the factored form of the polynomial helps us in graphing it.

5. If the graph of a polynomial just touches the x -axis and then changes direction, what can we conclude about the factored form of the polynomial?

There will be a factor raised to an even power.

For the following exercises, find the[latex]\,x\text{-}[/latex] or t -intercepts of the polynomial functions.

6. [latex]\,C\left(t\right)=2\left(t-4\right)\left(t+1\right)\left(t-6\right)[/latex]

7. [latex]\,C\left(t\right)=3\left(t+2\right)\left(t-3\right)\left(t+5\right)[/latex]

[latex]\left(-2,0\right),\left(3,0\right),\left(-5,0\right)[/latex]

8. [latex]\,C\left(t\right)=4t{\left(t-2\right)}^{2}\left(t+1\right)[/latex]

9. [latex]\,C\left(t\right)=2t\left(t-3\right){\left(t+1\right)}^{2}[/latex]

[latex]\,\left(3,0\right),\left(-1,0\right),\left(0,0\right)[/latex]

11. [latex]\,C\left(t\right)=4{t}^{4}+12{t}^{3}-40{t}^{2}[/latex]

[latex]\left(0,0\right),\text{ }\left(-5,0\right),\text{ }\left(2,0\right)[/latex]

12. [latex]\,f\left(x\right)={x}^{4}-{x}^{2}[/latex]

13. [latex]\,f\left(x\right)={x}^{3}+{x}^{2}-20x[/latex]

[latex]\left(0,0\right),\text{ }\left(-5,0\right),\text{ }\left(4,0\right)[/latex]

14. [latex]f\left(x\right)={x}^{3}+6{x}^{2}-7x[/latex]

15. [latex]f\left(x\right)={x}^{3}+{x}^{2}-4x-4[/latex]

[latex]\left(2,0\right),\text{ }\left(-2,0\right),\text{ }\left(-1,0\right)[/latex]

16. [latex]f\left(x\right)={x}^{3}+2{x}^{2}-9x-18[/latex]

17. [latex]f\left(x\right)=2{x}^{3}-{x}^{2}-8x+4[/latex]

[latex]\left(-2,0\right),\,\left(2,0\right),\,\left(\frac{1}{2},0\right)[/latex]

18. [latex]f\left(x\right)={x}^{6}-7{x}^{3}-8[/latex]

19. [latex]f\left(x\right)=2{x}^{4}+6{x}^{2}-8[/latex]

[latex]\left(1,0\right),\text{ }\left(-1,0\right)[/latex]

20. [latex]f\left(x\right)={x}^{3}-3{x}^{2}-x+3[/latex]

21. [latex]f\left(x\right)={x}^{6}-2{x}^{4}-3{x}^{2}[/latex]

[latex]\left(0,0\right),\,\left(\sqrt{3},0\right),\,\left(-\sqrt{3},0\right)[/latex]

22. [latex]f\left(x\right)={x}^{6}-3{x}^{4}-4{x}^{2}[/latex]

23. [latex]f\left(x\right)={x}^{5}-5{x}^{3}+4x[/latex]

[latex]\left(0,0\right),\text{ }\left(1,0\right)\text{, }\left(-1,0\right),\text{ }\left(2,0\right),\text{ }\left(-2,0\right)[/latex]

For the following exercises, use the Intermediate Value Theorem to confirm that the given polynomial has at least one zero within the given interval.

24. [latex]f\left(x\right)={x}^{3}-9x,\,[/latex] between[latex]\,x=-4\,[/latex] and[latex]\,x=-2.[/latex]

25. [latex]f\left(x\right)={x}^{3}-9x,\,[/latex] between[latex]\,x=2\,[/latex] and[latex]\,x=4.[/latex]

[latex]f\left(2\right)=–10\,[/latex] and[latex]\,f\left(4\right)=28.[/latex] Sign change confirms.

26. [latex]f\left(x\right)={x}^{5}-2x,\,[/latex] between[latex]\,x=1\,[/latex] and[latex]\,x=2.[/latex]

27. [latex]f\left(x\right)=-{x}^{4}+4,\,[/latex] between[latex]\,x=1\,[/latex] and[latex]\,x=3[/latex]

[latex]f\left(1\right)=3\,[/latex] and[latex]\,f\left(3\right)=–77.\,[/latex] Sign change confirms.

28. [latex]f\left(x\right)=-2{x}^{3}-x,\,[/latex] between[latex]\,x=–1\,[/latex] and[latex]\,x=1.[/latex]

29. [latex]f\left(x\right)={x}^{3}-100x+2,\,[/latex] between[latex]\,x=0.01\,[/latex] and[latex]\,x=0.1[/latex]

[latex]f\left(0.01\right)=1.000001\,[/latex] and[latex]\,f\left(0.1\right)=–7.999.\,[/latex] Sign change confirms.

For the following exercises, find the zeros and give the multiplicity of each.

30. [latex]f\left(x\right)={\left(x+2\right)}^{3}{\left(x-3\right)}^{2}[/latex]

31. [latex]f\left(x\right)={x}^{2}{\left(2x+3\right)}^{5}{\left(x-4\right)}^{2}[/latex]

0 with multiplicity 2,[latex]\,-\frac{3}{2}\,[/latex] with multiplicity 5, 4 with multiplicity 2

32. [latex]f\left(x\right)={x}^{3}{\left(x-1\right)}^{3}\left(x+2\right)[/latex]

33. [latex]f\left(x\right)={x}^{2}\left({x}^{2}+4x+4\right)[/latex]

0 with multiplicity 2, –2 with multiplicity 2

34. [latex]f\left(x\right)={\left(2x+1\right)}^{3}\left(9{x}^{2}-6x+1\right)[/latex]

35. [latex]f\left(x\right)={\left(3x+2\right)}^{5}\left({x}^{2}-10x+25\right)[/latex]

[latex]-\frac{2}{3}\,\text{with}\,\text{multiplicity}\,5\text{,}\,5\,\text{with}\,\text{multiplicity}\,\text{2}[/latex]

36. [latex]f\left(x\right)=x\left(4{x}^{2}-12x+9\right)\left({x}^{2}+8x+16\right)[/latex]

37. [latex]f\left(x\right)={x}^{6}-{x}^{5}-2{x}^{4}[/latex]

[latex]\text{0}\,\text{with}\,\text{multiplicity}\,4\text{,}\,2\,\text{with}\,\text{multiplicity}\,1\text{,}\,–\text{1}\,\text{with}\,\text{multiplicity}\,1[/latex]

38. [latex]f\left(x\right)=3{x}^{4}+6{x}^{3}+3{x}^{2}[/latex]

39. [latex]f\left(x\right)=4{x}^{5}-12{x}^{4}+9{x}^{3}[/latex]

[latex]\frac{3}{2}\,[/latex] with multiplicity 2, 0 with multiplicity 3

40. [latex]f\left(x\right)=2{x}^{4}\left({x}^{3}-4{x}^{2}+4x\right)[/latex]

41. [latex]f\left(x\right)=4{x}^{4}\left(9{x}^{4}-12{x}^{3}+4{x}^{2}\right)[/latex]

[latex]\text{0}\,\text{with}\,\text{multiplicity}\,6\text{,}\,\frac{2}{3}\,\text{with}\,\text{multiplicity}\,2[/latex]

For the following exercises, graph the polynomial functions. Note[latex]\,x\text{-}[/latex] and[latex]\,y\text{-}[/latex]intercepts, multiplicity, and end behavior.

42. [latex]f\left(x\right)={\left(x+3\right)}^{2}\left(x-2\right)[/latex]

43. [latex]g\left(x\right)=\left(x+4\right){\left(x-1\right)}^{2}[/latex]

44. [latex]h\left(x\right)={\left(x-1\right)}^{3}{\left(x+3\right)}^{2}[/latex]

45. [latex]k\left(x\right)={\left(x-3\right)}^{3}{\left(x-2\right)}^{2}[/latex]

46. [latex]m\left(x\right)=-2x\left(x-1\right)\left(x+3\right)[/latex]

47. [latex]n\left(x\right)=-3x\left(x+2\right)\left(x-4\right)[/latex]

x -intercepts[latex]\,\left(0, 0\right),\text{ }\left(–2, 0\right),\text{ }\left(4,0\right)\,[/latex] with multiplicity 1,[latex]\,y\text{-}[/latex]intercept[latex]\,\left(0, 0\right).[/latex] As [latex]x\to -\infty ,\phantom{\rule{0.2em}{0ex}}f\left(x\right)\to \infty ,\phantom{\rule{0.2em}{0ex}}\text{as}\phantom{\rule{0.2em}{0ex}}x\to \infty ,\phantom{\rule{0.2em}{0ex}}f\left(x\right)\to -\infty .[/latex]

For the following exercises, use the graphs to write the formula for a polynomial function of least degree.

[latex]f\left(x\right)=-\frac{2}{9}\left(x-3\right)\left(x+1\right)\left(x+3\right)[/latex]

[latex]f\left(x\right)=\frac{1}{4}{\left(x+2\right)}^{2}\left(x-3\right)[/latex]

For the following exercises, use the graph to identify zeros and multiplicity.

–4, –2, 1, 3 with multiplicity 1

–2, 3 each with multiplicity 2

56. Write the equation:

For the following exercises, use the given information about the polynomial graph to write the equation.

57. Degree 3. Zeros at[latex]\,x=–2,[/latex] [latex]\,x=1,\,[/latex]and[latex]\,x=3.\,[/latex] y -intercept at[latex]\,\left(0,–4\right).[/latex]

[latex]f\left(x\right)=-\frac{2}{3}\left(x+2\right)\left(x-1\right)\left(x-3\right)[/latex]

58. Degree 3. Zeros at[latex]\,x=\text{–5,}[/latex] [latex]\,x=–2,[/latex]and[latex]\,x=1.\,[/latex] y -intercept at[latex]\,\left(0,6\right)[/latex]

59. Degree 5. Roots of multiplicity 2 at[latex]\,x=3\,[/latex] and[latex]\,x=1\,[/latex] , and a root of multiplicity 1 at[latex]\,x=–3.\,[/latex] y -intercept at[latex]\,\left(0,9\right)[/latex]

[latex]f\left(x\right)=\frac{1}{3}{\left(x-3\right)}^{2}{\left(x-1\right)}^{2}\left(x+3\right)[/latex]

60. Degree 4. Root of multiplicity 2 at[latex]\,x=4,\,[/latex]and a roots of multiplicity 1 at[latex]\,x=1\,[/latex]and[latex]\,x=–2.\,[/latex] y -intercept at[latex]\,\left(0,\text{–}3\right).[/latex]

61. Degree 5. Double zero at[latex]\,x=1,\,[/latex]and triple zero at[latex]\,x=3.\,[/latex] Passes through the point[latex]\,\left(2,15\right).[/latex]

[latex]f\left(x\right)=-15{\left(x-1\right)}^{2}{\left(x-3\right)}^{3}[/latex]

62. Degree 3. Zeros at[latex]\,x=4,[/latex][latex]\,x=3,[/latex]and[latex] y -intercept at[latex]\,\left(0,-24\right).[/latex]

63. Degree 3. Zeros at[latex]\,x=-3,[/latex] [latex]\,x=-2\,[/latex] and[latex]\,x=1.\,[/latex] y -intercept at[latex]\,\left(0,12\right).[/latex]

[latex]f\left(x\right)=-2\left(x+3\right)\left(x+2\right)\left(x-1\right)[/latex]

64. Degree 5. Roots of multiplicity 2 at[latex]\,x=-3\,[/latex] and[latex]\,x=2\,[/latex] and a root of multiplicity 1 at[latex]\,x=-2.[/latex]  y -intercept at[latex]\,\left(0, 4\right).[/latex]

65. Degree 4. Roots of multiplicity 2 at[latex]\,x=\frac{1}{2}\,[/latex]and roots of multiplicity 1 at[latex]\,x=6\,[/latex]and[latex]\,x=-2.[/latex]  y -intercept at[latex]\,\left(0,18\right).[/latex]

[latex]f\left(x\right)=-\frac{3}{2}{\left(2x-1\right)}^{2}\left(x-6\right)\left(x+2\right)[/latex]

66. Double zero at[latex]\,x=-3\,[/latex] and triple zero at[latex]\,x=0.\,[/latex] Passes through the point[latex]\,\left(1,32\right).[/latex]

For the following exercises, use a calculator to approximate local minima and maxima or the global minimum and maximum.

67. [latex]f\left(x\right)={x}^{3}-x-1[/latex]

local max[latex]\,\left(–\text{.58, –}.62\right),\,[/latex] local min[latex]\,\left(\text{.58, –1}\text{.38}\right)\,[/latex]

68. [latex]f\left(x\right)=2{x}^{3}-3x-1[/latex]

69. [latex]f\left(x\right)={x}^{4}+x[/latex]

global min[latex]\,\left(–\text{.63, –}\text{.47}\right)\,[/latex]

70. [latex]f\left(x\right)=-{x}^{4}+3x-2[/latex]

71. [latex]f\left(x\right)={x}^{4}-{x}^{3}+1[/latex]

global min[latex]\,\text{(}\text{.75, }\text{.89)}[/latex]

For the following exercises, use the graphs to write a polynomial function of least degree.

[latex]f\left(x\right)={\left(x-500\right)}^{2}\left(x+200\right)[/latex]

Real-World Applications

For the following exercises, write the polynomial function that models the given situation.

75. A rectangle has a length of 10 units and a width of 8 units. Squares of[latex]\,x\,[/latex] by[latex]\,x\,[/latex] units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a polynomial function in terms of[latex]\,x.[/latex]

[latex]f\left(x\right)=4{x}^{3}-36{x}^{2}+80x[/latex]

76. Consider the same rectangle of the preceding problem. Squares of[latex]\,2x\,[/latex] by[latex]\,2x\,[/latex] units are cut out of each corner. Express the volume of the box as a polynomial in terms of[latex]\,x.[/latex]

77. A square has sides of 12 units. Squares[latex]\,x\text{ }+1\,[/latex] by[latex]\,x\text{ }+1\,[/latex] units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a function in terms of[latex]\,x.[/latex]

[latex]f\left(x\right)=4{x}^{3}-36{x}^{2}+60x+100[/latex]

78. A cylinder has a radius of[latex]\,x+2\,[/latex] units and a height of 3 units greater. Express the volume of the cylinder as a polynomial function.

79. A right circular cone has a radius of[latex]\,3x+6\,[/latex] and a height 3 units less. Express the volume of the cone as a polynomial function. The volume of a cone is[latex]\,V=\frac{1}{3}\pi {r}^{2}h\,[/latex] for radius[latex]\,r\,[/latex] and height[latex]\,h.[/latex]

[latex]f\left(x\right)= 9\pi \left({x}^{3}+5{x}^{2}+8x+4\right)[/latex]

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Understanding Graphs

Supercharge your thinking with the language of algebra.

Seeing Solutions

Plot solutions on the coordinate plane.

Practice Seeing Solutions

Cement your understanding of Seeing Solutions

Revealing Patterns

Graph solutions and look for patterns.

Practice Revealing Patterns

Cement your understanding of Revealing Patterns

Explore the continuity of solutions.

Practice Continuity

Cement your understanding of Continuity

Using Continuity

Practice Using Continuity

Cement your understanding of Using Continuity

Diminishing Returns

Graph and analyze an equation to learn how fast chargers work.

End of Unit 1

Complete all lessons above to reach this milestone.

0 of 9 lessons complete

Find where a graph touches the axes.

Practice Intercepts

Cement your understanding of Intercepts

Up and Down

Reason about graphs by finding increasing and decreasing patterns.

Practice Up and Down

Cement your understanding of Up and Down

Minimums and Maximums

Understand how to identify minimum and maximum values of solutions to an equation.

Practice Minimums and Maximums

Cement your understanding of Minimums and Maximums

Bounded Graphs

See how graphs look when some solutions don't exist.

Practice Bounded Graphs

Cement your understanding of Bounded Graphs

Graphs with Gaps

Practice Graphs with Gaps

Cement your understanding of Graphs with Gaps

Training an AI

Explore the error rate of an AI as it trains with graphs.

End of Unit 2

0 of 11 lessons complete

Rate of Change

Explore what makes a line a line.

Practice Rate of Change

Cement your understanding of Rate of Change

Slope and Parameters

Practice Slope and Parameters

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The y-Intercept

Investigate lines that don't go through the origin.

Practice The y-Intercept

Cement your understanding of The y-Intercept

Horizontal and Vertical Lines

Learn what it means to be linear.

Practice Horizontal and Vertical Lines

Cement your understanding of Horizontal and Vertical Lines

Rules of Thumb

Investigate approximating familiar relationships with graphs.

End of Unit 3

Multiplying Expressions

Discover parabolas and quadratic equations.

Practice Multiplying Expressions

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Quadratic Equations

Explore a parabola's vertex and symmetry.

Practice Quadratic Equations

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Practice Symmetry

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Shifting Parabolas

Investigate how to change a parabola's location and width.

Practice Shifting Parabolas

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Wide, Narrow, and Upside Down

Practice Wide, Narrow, and Upside Down

Cement your understanding of Wide, Narrow, and Upside Down

Building Quadratics

Combine different ways to change a parabola into a single equation.

Practice Building Quadratics

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Optimizing Revenue

Find the best price to set a coffee using graphs.

End of Unit 4

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Population Growth

End of Unit 5

0 of 1 lessons complete

Course description

This course explores the twin pillars of algebraic thinking—equations and graphs. With these two tools, we'll unpack algebra's big ideas and develop a powerful perspective to solve its essential problems. By the end of this course, you'll understand graphs and their relationship to the equations they represent, enabling you to answer questions involving equations even when it's impossible to solve them by hand. Though a familiarity with solving equations is helpful, we'll start with a review of the basics and build on them throughout the course. When you finish, you'll be ready to conquer a wide range of algebraic techniques in Algebra 1.

Topics covered

• Algebraic Expressions
• Coordinate Plane
• Linear Equations
• Ordered Pairs
• Simplifying
• Systems of Equations

Prerequisites and next steps

Though a familiarity with solving equations is helpful, we'll start with a review of the basics and build on them throughout the course.

Measurement

Start exploring geometry with an intuitive introduction to the essentials.

Module 3: Linear Functions and Their Graphs

3.3: Graphing Linear Functions

section 3.3 Learning Objectives

• Identify if an ordered pair is a solution to a linear equation
• Graph a linear equation by plotting points
• Identify the x-intercept and y-intercept of a linear equation

Use the intercepts to graph a linear equation

Graph horizontal and vertical lines, identifying ordered pairs as solutions to linear equations.

A line is a visual representation of a linear equation, and the line itself is made up of an infinite number of points (or ordered pairs). The picture below shows the line of the linear equation [latex]y=2x–5[/latex] with some of the specific points on the line.

Every point on the line is a solution to the equation [latex]y=2x–5[/latex]. You can try plugging in any of the points that are labeled, like the ordered pair, [latex](1,−3)[/latex].

[latex]\begin{array}{l}\,\,\,\,y=2x-5\\-3=2\left(1\right)-5\\-3=2-5\\-3=-3\\\text{This is true.}\end{array}[/latex]

You can also try ANY of the other points on the line. Every point on the line is a solution to the equation [latex]y=2x–5[/latex]. All this means is that determining whether an ordered pair is a solution of an equation is pretty straightforward. If the ordered pair is on the line created by the linear equation, then it is a solution to the equation. But if the ordered pair is not on the line—no matter how close it may look—then it is not a solution to the equation.

Identifying Solutions

Graphing a linear equations (using techniques learned later in this and upcoming sections) produces a line.  Every point on the line is a solution to the linear equation.  The line continues forever in both directions and has an infinite number of solutions.

To find out whether a specific ordered pair (x,y)  is a solution of a linear equation, you can do the following:

• Substitute the x  and  y  values into the equation. If the equation yields a true statement, then the ordered pair is a solution of the linear equation. If the ordered pair does not yield a true statement then it is not a solution.

Determine whether [latex](−2,4)[/latex] is a solution to the equation [latex]4y+5x=3[/latex].

[latex]\begin{array}{r}4y+5x=3\\4\left(4\right)+5\left(−2\right)=3\end{array}[/latex]

[latex]\begin{array}{r}16+\left(−10\right)=3\\6=3\end{array}[/latex]

The statement is not true, so [latex](−2,4)[/latex] is not a solution to the equation [latex]4y+5x=3[/latex].

[latex](−2,4)[/latex] is not a solution to the equation [latex]4y+5x=3[/latex].

Graphing lines using ordered pairs

Graphing ordered pairs helps us make sense of all kinds of mathematical relationships.

You can use a coordinate plane to plot points and to map various relationships, such as the relationship between an object’s distance and the elapsed time. Many mathematical relationships are linear relationships . Let’s look at what a linear relationship is.

A linear relationship is a relationship between variables such that when plotted on a coordinate plane, the points lie on a line. Let’s start by looking at a series of points in Quadrant I on the coordinate plane.

Look at the five ordered pairs (and their x – and y -coordinates) below. Do you see any pattern to the location of the points? If this pattern continued, what other points could be included?

(0,0) , (1,2) , (2,4) , (3,6) , (4, 8) , (?, ?)

You may have identified that if this pattern continued the next ordered pair would be at (5, 10).  Applying the same logic, you may identify that the ordered pairs (6, 12) and (7, 14) would also belong (if this coordinate plane were larger).

These series of points can also be represented in a table. In the table below, the x-  and y -coordinates of each ordered pair on the graph is recorded.

Notice that each y -coordinate is twice the corresponding x -value. All of these x-  and y -values follow the same pattern, and, when placed on a coordinate plane, they all line up!

Once you know the pattern that relates the x- and y- values, you can find a y -value for any x -value that lies on the line. So if the rule of this pattern is that each y -value is twice the corresponding x -value, then the ordered pairs (1.5, 3), (2.5, 5), and (3.5, 7) should all appear on the line too, correct? Look to see what happens.

If you were to keep adding ordered pairs ( x , y ) where the y -value was twice the x -value, you would end up with a graph like this.

Look at how all of the points blend together to create a line. You can think of a line, then, as a collection of an infinite number of individual points that share the same mathematical relationship. In this case, the relationship is that the y -value is twice the x -value.

There are multiple ways to represent a linear relationship—a table, a linear graph, and there is also a linear equation . A linear equation is an equation with two variables whose ordered pairs graph as a straight line.

There are several ways to create a graph from a linear equation. One way is to create a table of values for x and y , and then plot these ordered pairs on the coordinate plane. Two points are enough to determine a line. However, it’s always a good idea to plot more than two points to avoid possible errors.

Then you draw a line through the points to show all of the points that are on the line. The line continues endlessly in both directions. Every point on this line is a solution to the linear equation.

Graph the linear equation [latex]y=−\frac{3}{2}x[/latex].

Since the coefficient of x is [latex]−\frac{3}{2}[/latex], it is convenient to choose multiples of 2 for x . This ensures that y is an integer, and makes the line easier to graph.

Convert the table to ordered pairs. Then plot the ordered pairs.

[latex](0,0)[/latex]

[latex](2,−3)[/latex]

[latex](4,−6)[/latex]

[latex](6,−9)[/latex]

Draw a line through the points to indicate all of the points on the line.

Graph the linear equation [latex]y=2x+3[/latex].

Convert the table to ordered pairs. Plot the ordered pairs.

[latex](0, 3)[/latex]

[latex](1, 5)[/latex]

[latex](2, 7)[/latex]

[latex](3, 9)[/latex]

Identify the intercepts of a linear equation

The intercepts of a line are the points where the line intersects, or crosses, the horizontal and vertical axes. To help you remember what “intercept” means, think about the word “intersect.” The two words sound alike and in this case mean the same thing.

The straight line on the graph below intersects the two coordinate axes. The point where the line crosses the x -axis is called the x -intercept . The y -intercept is the point where the line crosses the y -axis.

The x -intercept above is the point [latex](−2,0)[/latex]. The y -intercept above is the point (0, 2).

Notice that the y -intercept always occurs where [latex]x=0[/latex], and the x -intercept always occurs where [latex]y=0[/latex].

To find the x – and y -intercepts of a linear equation, you can substitute 0 for y and for x, respectively.

For example, the linear equation [latex]3y+2x=6[/latex] has an x intercept when [latex]y=0[/latex], so [latex]3\left(0\right)+2x=6\\[/latex].

[latex]\begin{array}{r}2x=6\\x=3\end{array}[/latex]

The x -intercept is [latex](3,0)[/latex].

Likewise the y -intercept occurs when [latex]x=0[/latex].

[latex]\begin{array}{r}3y+2\left(0\right)=6\\3y=6\\y=2\end{array}[/latex]

The y -intercept is [latex](0,2)[/latex].

You can use intercepts to graph some linear equations. Once you have found the two intercepts, draw a line through them.  This method is often used when the equation is written in Standard Form.

Standard Form of a Line

One way that we can represent the equation of a line is in standard form . Standard form is given as

[latex]Ax+By=C[/latex]

where [latex]A[/latex], [latex]B[/latex], and [latex]C[/latex] are integers. The x  and y  terms are on one side of the equal sign, and the constant term is on the other side.

Let’s use the intercepts to graph the equation [latex]3y+2x=6[/latex]. You figured out that the intercepts of the line this equation represents are [latex](0,2)[/latex] and [latex](3,0)[/latex]. That’s all you need to know.

Graph [latex]3x+5y=30[/latex] using the x and y -intercepts.

[latex]\begin{array}{r}3x+5y=30\\3\left(0\right)+5y=30\\0+5y=30\\5y=30\\y=\,\,\,6\\y\text{-intercept}\,\left(0,6\right)\end{array}[/latex]

To find the y -intercept, set [latex]x=0[/latex] and solve for y .

[latex]\begin{array}{r}3x+5y=30\\3x+5\left(0\right)=30\\3x+0=30\\3x=30\\x=10\\x\text{-intercept}\left(10,0\right)\end{array}[/latex]

To find the x -intercept, set [latex]y=0[/latex] and solve for x .

As we saw in the previous example, when given an equation in the form [latex]Ax+By=C[/latex], it is often easy to find the x and y-intercepts. Note that because of the communitive property of addition, [latex]Ax+By=C[/latex] is equivalent to [latex]By+Ax=C[/latex].  In the example below you will see an example where the equation is in the form  [latex]By+Ax=C[/latex].

In the example below, the equation wasn’t given in the form  [latex]Ax + By=C[/latex],  but we can still find the x and y-intercepts in the form it is in.

Graph [latex]y=2x-4[/latex] using the x and y -intercepts.

[latex]\begin{array}{l}y=2x-4\\y=2\left(0\right)-4\\y=0-4\\y=-4\\y\text{-intercept}\left(0,-4\right)\end{array}[/latex]

To find the x -intercept, set [latex]y=0[/latex] and solve for x .

[latex]\begin{array}{l}y=2x-4\\0=2x-4\\4=2x\\x=2\\x\text{-intercept}\left(2,0\right)\end{array}[/latex]

We mentioned that we can graph  some linear equations using intercepts.  So, what is the exception.  In the next example, there is only one intercept; yet, we need two points to construct a graph!

Find the [latex]x[/latex] and [latex]y[/latex]-intercepts and sketch the graph.

[latex]2x-y=0[/latex]

First we find the [latex]y[/latex]-intercept by plugging in [latex]x=0[/latex].

[latex]2(0)-y=0[/latex]

[latex]\frac{-y}{-1}=\frac{0}{-1}[/latex]

[latex]y=0[/latex]

Therefore, the [latex]y[/latex]-intercept is the origin, [latex](0,0)[/latex]

Next, we find the [latex]x[/latex]-intercept by plugging in [latex]y=0[/latex].

[latex]2x-0=0[/latex]

[latex]\frac{2x}{2}=\frac{0}{2}[/latex]

[latex]x=0[/latex]

Hence, the [latex]x[/latex]-intercept is also [latex](0,0)[/latex].

But since this only produces one point on the graph, we will be unable to graph it using intercepts alone.  Instead, we can revert back to plugging in a different value for [latex]x[/latex] (or for [latex]y[/latex]) to find a second solution.

Suppose we plug in [latex]x=1[/latex]. We get

[latex]2(1)-y=0[/latex]

[latex]2-y=0[/latex]

[latex]\underline{-2\hspace{.35in}-2}\hspace{.1in}[/latex]

[latex]\frac{-y}{-1}=\frac{-2}{-1}[/latex]

[latex]y=2[/latex]

So, a second point on the graph is [latex](1,2)[/latex].  Of course, we can continue to find more solutions if we want, but two will be sufficient to produce the graph shown below.

Solve for y , then graph a linear equation

Often times, it is easier to make a table of values if the equation is in the form [latex]y=mx+b[/latex] where m and b are real numbers. But to take advantage of this, we often must first solve for [latex]y[/latex] in the equation.

Solve for [latex]y[/latex] and graph using a table of values.

[latex]3x+y=5[/latex].

First, isolate [latex]y[/latex] by subtracting [latex]3x[/latex] on both sides of the equation.

[latex]3x+y=5[/latex]

[latex]\hspace{.02in}\underline{-3x\hspace{.35in}-3x}[/latex]

[latex]y=-3x+5[/latex]

Plot the ordered pairs (shown below).

[latex](0,5)[/latex]

[latex](1,2)[/latex]

[latex](2,−1)[/latex]

[latex](3,−4)[/latex]

However, we must first sometimes solve for [latex]y[/latex] ourselves to put it into this form, as demonstrated in the next example.

Solve for [latex]y[/latex], then graph the equation using a table of values.

[latex]4x-3y=3[/latex]

We first isolate [latex]y[/latex] in the equation.

[latex]\underline{-4x\hspace{.42in}-4x}[/latex]

[latex]\hspace{.75in}-3y=-4x+3[/latex]

[latex]\frac{-3y}{-3}=\frac{-4x}{-3}+\frac{3}{-3}[/latex]

[latex]y=\frac{4}{3}x-1[/latex]

As we select values for [latex]x[/latex], we note that it would be wise to choose multiples of 3 in order to produce integer answers.  We will use -3, 0, and 3.

Plotting the points leads to the graph below.

Why not use intercepts?

You may have noticed that the original equation was in standard form.  We noted earlier that it is quick to find intercepts when dealing with this form.  In Example 6, we discussed one potential problem with graphing using intercepts (when there is only one intercept at the origin).  This example reveals another potential problem you may encounter.

The [latex]y[/latex]-intercept turns out nice here.  If we plug in [latex]x=0[/latex], we get

[latex]4(0)-3y=3[/latex]

[latex]\frac{-3y}{-3}=\frac{3}{-3}[/latex]

[latex]y=-1[/latex]

So, the [latex]y[/latex]-intercept is [latex](0,-1)[/latex].

However, if we plug in [latex]y=0[/latex] to find the [latex]x[/latex]-intercept, we get

[latex]4x-3(0)=3[/latex]

[latex]\frac{4x}{4}=\frac{3}{4}[/latex]

[latex]x=\frac{3}{4}[/latex]

Therefore, the [latex]x[/latex]-intercept is [latex]\left(\frac{3}{4},0\right)[/latex], which include a fractional value.  While there is nothing inherently wrong with this, it can be extremely difficult to accurately plot fraction on a graph, especially if you are using an online mathematics program.  So, it can be extremely helpful to have alternative techniques.  By solving for [latex]y[/latex] and carefully selecting [latex]x[/latex]-values to plug in, we were able to avoid having to plot any fractions on our graph.

The following video provides another example of solving for [latex]y[/latex] and then graphing using a table of values.

The linear equations [latex]x=2[/latex] and [latex]y=−3[/latex] only have one variable in each of them. However, because these are linear equations, then should graph on a coordinate plane as lines just as the linear equations above do. Just think of the equation [latex]x=2[/latex] as [latex]x=0y+2[/latex] and think of [latex]y=−3[/latex] as [latex]y=0x–3[/latex].

Graph [latex]y=−3[/latex].

Write [latex]y=−3[/latex] as [latex]y=0x–3[/latex], and evaluate y when x has several values. Or just realize that [latex]y=−3[/latex] means every y- value will be [latex]−3[/latex], no matter what x is.

[latex](0,−3)[/latex]

[latex](1,−3)[/latex]

[latex](3,−3)[/latex]

It is worth noting that there was nothing particularly special about the [latex]-3[/latex] in the above example.  What you want to take from this example is that if you encounter an equation of the form [latex]y=constant[/latex], regardless of the constant, it will always result in a horizontal line.

We can take a similar approach with equations of the form [latex]x=constant[/latex].

Graph [latex]x=2[/latex].

One difference with this example compared to all others we have seen up until now is that we cannot plug in values for [latex]x[/latex].  This is because we are told [latex]x[/latex] is  always 2.  However, we have always had the option of plugging in values for [latex]y[/latex] instead, and in this case, we must go that route.

As mentioned earlier, we could view this as [latex]x=0y+2[/latex].  However, applying what we learned from the previous example, this means that no matter what values we plug in for [latex]y[/latex], we will always get [latex]x=2[/latex].

Being careful with the order in our ordered pairs, we can now plot the points [latex](2,0)[/latex], [latex](2,1)[/latex], [latex](2,2)[/latex], and [latex](2,3)[/latex] to obtain the graph below.

Once again, there was nothing special about the particular constant.  We conclude that linear equations of the form [latex]x=constant[/latex] will always result in a vertical line.

HORIZONTAL AND VERTICAL LINES

• Horizontal Lines:  Any linear equation of the form [latex]y=a[/latex], where [latex]a[/latex] is any real number, is a horizontal line that crosses the [latex]y[/latex]-axis at the point [latex](0,a)[/latex]
• Vertical Lines:  Any linear equation of the form [latex]x=b[/latex], where [latex]b[/latex] is any real number, is a vertical line that crosses the [latex]x[/latex]-axis at the point [latex](b,0)[/latex]

In the following video you will see more examples of graphing horizontal and vertical lines.

Using function notation to express equations of lines

Because all non-vertical lines are functions, we often express the equation of a line using function notation. Recall from section 3.2, function notation  can written as [latex] f(x) = [/latex] . This is read  as “f of x”. It is important to note that [latex] f(x)[/latex]  does not mean [latex] f [/latex]  times [latex] x [/latex]    but  is merely a notation indicating a function using the variable [latex] x [/latex] . A function is not always indicated by [latex] f [/latex]    but can be any letter, often [latex] g [/latex]    or [latex] h [/latex]  . In linear equations the dependent variable [latex] y [/latex] is replaced with an  [latex] f(x) [/latex], as seen in the example below. 

Graph the line  [latex] f(x)= \frac{1}{2}x -3 [/latex]

Let’s make a table of values where we evaluate the function for different values for x.

If we plot these points and connect them, we will have the graph of our line. Notice the x-intercept is not showing on this graph, but we were still able to graph the line based on other points we found in our table of values.

A solution to a linear equation is an ordered pair [latex](x,y)[/latex] that, when substituted into the equation, results in a true statement.  Any linear equation in two variables has an infinite number of solutions.  Graphing these solution results in a line.

We can graph lines by finding several solutions, often organized in a table of values.  By convention, we typically plug in values for [latex]x[/latex], which can be made easier by first solving for [latex]y[/latex], expressing the equation in the form [latex]y=mx+b[/latex].

The intercepts of a graph are the points at which the line crosses the axes.  To find the [latex]y[/latex]-intercept, plug in [latex]x=0[/latex] and to find the [latex]x[/latex]-intercept, plug in [latex]y=0[/latex].  These intercepts are often quick to determine if the equation is in the standard form, [latex]Ax+By=C[/latex].  If this results in two distinct intercepts, we can also use these to help us graph the line.

There are two special cases to watch out for, where one of the variables is missing from the equation.  If an equation is of the form [latex]y=constant[/latex], it will correspond to a horizontal line.  If an equation is of the form [latex]x=constant[/latex], it will correspond to a vertical line.

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5.3: Graphs of Polynomial Functions

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Learning Objectives

• Recognize characteristics of graphs of polynomial functions.
• Use factoring to find zeros of polynomial functions.
• Identify zeros and their multiplicities.
• Determine end behavior.
• Understand the relationship between degree and turning points.
• Graph polynomial functions.
• Use the Intermediate Value Theorem.

The revenue in millions of dollars for a fictional cable company from 2006 through 2013 is shown in Table \(\PageIndex{1}\).

The revenue can be modeled by the polynomial function

\[R(t)=−0.037t^4+1.414t^3−19.777t^2+118.696t−205.332\]

where \(R\) represents the revenue in millions of dollars and \(t\) represents the year, with \(t=6\) corresponding to 2006. Over which intervals is the revenue for the company increasing? Over which intervals is the revenue for the company decreasing? These questions, along with many others, can be answered by examining the graph of the polynomial function. We have already explored the local behavior of quadratics, a special case of polynomials. In this section we will explore the local behavior of polynomials in general.

Recognizing Characteristics of Graphs of Polynomial Functions

Polynomial functions of degree 2 or more have graphs that do not have sharp corners; recall that these types of graphs are called smooth curves. Polynomial functions also display graphs that have no breaks. Curves with no breaks are called continuous. Figure \(\PageIndex{1}\) shows a graph that represents a polynomial function and a graph that represents a function that is not a polynomial.

Example \(\PageIndex{1}\): Recognizing Polynomial Functions

Which of the graphs in Figure \(\PageIndex{2}\) represents a polynomial function?

• The graphs of \(f\) and \(h\) are graphs of polynomial functions. They are smooth and continuous .
• The graphs of \(g\) and \(k\) are graphs of functions that are not polynomials. The graph of function \(g\) has a sharp corner. The graph of function \(k\) is not continuous.

Do all polynomial functions have as their domain all real numbers?

• Yes. Any real number is a valid input for a polynomial function.

Using Factoring to Find Zeros of Polynomial Functions

Recall that if \(f\) is a polynomial function, the values of \(x\) for which \(f(x)=0\) are called zeros of \(f\). If the equation of the polynomial function can be factored, we can set each factor equal to zero and solve for the zeros.

We can use this method to find x-intercepts because at the x-intercepts we find the input values when the output value is zero. For general polynomials, this can be a challenging prospect. While quadratics can be solved using the relatively simple quadratic formula, the corresponding formulas for cubic and fourth-degree polynomials are not simple enough to remember, and formulas do not exist for general higher-degree polynomials. Consequently, we will limit ourselves to three cases in this section:

The polynomial can be factored using known methods: greatest common factor and trinomial factoring. The polynomial is given in factored form. Technology is used to determine the intercepts.

HOwTO: Given a polynomial function \(f\), find the x-intercepts by factoring

• Set \(f(x)=0\).
• Factor out any common monomial factors.
• Factor any factorable binomials or trinomials.
• Set each factor equal to zero and solve to find the x-intercepts.

Example \(\PageIndex{2}\): Finding the x-Intercepts of a Polynomial Function by Factoring

Find the x-intercepts of \(f(x)=x^6−3x^4+2x^2\).

We can attempt to factor this polynomial to find solutions for \(f(x)=0\).

\[\begin{align*} x^6−3x^4+2x^2&=0 & &\text{Factor out the greatest common factor.} \\ x^2(x^4−3x^2+2)&=0 & &\text{Factor the trinomial.} \\ x^2(x^2−1)(x^2−2)&=0 & &\text{Set each factor equal to zero.} \end{align*}\]

\[\begin{align*} x^2&=0 & & & (x^2−1)&=0 & & & (x^2−2)&=0 \\ x^2&=0 & &\text{ or } & x^2&=1 & &\text{ or } & x^2&=2 \\ x&=0 &&& x&={\pm}1 &&& x&={\pm}\sqrt{2} \end{align*}\] .

This gives us five x-intercepts: \((0,0)\), \((1,0)\), \((−1,0)\), \((\sqrt{2},0)\),and \((−\sqrt{2},0)\) (Figure \(\PageIndex{3}\)). We can see that this is an even function.

Example \(\PageIndex{3}\): Finding the x-Intercepts of a Polynomial Function by Factoring

Find the x-intercepts of \(f(x)=x^3−5x^2−x+5\).

Find solutions for \(f(x)=0\) by factoring.

\[\begin{align*} x^3−5x^2−x+5&=0 &\text{Factor by grouping.} \\ x^2(x−5)−(x−5)&=0 &\text{Factor out the common factor.} \\ (x^2−1)(x−5)&=0 &\text{Factor the difference of squares.} \\ (x+1)(x−1)(x−5)&=0 &\text{Set each factor equal to zero.} \end{align*}\]

\[\begin{align*} x+1&=0 & &\text{or} & x−1&=0 & &\text{or} & x−5&=0 \\ x&=−1 &&& x&=1 &&& x&=5\end{align*}\]

There are three x-intercepts: \((−1,0)\), \((1,0)\), and \((5,0)\) (Figure \(\PageIndex{4}\)).

Example \(\PageIndex{4}\): Finding the y- and x-Intercepts of a Polynomial in Factored Form

Find the y- and x-intercepts of \(g(x)=(x−2)^2(2x+3)\).

The y-intercept can be found by evaluating \(g(0)\).

\[\begin{align*} g(0)&=(0−2)^2(2(0)+3) \\ &=12 \end{align*}\]

So the y-intercept is \((0,12)\).

The x-intercepts can be found by solving \(g(x)=0\).

\[(x−2)^2(2x+3)=0\]

\[\begin{align*} (x−2)^2&=0 & & & (2x+3)&=0 \\ x−2&=0 & &\text{or} & x&=−\dfrac{3}{2} \\ x&=2 \end{align*}\]

So the x-intercepts are \((2,0)\) and \(\left(−\dfrac{3}{2},0\right)\).

We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial as shown in Figure \(\PageIndex{5}\).

Example \(\PageIndex{5}\): Finding the x-Intercepts of a Polynomial Function Using a Graph

Find the x-intercepts of \(h(x)=x^3+4x^2+x−6\).

This polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities.

Looking at the graph of this function, as shown in Figure \(\PageIndex{6}\), it appears that there are x-intercepts at \(x=−3,−2, \text{ and }1\).

We can check whether these are correct by substituting these values for \(x\) and verifying that \[h(−3)=h(−2)=h(1)=0. \nonumber\]

Since \(h(x)=x^3+4x^2+x−6\), we have:

\[ \begin{align*} h(−3)&=(−3)^3+4(−3)^2+(−3)−6=−27+36−3−6=0 \\[4pt] h(−2) &=(−2)^3+4(−2)^2+(−2)−6 =−8+16−2−6=0 \\[4pt] h(1)&=(1)^3+4(1)^2+(1)−6=1+4+1−6=0 \end{align*}\]

Each x-intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now write the polynomial in factored form.

\[\begin{align*} h(x)&=x^3+4x^2+x−6 \\ &=(x+3)(x+2)(x−1) \end{align*}\]

Exercise \(\PageIndex{1}\)

Find the y-and x-intercepts of the function \(f(x)=x^4−19x^2+30x\).

• y-intercept \((0,0)\);
• x-intercepts \((0,0)\), \((–5,0)\), \((2,0)\), and \((3,0)\)

Identifying Zeros and Their Multiplicities

Graphs behave differently at various x-intercepts. Sometimes, the graph will cross over the horizontal axis at an intercept. Other times, the graph will touch the horizontal axis and bounce off. Suppose, for example, we graph the function

\[f(x)=(x+3)(x−2)^2(x+1)^3.\]

Notice in Figure \(\PageIndex{7}\) that the behavior of the function at each of the x-intercepts is different.

The x-intercept −3 is the solution of equation \((x+3)=0\). The graph passes directly through thex-intercept at \(x=−3\). The factor is linear (has a degree of 1), so the behavior near the intercept is like that of a line—it passes directly through the intercept. We call this a single zero because the zero corresponds to a single factor of the function.

The x-intercept 2 is the repeated solution of equation \((x−2)^2=0\). The graph touches the axis at the intercept and changes direction. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept.

\[(x−2)^2=(x−2)(x−2)\]

The factor is repeated, that is, the factor \((x−2)\) appears twice. The number of times a given factor appears in the factored form of the equation of a polynomial is called the multiplicity . The zero associated with this factor, \(x=2\), has multiplicity 2 because the factor \((x−2)\) occurs twice.

The x-intercept −1 is the repeated solution of factor \((x+1)^3=0\).The graph passes through the axis at the intercept, but flattens out a bit first. This factor is cubic (degree 3), so the behavior near the intercept is like that of a cubic—with the same S-shape near the intercept as the toolkit function \(f(x)=x^3\). We call this a triple zero, or a zero with multiplicity 3.

For zeros with even multiplicities, the graphs touch or are tangent to the x-axis. For zeros with odd multiplicities, the graphs cross or intersect the x-axis. See Figure \(\PageIndex{8}\) for examples of graphs of polynomial functions with multiplicity 1, 2, and 3.

For higher even powers, such as 4, 6, and 8, the graph will still touch and bounce off of the horizontal axis but, for each increasing even power, the graph will appear flatter as it approaches and leaves the x-axis.

For higher odd powers, such as 5, 7, and 9, the graph will still cross through the horizontal axis, but for each increasing odd power, the graph will appear flatter as it approaches and leaves the x-axis.

Graphical Behavior of Polynomials at x-Intercepts

If a polynomial contains a factor of the form \((x−h)^p\), the behavior near the x-intercepth is determined by the power \(p\). We say that \(x=h\) is a zero of multiplicity \(p\).

The graph of a polynomial function will touch the x-axis at zeros with even multiplicities. The graph will cross the x-axis at zeros with odd multiplicities.

The sum of the multiplicities is the degree of the polynomial function.

HOWTO: Given a graph of a polynomial function of degree \(n\), identify the zeros and their multiplicities

• If the graph crosses the x-axis and appears almost linear at the intercept, it is a single zero.
• If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity.
• If the graph crosses the x-axis at a zero, it is a zero with odd multiplicity.
• The sum of the multiplicities is \(n\).

Example \(\PageIndex{6}\): Identifying Zeros and Their Multiplicities

Use the graph of the function of degree 6 in Figure \(\PageIndex{9}\) to identify the zeros of the function and their possible multiplicities.

The polynomial function is of degree \(n\). The sum of the multiplicities must be \(n\).

Starting from the left, the first zero occurs at \(x=−3\). The graph touches the x-axis, so the multiplicity of the zero must be even. The zero of −3 has multiplicity 2.

The next zero occurs at \(x=−1\). The graph looks almost linear at this point. This is a single zero of multiplicity 1.

The last zero occurs at \(x=4\).The graph crosses the x-axis, so the multiplicity of the zero must be odd. We know that the multiplicity is likely 3 and that the sum of the multiplicities is likely 6.

Exercise \(\PageIndex{2}\)

Use the graph of the function of degree 5 in Figure \(\PageIndex{10}\) to identify the zeros of the function and their multiplicities.

Figure \(\PageIndex{10}\): Graph of a polynomial function with degree 5.

The graph has a zero of –5 with multiplicity 1, a zero of –1 with multiplicity 2, and a zero of 3 with even multiplicity.

Determining End Behavior

As we have already learned, the behavior of a graph of a polynomial function of the form

\[f(x)=a_nx^n+a_{n−1}x^{n−1}+...+a_1x+a_0\]

will either ultimately rise or fall as \(x\) increases without bound and will either rise or fall as \(x\) decreases without bound. This is because for very large inputs, say 100 or 1,000, the leading term dominates the size of the output. The same is true for very small inputs, say –100 or –1,000.

Recall that we call this behavior the end behavior of a function. As we pointed out when discussing quadratic equations, when the leading term of a polynomial function, \(a_nx^n\), is an even power function, as \(x\) increases or decreases without bound, \(f(x)\) increases without bound. When the leading term is an odd power function, as \(x\) decreases without bound, \(f(x)\) also decreases without bound; as \(x\) increases without bound, \(f(x)\) also increases without bound. If the leading term is negative, it will change the direction of the end behavior. Figure \(\PageIndex{11}\) summarizes all four cases.

Understanding the Relationship between Degree and Turning Points

In addition to the end behavior, recall that we can analyze a polynomial function’s local behavior. It may have a turning point where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). Look at the graph of the polynomial function \(f(x)=x^4−x^3−4x^2+4x\) in Figure \(\PageIndex{12}\). The graph has three turning points.

This function \(f\) is a 4th degree polynomial function and has 3 turning points. The maximum number of turning points of a polynomial function is always one less than the degree of the function.

Definition: Interpreting Turning Points

A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). A polynomial of degree \(n\) will have at most \(n−1\) turning points.

Example \(\PageIndex{7}\): Finding the Maximum Number of Turning Points Using the Degree of a Polynomial Function

Find the maximum number of turning points of each polynomial function.

• \(f(x)=−x^3+4x^5−3x^2+1\)
• \(f(x)=−(x−1)^2(1+2x^2)\)

a. \(f(x)=−x^3+4x^5−3x^2+1\)

First, rewrite the polynomial function in descending order: \(f(x)=4x^5−x^3−3x^2+1\)

Identify the degree of the polynomial function. This polynomial function is of degree 5.

The maximum number of turning points is \(5−1=4\).

b. \(f(x)=−(x−1)^2(1+2x^2)\)

First, identify the leading term of the polynomial function if the function were expanded.

Then, identify the degree of the polynomial function. This polynomial function is of degree 4.

The maximum number of turning points is \(4−1=3\).

Graphing Polynomial Functions

We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph polynomial functions.

Howto: Given a polynomial function, sketch the graph

• Find the intercepts.
• Check for symmetry. If the function is an even function, its graph is symmetrical about the y-axis, that is, \(f(−x)=f(x)\). If a function is an odd function, its graph is symmetrical about the origin, that is, \(f(−x)=−f(x)\).
• Use the multiplicities of the zeros to determine the behavior of the polynomial at the x-intercepts.
• Determine the end behavior by examining the leading term.
• Use the end behavior and the behavior at the intercepts to sketch a graph.
• Ensure that the number of turning points does not exceed one less than the degree of the polynomial.
• Optionally, use technology to check the graph.

Example \(\PageIndex{8}\): Sketching the Graph of a Polynomial Function

Sketch a graph of \(f(x)=−2(x+3)^2(x−5)\).

This graph has two x-intercepts. At \(x=−3\), the factor is squared, indicating a multiplicity of 2. The graph will bounce at this x-intercept. At \(x=5\),the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept.

The y-intercept is found by evaluating \(f(0)\).

\[\begin{align*} f(0)&=−2(0+3)^2(0−5) \\ &=−2⋅9⋅(−5) \\ &=90 \end{align*}\]

The y-intercept is \((0,90)\).

Additionally, we can see the leading term, if this polynomial were multiplied out, would be \(−2x3\), so the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity. See Figure \(\PageIndex{13}\).

To sketch this, we consider that:

• As \(x{\rightarrow}−{\infty}\) the function \(f(x){\rightarrow}{\infty}\),so we know the graph starts in the second quadrant and is decreasing toward the x-axis.
• Since \(f(−x)=−2(−x+3)^2(−x–5)\) is not equal to \(f(x)\), the graph does not display symmetry.
• At \((−3,0)\), the graph bounces off of thex-axis, so the function must start increasing.
• At \((0,90)\), the graph crosses the y-axis at the y-intercept. See Figure \(\PageIndex{14}\).

Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at \((5,0)\). See Figure \(\PageIndex{15}\).

As \(x{\rightarrow}{\infty}\) the function \(f(x){\rightarrow}−{\infty}\),

so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant.

Using technology, we can create the graph for the polynomial function, shown in Figure \(\PageIndex{16}\), and verify that the resulting graph looks like our sketch in Figure \(\PageIndex{15}\).

Exercise \(\PageIndex{8}\)

Sketch a graph of \(f(x)=\dfrac{1}{4}x(x−1)^4(x+3)^3\).

Using the Intermediate Value Theorem

In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the x-axis, we can confirm that there is a zero between them. Consider a polynomial function \(f\) whose graph is smooth and continuous. The Intermediate Value Theorem states that for two numbers \(a\) and \(b\) in the domain of \(f\),if \(a<b\) and \(f(a){\neq}f(b)\),then the function \(f\) takes on every value between \(f(a)\) and \(f(b)\). We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function \(f\) at \(x=a\) lies above the x-axis and another point at \(x=b\) lies below thex-axis, there must exist a third point between \(x=a\) and \(x=b\) where the graph crosses the x-axis. Call this point \((c,f(c))\).This means that we are assured there is a solution \(c\) where \(f(c)=0\).

In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the x-axis. Figure \(\PageIndex{18}\) shows that there is a zero between \(a\) and \(b\).

Definition: Intermediate Value Theorem

Let \(f\) be a polynomial function. The Intermediate Value Theorem states that if \(f(a)\) and \(f(b)\) have opposite signs, then there exists at least one value \(c\) between \(a\) and \(b\) for which \(f(c)=0\).

Example \(\PageIndex{9}\): Using the Intermediate Value Theorem

Show that the function \(f(x)=x^3−5x^2+3x+6\) has at least two real zeros between \(x=1\) and \(x=4\).

As a start, evaluate \(f(x)\) at the integer values \(x=1,\;2,\;3,\; \text{and }4\) (Table \(\PageIndex{2}\)).

We see that one zero occurs at \(x=2\). Also, since \(f(3)\) is negative and \(f(4)\) is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4.

We have shown that there are at least two real zeros between \(x=1\) and \(x=4\).

We can also see on the graph of the function in Figure \(\PageIndex{19}\) that there are two real zeros between \(x=1\) and \(x=4\).

Exercise \(\PageIndex{4}\)

Show that the function \(f(x)=7x^5−9x^4−x^2\) has at least one real zero between \(x=1\) and \(x=2\).

Because \(f\) is a polynomial function and since \(f(1)\) is negative and \(f(2)\) is positive, there is at least one real zero between \(x=1\) and \(x=2\).

Writing Formulas for Polynomial Functions

Now that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. Because a polynomial function written in factored form will have an x-intercept where each factor is equal to zero, we can form a function that will pass through a set of x-intercepts by introducing a corresponding set of factors.

Note: Factored Form of Polynomials

If a polynomial of lowest degree \(p\) has horizontal intercepts at \(x=x_1,x_2,…,x_n\), then the polynomial can be written in the factored form: \(f(x)=a(x−x_1)^{p_1}(x−x_2)^{p_2}⋯(x−x_n)^{p_n}\) where the powers \(p_i\) on each factor can be determined by the behavior of the graph at the corresponding intercept, and the stretch factor \(a\) can be determined given a value of the function other than the x-intercept.

• Identify the x-intercepts of the graph to find the factors of the polynomial.
• Examine the behavior of the graph at the x-intercepts to determine the multiplicity of each factor.
• Find the polynomial of least degree containing all the factors found in the previous step.
• Use any other point on the graph (the y-intercept may be easiest) to determine the stretch factor.

Example \(\PageIndex{10}\): Writing a Formula for a Polynomial Function from the Graph

Write a formula for the polynomial function shown in Figure \(\PageIndex{20}\).

This graph has three x-intercepts: \(x=−3,\;2,\text{ and }5\). The y-intercept is located at \((0,2)\).At \(x=−3\) and \( x=5\), the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. At \(x=2\), the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). Together, this gives us

\[f(x)=a(x+3)(x−2)^2(x−5)\]

To determine the stretch factor, we utilize another point on the graph. We will use the y-intercept \((0,–2)\), to solve for \(a\).

\[\begin{align*} f(0)&=a(0+3)(0−2)^2(0−5) \\ −2&=a(0+3)(0−2)^2(0−5) \\ −2&=−60a \\ a&=\dfrac{1}{30} \end{align*}\]

The graphed polynomial appears to represent the function \(f(x)=\dfrac{1}{30}(x+3)(x−2)^2(x−5)\).

Exercise \(\PageIndex{5}\)

Given the graph shown in Figure \(\PageIndex{21}\), write a formula for the function shown.

\(f(x)=−\frac{1}{8}(x−2)^3(x+1)^2(x−4)\)

Using Local and Global Extrema

With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph.

Each turning point represents a local minimum or maximum. Sometimes, a turning point is the highest or lowest point on the entire graph. In these cases, we say that the turning point is a global maximum or a global minimum . These are also referred to as the absolute maximum and absolute minimum values of the function.

Note: Local and Global Extrema

A local maximum or local minimum at \(x=a\) (sometimes called the relative maximum or minimum, respectively) is the output at the highest or lowest point on the graph in an open interval around \(x=a\).If a function has a local maximum at \(a\), then \(f(a){\geq}f(x)\)for all \(x\) in an open interval around \(x=a\). If a function has a local minimum at \(a\), then \(f(a){\leq}f(x)\)for all \(x\) in an open interval around \(x=a\).

A global maximum or global minimum is the output at the highest or lowest point of the function. If a function has a global maximum at \(a\), then \(f(a){\geq}f(x)\) for all \(x\). If a function has a global minimum at \(a\), then \(f(a){\leq}f(x)\) for all \(x\).

We can see the difference between local and global extrema in Figure \(\PageIndex{22}\).

No. Only polynomial functions of even degree have a global minimum or maximum. For example, \(f(x)=x\) has neither a global maximum nor a global minimum.

Example \(\PageIndex{11}\): Using Local Extrema to Solve Applications

An open-top box is to be constructed by cutting out squares from each corner of a 14 cm by 20 cm sheet of plastic then folding up the sides. Find the size of squares that should be cut out to maximize the volume enclosed by the box.

We will start this problem by drawing a picture like that in Figure \(\PageIndex{23}\), labeling the width of the cut-out squares with a variable, \(w\).

Notice that after a square is cut out from each end, it leaves \(a(14−2w)\) cm by \((20−2w)\) cm rectangle for the base of the box, and the box will be \(w\) cm tall. This gives the volume

\[\begin{align*} V(w)&=(20−2w)(14−2w)w \\ &=280w−68w^2+4w^3 \end{align*}\]

Notice, since the factors are \(20–2w\), \(14–2w\), and \(w\), the three zeros are \(10, \,7,\) and \(0,) respectively. Because a height of \(0\) cm is not reasonable, we consider the only the zeros \(10\) and \(7.\) The shortest side is \(14\) and we are cutting off two squares, so values \(w\) may take on are greater than zero or less than \(7.\) This means we will restrict the domain of this function to \(0<w<7\).Using technology to sketch the graph of \(V(w)\) on this reasonable domain, we get a graph like that in Figure \(\PageIndex{24}\). We can use this graph to estimate the maximum value for the volume, restricted to values for \(w\) that are reasonable for this problem—values from 0 to 7.

From this graph, we turn our focus to only the portion on the reasonable domain, \([0, 7]\). We can estimate the maximum value to be around 340 cubic cm, which occurs when the squares are about 2.75 cm on each side. To improve this estimate, we could use advanced features of our technology, if available, or simply change our window to zoom in on our graph to produce Figure \(\PageIndex{25}\).

From this zoomed-in view, we can refine our estimate for the maximum volume to about 339 cubic cm, when the squares measure approximately 2.7 cm on each side.

Use technology to find the maximum and minimum values on the interval \([−1,4]\) of the function \(f(x)=−0.2(x−2)^3(x+1)^2(x−4)\).

The minimum occurs at approximately the point \((0,−6.5)\), and the maximum occurs at approximately the point \((3.5,7)\).

Key Concepts

• Polynomial functions of degree 2 or more are smooth, continuous functions.
• To find the zeros of a polynomial function, if it can be factored, factor the function and set each factor equal to zero.
• Another way to find the x-intercepts of a polynomial function is to graph the function and identify the points at which the graph crosses the x-axis.
• The multiplicity of a zero determines how the graph behaves at the x-intercepts.
• The graph of a polynomial will cross the horizontal axis at a zero with odd multiplicity.
• The graph of a polynomial will touch the horizontal axis at a zero with even multiplicity.
• The end behavior of a polynomial function depends on the leading term.
• The graph of a polynomial function changes direction at its turning points.
• A polynomial function of degree \(n\) has at most \(n−1\) turning points.
• To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at most \(n−1\) turning points.
• Graphing a polynomial function helps to estimate local and global extremas.
• The Intermediate Value Theorem tells us that if \(f(a)\) and \(f(b)\) have opposite signs, then there exists at least one value \(c\) between \(a\) and \(b\) for which \(f(c)=0\).

global maximum highest turning point on a graph; \(f(a)\) where \(f(a){\geq}f(x)\) for all \(x\).

global minimum lowest turning point on a graph; \(f(a)\) where \(f(a){\leq}f(x)\) for all \(x\).

Intermediate Value Theorem for two numbers \(a\) and \(b\) in the domain of \(f\), if \(a<b\) and \(f(a){\neq}f(b)\), then the functionf takes on every value between \(f(a)\) and \(f(b)\); specifically, when a polynomial function changes from a negative value to a positive value, the function must cross the x-axis

multiplicity the number of times a given factor appears in the factored form of the equation of a polynomial; if a polynomial contains a factor of the form \((x−h)^p\), \(x=h\) is a zero of multiplicity \(p\).