Linear Inequalities in One Variable with Applications These poll results, showing a margin of error at 4 percent, are an example of a real-world scenario that can be represented by linear inequalities. Learning Objectives After completing this section, you should be able to: Graph inequalities in one variable. Solve linear inequalities in one variable. Construct a linear inequality to solve applications. In this section, we will study linear inequalities in one variable. Inequalities can be used when the possible values (answers) in a certain situation are numerous, not just a few, or when the exact value (answer) is not known but it is known to be within a range of possible values. There are many real-world scenarios that can be represented by linear inequalities. For example, consider the survey of the mayoral election in Surveys and polls are usually conducted with only a small group of people. The margin of error indicates a range of how the actual group of voters would vote given the results of the survey. This range can be expressed using inequalities. Another example involves college tuition. Say a local community college charges $113 per credit hour. You budget $1,500 for tuition this fall semester. What are the number of credit hours that you could take this fall? Since this answer could be many different values, it can be expressed as an inequality. Graphing Inequalities on the Number Line In Algebraic Expressions , we introduced equality and the = symbol. In this section, we look at inequality and the symbols < , > , ≤ , and ≥ . The table below summarizes the symbols and their meaning. Symbol Meaning < less than > greater than ≤ less than or equal to ≥ greater than or equal to Suppose you had the inequality statement x > 3 . What possible number or numbers would make the inequality x > 3 true? If you are thinking, " x could be 4," that's correct, but x could also be 5, 6, 37, 1 million, or even 3.001. The number of solutions is infinite; any number greater than 3 is a solution to the inequality x > 3 . Rather than trying to list all possible solutions, we show all the solutions to the inequality x > 3 on the number line. All the numbers to the right of 3 on the number line are shaded, to show that all numbers greater than 3 are solutions. At the number 3 itself, an open parenthesis is drawn, since the number 3 is not part of the solutions of x > 3 . We can also represent inequalities using interval notation. There is no upper end to the solution to this inequality. In interval notation, we express x > 3 as ( 3 , ∞ ) . The symbol ∞ is read as "infinity." Infinity is not an actual number. shows both the number line and the interval notation for x > 3 . The inequality x > 3 is graphed on this number line and written in interval notation. We used the left parenthesis symbol to show that the endpoint of the inequality is not included. Parentheses are used when the endpoints are not included as a possible answer to the inequality. The notation for inequalities on a number line and in interval notation use the same symbols to express the endpoints of intervals. The inequality x ≤ 1 means all numbers less than or equal to 1. To illustrate that solution on a number line, we first put a bracket at x = 1 ; brackets are used when the endpoint is included. We then shade in all the numbers to the left of 1, to show that all numbers less than one are solutions. There is no lower end to those numbers. We write x ≤ 1 in interval notation as ( − ∞ , 1 ] . The symbol − ∞ is read as "negative infinity." shows both the number line and interval notation for x = 1 . The inequality x ≤ 1 is graphed on this number line and written in interval notation. summarizes the general representations in both number line form and interval notation of solutions for x > a , x < a , x ≥ a , and x ≤ a . Summary of representations in number line form and interval notation. Graphing an Inequality Graph the inequality x ≥ − 3 and write the solution in interval notation. Shade to the right of − 3 to show all the numbers greater than − 3 , and put a bracket at − 3 to show that the numbers are greater than or equal to − 3 ( ) Write in interval notation starting at − 3 with a bracket to show that − 3 is included in the solution and then infinity because the solution includes all the numbers greater than or equal to − 3 : [ − 3 , ∞ ) Graphing a Compound Inequality Graph the inequality x > − 3 and x < 4 and write the solution in interval notation. Step 1: Graph x > − 3 ( ). Step 2: Graph x < 4 ( ). Step 3: Graph both on the same number line and think of where the solutions are to BOTH inequalities . This will be where BOTH are shaded. Step 4: Write the solution in interval notation: ( − 3 , 4 ) Where Did the Inequality Symbols Come From? The first use of the < symbol to represent "less than" and > to represent "greater than" appeared in a mathematics book written by Englishman Thomas Harriot that was published in 1631. However, Harriot did not invent the symbols…the editor of the book did! Harriot used triangular symbols to represent less than and greater than; the editor, for reasons unknown, changed to symbols that are similar to the ones we use today. The symbols used to represent less than or equal to, and greater than or equal to ( ≤ and ≥ ) were first used in 1731 by French hydrologist and surveyor Pierre Bouguer. Interestingly, English mathematician John Wallis had used similar symbols as early as 1670, but he put the bar above the less than and greater than symbols instead of below them. Solving Linear Inequalities A linear inequality is much like a linear equation—but the equal sign is replaced with an inequality sign. A linear inequality is an inequality in one variable that can be written in one of the forms a x + b < c , a x + b ≤ c , a x + b ≥ c , or a x + b > c , where a , b , and c are all real numbers. When we solved linear equations, we were able to use the properties of equality to add, subtract, multiply, or divide both sides and still keep the equality. Similar properties hold true for inequalities. We can add or subtract the same quantity from both sides of an inequality and still keep the inequality. For example, we know that 2 is less than 4, i.e., 2 < 4 . If we add 6 to both sides of this inequality, we still have a true statement: 2 + 6 < 4 + 6 8 < 10 The same would happen if we subtracted 6 from both sides of the inequality; the statement would stay true: 2 − 6 < 4 − 6 − 4 < − 2 Notice that the inequality signs stayed the same. This leads us to the Addition and Subtraction Properties of Inequality. For any numbers a , b , and c , if a < b , then a + c < b + c and a − c < b − c . For any numbers a , b , and c , if a > b , then a + c > b + c and a - c > b - c . We can add or subtract the same quantity from both sides of an inequality and still keep the inequality the same. But what happens to an inequality when we divide or multiply both sides by a number? Let's first multiply and divide both sides by a positive number, starting with an inequality we know is true, 10 < 15 . We will multiply and divide this inequality by 5: 10 < 15 10 < 15 10 ( 5 ) ? 15 ( 5 ) 10 5 ? 15 5 50 ? 75 2 ? 3 50 < 75 ( true ) 2 < 3 ( true ) The inequality signs stayed the same. Does the inequality stay the same when we divide or multiply by a negative number? Let's use our inequality 10 < 15 to find out, multiplying it and dividing it by − 5 : 10 < 15 10 < 15 10 ( − 5 ) ? 15 ( − 5 ) 10 − 5 ? 15 − 5 − 50 ? − 75 − 2 ? − 3 − 50 > − 75 ( true ) − 2 > − 3 ( true ) Notice that when we filled in the inequality signs, the inequality signs reversed their direction in order to make it true! To summarize, when we divide or multiply an inequality by a positive number, the inequality sign stays the same. When we divide or multiply an inequality by a negative number, the inequality sign reverses. This gives us the Multiplication and Division Property of Inequality. For any numbers a , b , and c , multiply or divide by a positive: if a < b and c > 0 , then a c < b c and a c < b c if a > b and c > 0 , then a c > b c and a c > b c multiply or divide by a negative: if a < b and c < 0 , then a c > b c and a c > b c if a > b and c < 0 , then a c < b c and a c < b c To summarize, when we divide or multiply an inequality by: a positive number, the inequality sign stays the same. a negative number, the inequality sign reverses. Be careful to only reverse the inequality sign when you are multiplying and dividing by a negative. You do NOT reverse the inequality sign when you add or subtract a negative. For example, 2 x < − 4 is solved by dividing both sides of the inequality by 2 to get x < − 2 . You do NOT reverse the inequality sign because there is a negative 4. As another example, − 2 x + 5 < 3 x is solved by adding − 2 x to both sides to get 5 < 5 x . This does not reverse the inequality sign because we were not multiplying or dividing by a negative. We then divide both sides by 5 and get 1 < x . Solving a Linear Inequality Using One Operation Solve 9 y < 54 , graph the solution on the number line, and write the solution in interval notation. 9 y < 54 9 y 9 < 54 9 y < 6 Solving a Linear Inequality Using Multiple Operations Solve the inequality 6 y ≤ 11 y + 17 , graph the solution on the number line, and write the solution in interval notation. 6 y ≤ 11 y + 17 6 y − 11 y ≤ 11 y + 17 − 11 y − 5 y ≤ 17 − 5 y − 5 ≥ − 17 5 y ≥ − 17 5 Solving Applications with Linear Inequalities Many real-life situations require us to solve inequalities. The method we will use to solve applications with linear inequalities is very much like the one we used when we solved applications with equations. We will read the problem and make sure all the words are understood. Next, we will identify what we are looking for and assign a variable to represent it. We will restate the problem in one sentence to make it easy to translate into an inequality. Then, we will solve the inequality. Sometimes an application requires the solution to be a whole number, but the algebraic solution to the inequality is not a whole number. In that case, we must round the algebraic solution to a whole number. The context of the application will determine whether we round up or down. Constructing a Linear Inequality to Solve an Application with Tablet Computers A teacher won a mini grant of $4,000 to buy tablet computers for their classroom. The tablets they would like to buy cost $254.12 each, including tax and delivery. What is the maximum number of tablets the teacher can buy? Let t = the number of tablets . t times $254.12 has to be less than $4,000, so 254.12 t ≤ 4 , 000 . Solve for t : 254.12 t 254.12 ≤ 4,000 254.12 t ≤ 15.74 The teacher can buy 15 tablets and stay under $4,000. Constructing a Linear Inequality to Solve a Tuition Application The local community college charges $113 per credit hour. Your budget is $1,500 for tuition this fall semester. What number of credit hours could you take this fall? Let c = the number of credit hours you could take. c times $113 has to be less than $1,500, so 113 c ≤ 1 , 500 . Solve for c : 113 c 113 ≤ 1500 113 c ≤ 13.27 You can take up to 13 credits and stay under $1,500. Constructing a Linear Inequality to Solve an Application with Travel Costs Brenda’s best friend is having a destination wedding and the event will last 3 days and 3 nights. Brenda has $500 in savings and can earn $15 an hour babysitting. She expects to pay $350 for airfare, $375 for food and entertainment, and $60 a night for her share of a hotel room. How many hours must she babysit to have enough money to pay for the trip? Let b = number of babysitting hours. b times $15 plus $500 has to be more than $ 350 + $ 375 + $ 60 / night , so 15 b + 500 ≥ 350 + 375 + 60 ( 3 ) . Solve for b : 15 b + 500 − 500 ≥ 905 − 500 15 b ≥ 405 15 b 15 ≥ 405 15 b ≥ 27 Brenda must babysit at least 27 hours. The Desmos activities called "Inequalities on a Number Line" and "Compound Inequalities on a Number Line" are ways for students to develop and deepen their understanding of inequalities. Teachers will need a Desmos account to assign the activity for student use. Once they have assigned the activity to their students, teachers need to share the code for the activity with their students. Students will input the code to work on the activity. Check Your Understanding Key Terms linear inequality Addition and Subtraction Property of Linear Inequalities Multiplication and Division Property of Linear Inequalities Key Concepts Inequalities can be used when the possible values (answers) in a certain situation are numerous, or when the exact value (answer) is not known, but it is known to be within a range of possible values. Linear inequalities can be represented using a number line or using interval notation. Formulas For any numbers a , b , and , if a < b , then a + c < b + c and a − c < b − c . For any numbers a , b , and c , if a > b , then a + c > b + c and a − c > b − c . For any numbers a , b , and c , multiply or divide by a positive: if a < b and c > 0 , then a c < b c and a c < b c if a > b and c > 0 , then a c > b c and a c > b c multiply or divide by a negative: if a < b and c < 0 , then a c > b c and a c > b c if a > b and c < 0 , then a c < b c and a c < b c

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Ratios and Proportions This bar graph shows popular social media app usage. ( Source ) Learning Objectives After completing this section, you should be able to: Construct ratios to express comparison of two quantities. Use and apply proportional relationships to solve problems. Determine and apply a constant of proportionality. Use proportions to solve scaling problems. Ratios and proportions are used in a wide variety of situations to make comparisons. For example, using the information from , we can see that the number of Facebook users compared to the number of Twitter users is 2,006 M to 328 M. Note that the "M" stands for million, so 2,006 million is actually 2,006,000,000 and 328 million is 328,000,000. Similarly, the number of Qzone users compared to the number of Pinterest users is in a ratio of 632 million to 175 million. These types of comparisons are ratios. Constructing Ratios to Express Comparison of Two Quantities Note there are three different ways to write a ratio , which is a comparison of two numbers that can be written as: a to b OR a : b OR the fraction a / b . Which method you use often depends upon the situation. For the most part, we will want to write our ratios using the fraction notation. Note that, while all ratios are fractions, not all fractions are ratios. Ratios make part to part, part to whole, and whole to part comparisons. Fractions make part to whole comparisons only. Expressing the Relationship between Two Currencies as a Ratio The Euro (€) is the most common currency used in Europe. Twenty-two nations, including Italy, France, Germany, Spain, Portugal, and the Netherlands use it. On June 9, 2021, 1 U.S. dollar was worth 0.82 Euros. Write this comparison as a ratio. Using the definition of ratio, let a = 1 U.S. dollar and let b = 0.82 Euros. Then the ratio can be written as either 1 to 0.82; or 1:0.82; or 1 0.82 . Expressing the Relationship between Two Weights as a Ratio The gravitational pull on various planetary bodies in our solar system varies. Because weight is the force of gravity acting upon a mass, the weights of objects is different on various planetary bodies than they are on Earth. For example, a person who weighs 200 pounds on Earth would weigh only 33 pounds on the moon! Write this comparison as a ratio. Using the definition of ratio, let a = 200 pounds on Earth and let b = 33 pounds on the moon. Then the ratio can be written as either 200 to 33; or 200:33; or 200 33 . Using and Applying Proportional Relationships to Solve Problems Using proportions to solve problems is a very useful method. It is usually used when you know three parts of the proportion, and one part is unknown. Proportions are often solved by setting up like ratios. If a b and c d are two ratios such that a b = c d , then the fractions are said to be proportional . Also, two fractions a b and c d are proportional ( a b = c d ) if and only if a × d = b × c . Solving a Proportion Involving Two Currencies You are going to take a trip to France. You have $520 U.S. dollars that you wish to convert to Euros. You know that 1 U.S. dollar is worth 0.82 Euros. How much money in Euros can you get in exchange for $520? Step 1: Set up the two ratios into a proportion; let x be the variable that represents the unknown. Notice that U.S. dollar amounts are in both numerators and Euro amounts are in both denominators. 1 0.82 = 520 x Step 2: Cross multiply, since the ratios a b and c d are proportional, then a × d = b × c . 520 ( 0.82 ) = 1 ( x ) 426.4 = x You should receive 426.4 Euros ( 426.4 € ) . Solving a Proportion Involving Weights on Different Planets A person who weighs 170 pounds on Earth would weigh 64 pounds on Mars. How much would a typical racehorse (1,000 pounds) weigh on Mars? Round your answer to the nearest tenth. Step 1: Set up the two ratios into a proportion. Notice the Earth weights are both in the numerator and the Mars weights are both in the denominator. 170 64 = 1,000 x Step 2: Cross multiply, and then divide to solve. 170 x = 1,000 ( 64 ) 170 x = 64,000 170 x 170 = 64,000 170 x = 376.5 So the 1,000-pound horse would weigh about 376.5 pounds on Mars. Solving a Proportion Involving Baking A cookie recipe needs 2 1 4 cups of flour to make 60 cookies. Jackie is baking cookies for a large fundraiser; she is told she needs to bake 1,020 cookies! How many cups of flour will she need? Step 1: Set up the two ratios into a proportion. Notice that the cups of flour are both in the numerator and the amounts of cookies are both in the denominator. To make the calculations more efficient, the cups of flour ( 2 1 4 ) is converted to a decimal number (2.25). 2.25 60 = x 1020 Step 2: Cross multiply, and then simplify to solve. 2.25 ( 1,020 ) = 60 x 2,295 = 60 x 38.25 = x Jackie will need 38.25, or 38 1 4 , cups of flour to bake 1,020 cookies. Part of the definition of proportion states that two fractions a b and c d are proportional if a × d = b × c . This is the "cross multiplication" rule that students often use (and unfortunately, often use incorrectly). The only time cross multiplication can be used is if you have two ratios (and only two ratios) set up in a proportion. For example, you cannot use cross multiplication to solve for x in an equation such as 2 5 = x 8 + 3 x because you do not have just the two ratios. Of course, you could use the rules of algebra to change it to be just two ratios and then you could use cross multiplication, but in its present form, cross multiplication cannot be used. Eudoxus was born around 408 BCE in Cnidus (now known as Knidos) in modern-day Turkey. As a young man, he traveled to Italy to study under Archytas, one of the followers of Pythagoras. He also traveled to Athens to hear lectures by Plato and to Egypt to study astronomy. He eventually founded a school and had many students. Eudoxus made many contributions to the field of mathematics. In mathematics, he is probably best known for his work with the idea of proportions. He created a definition of proportions that allowed for the comparison of any numbers, even irrational ones. His definition concerning the equality of ratios was similar to the idea of cross multiplying that is used today. From his work on proportions, he devised what could be described as a method of integration, roughly 2000 years before calculus (which includes integration) would be fully developed by Isaac Newton and Gottfried Leibniz. Through this technique, Eudoxus became the first person to rigorously prove various theorems involving the volumes of certain objects. He also developed a planetary theory, made a sundial still usable today, and wrote a seven volume book on geography called Tour of the Earth , in which he wrote about all the civilizations on the Earth, and their political systems, that were known at the time. While this book has been lost to history, over 100 references to it by different ancient writers attest to its usefulness and popularity. Determining and Applying a Constant of Proportionality In the last example, we were given that 2 1 4 cups of flour could make 60 cookies; we then calculated that 38 1 4 cups of flour would make 1,020 cookies, and 720 cookies could be made from 27 cups of flour. Each of those three ratios is written as a fraction below (with the fractions converted to decimals). What happens if you divide the numerator by the denominator in each? 2.25 60 = 0.0375 38.25 1,020 = 0.0375 27 720 = 0.0375 . The quotients in each are exactly the same! This number, determined from the ratio of cups of flour to cookies, is called the constant of proportionality . If the values a and b are related by the equality a b = k , then k is the constant of proportionality between a and b . Note since a b = k , then b = a k . and b = a k . One piece of information that we can derive from the constant of proportionality is a unit rate. In our example (cups of flour divided by cookies), the constant of proportionality is telling us that it takes 0.0375 cups of flour to make one cookie. What if we had performed the calculation the other way (cookies divided by cups of flour)? 60 2.25 = 26.66666... 1,020 38.25 = 26.66666... 720 27 = 26.66666... In this case, the constant of proportionality ( 26.66666 … = 26 2 3 ) is telling us that 26 2 3 cookies can be made with one cup of flour. Notice in both cases, the "one" unit is associated with the denominator. The constant of proportionality is also useful in calculations if you only know one part of the ratio and wish to find the other. Finding a Constant of Proportionality Isabelle has a part-time job. She kept track of her pay and the number of hours she worked on four different days, and recorded it in the table below. What is the constant of proportionality, or pay divided by hours? What does the constant of proportionality tell you in this situation? Pay $87.50 $50.00 $37.50 $100.00 Hours 7 4 3 8 To find the constant of proportionality, divide the pay by hours using the information from any of the four columns. For example, 87.5 7 = 12.5 . The constant of proportionality is 12.5, or $12.50. This tells you Isabelle's hourly pay: For every hour she works, she gets paid $12.50. Applying a Constant of Proportionality: Running mph Zac runs at a constant speed: 4 miles per hour (mph). One day, Zac left his house at exactly noon (12:00 PM) to begin running; when he returned, his clock said 4:30 PM. How many miles did he run? The constant of proportionality in this problem is 4 miles per hour (or 4 miles in 1 hour). Since a b = k , where k is the constant of proportionality, we have a miles b hours = k a 4 .5 = 4 (30 minutes is ½ , or 0.5 , hours) a = 4 ( 4.5 ) , since from the definition we know a = k b a = 18 Zac ran 18 miles. Applying a Constant of Proportionality: Filling Buckets Joe had a job where every time he filled a bucket with dirt, he was paid $2.50. One day Joe was paid $337.50. How many buckets did he fill that day? The constant of proportionality in this situation is $2.50 per bucket (or $2.50 for 1 bucket). Since a b = k , where k is the constant of proportionality, we have a dollars b buckets = k 337.50 b = 2.50 Since we are solving for b , and we know from the definition that b = a k : b = 337.50 2.50 b = 135 Joe filled 135 buckets. Applying a Constant of Proportionality: Miles vs. Kilometers While driving in Canada, Mabel quickly noticed the distances on the road signs were in kilometers, not miles. She knew the constant of proportionality for converting kilometers to miles was about 0.62—that is, there are about 0.62 miles in 1 kilometer. If the last road sign she saw stated that Montreal is 104 kilometers away, about how many more miles does Mabel have to drive? Round your answer to the nearest tenth. The constant of proportionality in this situation is 0.62 miles per 1 kilometer. Since a b = k , where k is the constant of proportionality, we have a miles b kilometers = k a 104 = 0.62 a = 0.62 ( 104 ) a = 64.48 Rounding the answer to the nearest tenth, Mabel has to drive 64.5 miles. Using Proportions to Solve Scaling Problems A map of the northeastern United States Ratio and proportions are used to solve problems involving scale . One common place you see a scale is on a map (as represented in ). In this image, 1 inch is equal to 200 miles. This is the scale. This means that 1 inch on the map corresponds to 200 miles on the surface of Earth. Another place where scales are used is with models: model cars, trucks, airplanes, trains, and so on. A common ratio given for model cars is 1:24—that means that 1 inch in length on the model car is equal to 24 inches (2 feet) on an actual automobile. Although these are two common places that scale is used, it is used in a variety of other ways as well. Solving a Scaling Problem Involving Maps is an outline map of the state of Colorado and its counties. If the distance of the southern border is 380 miles, determine the scale (i.e., 1 inch = how many miles). Then use that scale to determine the approximate lengths of the other borders of the state of Colorado. Outline Map of Colorado (credit: "Map of Colorado Counties" by David Benbennick/Wikimedia Commons, Public Domain) When the southern border is measured with a ruler, the length is 4 inches. Since the length of the border in real life is 380 miles, our scale is 1 inch = 95 miles. The eastern and western borders both measure 3 inches, so their lengths are about 285 miles. The northern border measures the same as the southern border, so it has a length of 380 miles. Solving a Scaling Problem Involving Model Cars Die-cast NASCAR model cars are said to be built on a scale of 1:24 when compared to the actual car. If a model car is 9 inches long, how long is a real NASCAR automobile? Write your answer in feet. The scale tells us that 1 inch of the model car is equal to 24 inches (2 feet) on the real automobile. So set up the two ratios into a proportion. Notice that the model lengths are both in the numerator and the NASCAR automobile lengths are both in the denominator. 1 24 = 9 x 24 ( 9 ) = x 216 = x This amount (216) is in inches. To convert to feet, divide by 12, because there are 12 inches in a foot (this conversion from inches to feet is really another proportion!). The final answer is: 216 12 = 18 The NASCAR automobile is 18 feet long. Check Your Understanding Key Terms ratio proportion constant of proportionality scale construct ratios solve proportions use proportions to solve scaling problems Key Concepts A ratio is a comparison of two numbers. The ratio of two numbers a and b can be written as: a to b OR a : b OR the fraction a / b . All fractions are ratios, but not all ratios are fractions. Ratios make part to part, part to whole, and whole to part comparisons. Fractions make part to whole comparisons only. When two ratios are equal, we say they are in proportion or are proportional. Setting up proportions allows us to solve many various situations where three of the four values of the proportion are known.

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Graphing Linear Equations and Inequalities How much would it cost to fill up your gas tank? (credit: "Gas Under 4 Bucks" by Mark Turnauckas, Flickr/CC BY 2.0) Learning Objectives After completing this section, you should be able to: Graph linear equations and inequalities in two variables. Solve applications of linear equations and inequalities. In this section, we will learn how to graph linear equations and inequalities. There are several real-world scenarios that can be represented by graphs of linear inequalities. Think of filling your car up with gasoline. If gasoline is $3.99 per gallon and you put 10 gallons in your car, you will pay $39.90. Your friend buys 15 gallons of gasoline and pays $59.85. You can plot these points on a coordinate system and connect the points with a line to create the graph of a line. You'll learn to do both in this section. Plotting Points on a Rectangular Coordinate System Just like maps use a grid system to identify locations, a grid system is used in algebra to show a relationship between two variables in a rectangular coordinate system. The rectangular coordinate system is also called the x y -plane or the “coordinate plane.” The rectangular coordinate system is formed by two intersecting number lines, one horizontal and one vertical. The horizontal number line is called the x -axis. The vertical number line is called the y -axis. These axes divide a plane into four regions, called quadrants. The quadrants are identified by Roman numerals, beginning on the upper right and proceeding counterclockwise. See . Quadrants on the Coordinate Plane In the rectangular coordinate system, every point is represented by an ordered pair ( ). The first number in the ordered pair is the x -coordinate of the point, and the second number is the y -coordinate of the point. The phrase "ordered pair" means that the order is important. At the point where the axes cross and where both coordinates are zero, the ordered pair is ( 0 , 0 ) . The point ( 0 , 0 ) has a special name. It is called the origin . Ordered Pair We use the coordinates to locate a point on the xy -plane. Let's plot the point ( 1 , 3 ) as an example. First, locate 1 on the x -axis and lightly sketch a vertical line through x = 1 . Then, locate 3 on the y -axis and sketch a horizontal line through y = 3 . Now, find the point where these two lines meet—that is the point with coordinates ( 1 , 3 ) . See . Point ( 1 , 3 ) Plotted on the Coordinate Plane Notice that the vertical line through x = 1 and the horizontal line through y = 3 are not part of the graph. The dotted lines are just used to help us locate the point ( 1 , 3 ) . When one of the coordinates is zero, the point lies on one of the axes. In , the point ( 0 , 4 ) is on the y -axis and the point (−2, 0) is on the x -axis. Points ( − 2 , 0 ) and ( 0 , 4 ) Plotted on the Coordinate Plane Plotting Points on a Coordinate System Plot the following points in the rectangular coordinate system and identify the quadrant in which the point is located: ( − 5 , 4 ) ( − 3 , − 4 ) ( 2 , − 3 ) ( 0 , − 1 ) ( 3 , 5 2 ) The first number of the coordinate pair is the x -coordinate, and the second number is the y -coordinate. To plot each point, sketch a vertical line through the x -coordinate and a horizontal line through the y -coordinate ( ). Their intersection is the point. Since x = − 5 , the point is to the left of the y -axis. Also, since y = 4 , the point is above the x -axis. The point ( − 5 , 4 ) is in quadrant II. Since x = − 3 , the point is to the left of the y -axis. Also, since y = − 4 , the point is below the x -axis. The point ( − 3 , − 4 ) is in quadrant III. Since x = 2 , the point is to the right of the y -axis. Since y = − 3 , the point is below the x -axis. The point ( 2 , − 3 ) is in quadrant IV. Since x = 0 , the point whose coordinates are ( 0 , − 1 ) is on the y -axis. Since x = 3 , the point is to the right of the y -axis. Since y = 5 2 , which is equal to 2.5, the point is above the x -axis. The point ( 3 , 5 2 ) is in quadrant I. Graphing Linear Equations in Two Variables Up to now, all the equations you have solved were equations with just one variable. In almost every case, when you solved the equation, you got exactly one solution. But equations can have more than one variable. Equations with two variables may be of the form A x + B y = C . An equation of this form, where A and B are both not zero, is called a linear equation in two variables . Here is an example of a linear equation in two variables, x and y . A x + B y = C A + 4 y = 8 A = 1 , B = 4 , C = 8 The equation y = − 3 x + 5 is also a linear equation. But it does not appear to be in the form A x + B y = C . We can use the addition property of equality and rewrite it in A x + B y = C form. Step 1: Add 3 x to both sides. y + 3 x = − 3 x + 5 + 3 x Step 2: Simplify. y + 3 x = 5 Step 3: Put it in A x + B y = C form. 3 x + y = 5 By rewriting y = − 3 x + 5 as 3 x + y = 5 , we can easily see that it is a linear equation in two variables because it is of the form A x + B y = C . When an equation is in the form A x + B y = C , we say it is in standard form of a linear equation . Most people prefer to have A , B , and C be integers and A ≥ 0 when writing a linear equation in standard form, although it is not strictly necessary. Linear equations have infinitely many solutions. For every number that is substituted for x there is a corresponding y value. This pair of values is a solution to the linear equation and is represented by the ordered pair ( x , y ). When we substitute these values of x and y into the equation, the result is a true statement, because the value on the left side is equal to the value on the right side. We can plot these solutions in the rectangular coordinate system. The points will line up perfectly in a straight line. We connect the points with a straight line to get the graph of the linear equation. We put arrows on the ends of each side of the line to indicate that the line continues in both directions. A graph is a visual representation of all the solutions of a linear equation. The line shows you all the solutions to that linear equation. Every point on the line is a solution of that linear equation. And every solution of the linear equation is on this line. This line is called the graph of the equation. Points not on the line are not solutions! The graph of a linear equation A x + B y = C is a straight line. Every point on the line is a solution of the equation. Every solution of this equation is a point on this line. Determining Points on a Line is the graph of y = 2 x − 3 . Graph of y = 2 x – 3 For each ordered pair, decide: Is the ordered pair a solution to the equation? Is the point on the line? A: ( 0 , – 3 ) B: ( 3 , 3 ) C: ( 2 , – 3 ) D: ( – 1 , – 5 ) Substitute the x - and y -values into the equation to check if the ordered pair is a solution to the equation. A : ( 0 , − 3 ) y = 2 x − 3 − 3 = ? 2 ( 0 ) − 3 − 3 = − 3 ✓ ( 0 , − 3 ) is a solution. B : ( 3 , 3 ) y = 2 x − 3 3 = ? 2 ( 3 ) − 3 3 = 3 ✓ ( 3 , 3 ) is a solution. C : ( 2 , − 3 ) y = 2 x − 3 − 3 = ? 2 ( 2 ) − 3 − 3 ≠ 1 ( 2 , − 3 ) is not a solution. D : ( − 1 , − 5 ) y = 2 x − 3 − 5 = ? 2 ( − 1 ) − 3 − 5 = − 5 ✓ ( − 1 , − 5 ) is a solution. Plot the points ( 0 , - 3 ) , ( 3 , 3 ) , ( 2 , - 3 ) , and ( - 1 , - 5 ) . In , the points ( 0 , 3 ) , ( 3 , - 3 ) , and ( − 1 , − 5 ) are on the line y = 2 x − 3 , and the point ( 2 , − 3 ) is not on the line. The points that are solutions to y = 2 x − 3 are on the line, but the point that is not a solution is not on the line. The steps to take when graphing a linear equation by plotting points are: Step 1: Find three points whose coordinates are solutions to the equation. Organize them in a table. Step 2: Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work. Step 3: Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line. It is true that it only takes two points to determine a line, but it is a good habit to use three points. If you only plot two points and one of them is incorrect, you can still draw a line, but it will not represent the solutions to the equation. It will be the wrong line. If you use three points, and one is incorrect, the points will not line up. This tells you something is wrong, and you need to check your work. Graphing a Line by Plotting Points Graph the equation: y = 1 2 x + 3 . Find three points that are solutions to the equation. Since this equation has the fraction 1 2 as a coefficient of x , we will choose values of x carefully. We will use zero as one choice and multiples of 2 for the other choices. Why are multiples of two a good choice for values of x ? By choosing multiples of 2, the multiplication by 1 2 simplifies to a whole number. x = 0 x = 2 x = 4 y = 1 2 x + 3 y = 1 2 x + 3 y = 1 2 x + 3 y = 1 2 ( 0 ) + 3 y = 1 2 ( 2 ) + 3 y = 1 2 ( 4 ) + 3 y = 0 + 3 y = 1 + 3 y = 2 + 3 y = 3 y = 4 y = 5 x y ( x , y ) 0 3 ( 0 , 3 ) 2 4 ( 2 , 4 ) 4 5 ( 4 , 5 ) Plot the points, check that they line up, and draw the line ( ). Solving Applications Using Linear Equations in Two Variables Many fields use linear equalities to model a problem. While our examples may be about simple situations, they give us an opportunity to build our skills and to get a feel for how they might be used. Pumping Gas Gasoline costs $3.53 per gallon. You put 10 gallons of gasoline in your car, and pay $35.30. Your friend puts 15 gallons of gasoline in their car and pays $52.95. Your neighbor needs 5 gallons of gasoline, how much will they pay? Let x = the number of gallons of gasoline and let y = the total cost . If gas is $3.53 per gallon, then y = 3.53 x . The two points given are (10, 35.30) and (15, 52.95). Plot the points, check that they line up, and draw the line ( ). We can see the point at x = 5 . The y -value is found by multiplying 5 by $3.53 to get $17.65. Your neighbor will pay $17.65. René Descartes René Descartes (credit: Flickr, Public Domain) René Descartes was born in 1596 in La Haye, France. He was sickly as a child, so much so that he was allowed to stay in bed until 11:00 AM rather than get up at 5:00 AM like the other school children. He kept this habit of rising late for most of the rest of his life. After his primary schooling, Descartes attended the University of Poitiers, receiving a law degree in 1616. He then embarked on a myriad of journeys, joining two different militaries (one in the Netherlands, the other in Bavaria) and generally travelling around Europe until 1628, when he settled in the Netherlands. It was here that he began to delve deeply into his ideas of science, mathematics, and philosophy. In 1637, at the urging of his friends, Descartes published Discourse on the Method for Conducting One's Reason Well and Seeking the Truth in the Sciences . The book had three appendices: La Dioptrique , a work on optics; Les Météores , which pertained to meteorology; and La Géométrie , a work on mathematics. It was in this appendix that he proposed a geometric way of representing many different algebraic expressions and equations. It is this system of representation that almost all mathematical textbooks use today. These publications (along with several others) brought much fame to Descartes. So renowned was his reputation that late in 1649, Queen Christina of Sweden asked Descartes to come to Sweden to tutor her. However, she wished to do her studies at 5:00 in the morning; Descartes had to break his lifelong habit of sleeping in late. A few months later, in February 1650, Descartes died of pneumonia. Graphing Linear Inequalities Previously we learned to solve inequalities with only one variable. We will now learn about inequalities containing two variables that can be written in one of the following forms: A x + B y ≥ C , A x + B y > C , A x + B y ≤ C , and A x + B y < C where A and B are not both zero. We will look at linear inequalities in two variables, which are very similar to linear equations in two variables. Like linear equations, linear inequalities in two variables have many solutions. Any ordered pair ( x , y ) that makes an inequality true when we substitute in the values is a solution to a linear inequality . Determining Solutions to an Inequality Determine whether each ordered pair is a solution to the inequality y > x + 4 : ( 0 , 0 ) ( 1 , 6 ) ( 2 , 6 ) ( – 5 , – 15 ) ( − 8 , 12 ) ( 0 , 0 ) y > x + 4 Substitute 0 for x and 0 for y 0 > ? 0 + 4 Simplify . 0 > 4 ( 0 , 0 ) is not a solution to y > x + 4. ( 1 , 6 ) y > x + 4 Substitute 1 for x and 6 for y 6 > ? 1 + 4 Simplify . 6 > 5 ( 1 , 6 ) is a solution to y > x + 4. ( 2 , 6 ) y > x + 4 Substitute 2 for x and 6 for y 6 > ? 2 + 4 Simplify . 6 > 6 ( 2 , 6 ) is not a solution to y > x + 4. ( − 5 , − 15 ) y > x + 4 Substitute − 5 for x and − 15 for y − 15 > ? − 5 + 4 Simplify . − 15 > − 1 So , ( − 5 , − 15 ) is not a solution to y > x + 4. ( − 8 , 12 ) y > x + 4 Substitute − 8 for x and 12 for y 12 > ? − 8 + 4 Simplify . 12 > − 4 So , ( − 8 , 12 ) is a solution to y > x + 4. Let us think about x > 3 . The point x = 3 separated that number line into two parts. On one side of 3 are all the numbers less than 3. On the other side of 3 all the numbers are greater than 3. See . Solution to x > 3 on a Number Line Similarly, the line y = x + 4 separates the plane into two regions. On one side of the line are points with y < x + 4 . On the other side of the line are the points with y > x + 4 . We call the line y = x + 4 a boundary line . For an inequality in one variable, the endpoint is shown with a parenthesis ( ) or a bracket ( ) depending on whether or not a is included in the solution: Endpoint with Parenthesis Endpoint with Bracket Similarly, for an inequality in two variables, the boundary line is shown with a solid or dashed line to show whether or not it the line is included in the solution. A x + B y < C A x + B y ≤ C A x + B y > C A x + B y ≥ C Boundary line is A x + B y = C Boundary line is A x + B y = C Boundary line is not included in solution. Boundary line is included in solution. Boundary line is dashed. Boundary line is solid. Now, let us take a look at what we found in Example 5.43. We will start by graphing the line y = x + 4 , and then we will plot the five points we tested, as graphed in . We found that some of the points were solutions to the inequality y > x + 4 and some were not. Which of the points we plotted are solutions to the inequality y > x + 4 ? The points ( 1 , 6 ) and ( − 8 , 12 ) are solutions to the inequality y > x + 4 . Notice that they are both on the same side of the boundary line y = x + 4 . The two points ( 0 , 0 ) and ( − 5 , − 15 ) are on the other side of the boundary line y = x + 4 , and they are not solutions to the inequality y > x + 4 . For those two points, y < x + 4 . What about the point ( 2 , 6 ) ? Because 6 = 2 + 4 , the point is a solution to the equation y = x + 4 , but not a solution to the inequality y > x + 4 . So, the point ( 2 , 6 ) is on the boundary line. Graph of y = x + 4 Let us take another point above the boundary line and test whether or not it is a solution to the inequality y > x + 4 . The point ( 0 , 10 ) clearly looks to be above the boundary line, doesn’t it? Is it a solution to the inequality? y > x + 4 10 > ? 0 + 4 10 > 4 ✓ Yes, ( 0 , 10 ) is a solution to y > x + 4 . Any point you choose above the boundary line is a solution to the inequality y > x + 4 . All points above the boundary line are solutions. Similarly, all points below the boundary line, the side with ( 0 , 0 ) and ( − 5 , − 15 ) , are not solutions to y > x + 4 , as shown in . Graph of y = x + 4 , with y > x + 4 Above the Boundary Line and y < x + 4 Below the Boundary Line The graph of the inequality y > x + 4 is shown in . The line y = x + 4 divides the plane into two regions. The shaded side shows the solutions to the inequality y > x + 4 . The points on the boundary line, those where y = x + 4 , are not solutions to the inequality y > x + 4 , so the line itself is not part of the solution. We show that by making the boundary line dashed, not solid. Graph of y > x + 4 Writing a Linear Inequality Shown by a Graph The boundary line shown in this graph is y = 2 x − 1 . Write the inequality shown in . The line y = 2 x − 1 is the boundary line. On one side of the line are the points with y > 2 x − 1 and on the other side of the line are the points with y < 2 x − 1 . Let us test the point ( 0 , 0 ) and see which inequality describes its position relative to the boundary line. At ( 0 , 0 ) , which inequality is true: y > 2 x − 1 or y < 2 x − 1 ? 0 > 2 ( 0 ) - 1 0 < 2 ( 0 ) - 1 0 > 0 - 1 0 < 0 - 1 0 > - 1 0 < − 1 True False Since y > 2 x − 1 is true, the side of the line with ( 0 , 0 ) , is the solution. The shaded region shows the solution of the inequality y > 2 x − 1 . Since the boundary line is graphed with a dashed line, the inequality does not include the equal sign. The graph shows the inequality y > 2 x − 1 . We could use an y point as a test point, provided it is not on the line. Why did we choose ( 0 , 0 ) ? Because it is the easiest to evaluate. You may want to pick a point on the other side of the boundary line and check that y < 2 x − 1 . Graphing a Linear Inequality Graph the linear inequality y ≥ 3 4 x − 2 . Step 1. Identify and graph the boundary line ( ). If the inequality is ≤ or ≥, the boundary line is solid. If the inequality is < or >, the boundary line is dashed. Replace the inequality sign with an equal sign to find the boundary line. Graph the boundary line y = 3 4 x − 2 . The inequality sign is ≥, so we draw a solid line. Step 2. Test a point that is not on the boundary line. Is it a solution of the inequality? We’ll test ( 0 , 0 ) . Is it a solution of the inequality? At ( 0 , 0 ) , is y ≥ 3 4 x − 2 ? 0 ≥ ? 3 4 ( 0 ) − 2 0 ≥ − 2 So, ( 0 , 0 ) is a solution. Step 3. Shade in one side of the boundary line ( ). If the test point is a solution, shade in the side that includes the point. If the test point is not a solution, shade in the opposite side. The test point ( 0 , 0 ) is a solution to y ≥ 3 4 x − 2 . So we shade in that side. All points in the shaded region and on the boundary line represent the solution to y ≥ 3 4 x − 2 . Graphing Linear Inequalities in Two Variables Solving Applications Using Linear Inequalities in Two Variables Many fields use linear inequalities to model a problem. While our examples may be about simple situations, they give us an opportunity to build our skills and to get a feel for how they might be used. Working Multiple Jobs Hilaria works two part time jobs to earn enough money to meet her obligations of at least $240 a week. Her job in food service pays $10 an hour and her tutoring job on campus pays $15 an hour. How many hours does Hilaria need to work at each job to earn at least $240? Let x be the number of hours she works at the job in food service and let y be the number of hours she works tutoring. Write an inequality that would model this situation. Graph the inequality. Find three ordered pairs ( x,y ) that would be solutions to the inequality. Then, explain what that means for Hilaria. Let x be the number of hours she works at the job in food service and let y be the number of hours she works tutoring. She earns $10 per hour at the job in food service and $15 an hour tutoring. At each job, the number of hours multiplied by the hourly wage will give the amount earned at that job. Amount earned at the food service job ︸ 10 x plus + the amount earned tutoring ︸ 15 y is at least ︸ ≥ 240 Graph the inequality: Step 1: Graph the boundary line 10 x + 15 y = 240 Create a table of values x y 0 10 ( 0 ) + 15 y = 240 − > y = 16 6 10 ( 6 ) + 15 y = 240 − > y = 12 12 10 ( 12 ) + 15 y = 240 − > y = 8 Step 2: Pick a test point. Let us pick ( 0 , 0 ) again: 10 ( 0 ) + 15 ( 0 ) ≥ 240 ? 0 ≥ 240 is false and not a solution so the shading happens on the other side of the boundary line ( ). From the graph, we see that the ordered pairs ( 15 , 10 ) , ( 0 , 16 ) , ( 24 , 0 ) represent three of infinitely many solutions. Check the values in the inequality. ( 15 , 10 ) 10 x + 15 y ≥ 240 10 ( 15 ) + 15 ( 10 ) ≥ ? 240 300 ≥ 240 True ( 0 , 16 ) 10 x + 15 y ≥ 240 10 ( 0 ) + 15 ( 16 ) ≥ ? 240 240 ≥ 240 True ( 24 , 0 ) 10 x + 15 y ≥ 240 10 ( 24 ) + 15 ( 0 ) ≥ ? 240 240 ≥ 240 True For Hilaria, it means that to earn at least $240, she can work 15 hours tutoring and 10 hours at her food service job, earn all her money tutoring for 16 hours, or earn all her money while working 24 hours at the job in food service. Check Your Understanding Key Terms ordered pair origin points on the axes linear equation in two variables standards form of a linear equation solution linear inequality in two variables solution to a linear inequality boundary line Key Concepts Linear equations can be represented graphically on a rectangular coordinate system. Solving linear equations in two variables means finding the point where two lines intersect. There are three possibilities: The lines intersect at exactly one point; the lines do not intersect (they are parallel); or the lines intersect everywhere (they are the same line). Solving linear inequalities in two variables means finding a region of possible answers. Every point in this region will make both inequalities true statements. Plotting points is a standard way to help graph linear equations and linear inequalities. Videos Graphing Linear Inequalities in Two Variables

Math


Quadratic Equations In One Variable with Applications The Gateway Arch in St. Louis, Missouri (credit: modification of work "Gateway Arch - St. Louis - Missouri" by Sam valadi/Flickr, CC BY 2.0) Learning Objectives After completing this section, you should be able to: Multiply binomials. Factor trinomials. Solve quadratic equations by graphing. Solve quadratic equations by factoring. Solve quadratic equations using square root method. Solve quadratic equations using the quadratic formula. Solve real world applications modeled by quadratic equations. In this section, we will discuss quadratic equations. There are several real-world scenarios that can be represented by the graph of a quadratic equation. Think of the Gateway Arch in St. Louis, Missouri. Both ends of the arch are 630 feet apart and the arch is 630 feet tall. You can plot these points on a coordinate system and create a parabola to graph the quadratic equation. Identify Polynomials, Monomials, Binomials and Trinomials You have learned that a term is a constant, or the product of a constant and one or more variables. When it is of the form a x m , where a is a constant and x m is a positive whole number, it is called a monomial . Some examples of monomial are 8, - 2 x 2 , 4 y 3 , and 11 z . A monomial or two or more monomials combined by addition or subtraction is a polynomial . Some examples include: b + 11 , 4 y 2 − 7 y + 2 , and 4 x 4 + x 3 + 8 x 2 − 9 x + 1 . Some polynomials have special names, based on the number of terms. A monomial is a polynomial with exactly one term (examples: 14, 8 y 2 , - 9 x 3 y 5 , and − 13 ). A binomial has exactly two terms (examples: a + 7 , 4 b − 5 , y 2 − 16 , and 3 x 3 − 9 x 2 ), and a trinomial has exactly three terms (examples: x 2 − 7 x + 12 , 9 y 2 + 2 y − 8 , 6 m 4 − m 3 + 8 m , and x 4 + 3 x 2 − 1 ). Notice that every monomial, binomial, and trinomial is also a polynomial. They are just special members of the “family” of polynomials and so they have special names. We use the words monomial, binomial, and trinomial when referring to these special polynomials and just call all the rest polynomials. Multiply Binomials Recall multiplying algebraic expressions from Algebraic Expressions . In this section, we will continue that work and multiply binomials as well. We can use an area model to do multiplication. Multiply Binomials Multiply ( x + 2 ) ( x + 3 ) . Step 1: Use the distributive property: ( x ) ( x ) + ( x ) ( 3 ) + ( 2 ) ( x ) + ( 2 ) ( 3 ) In the area model ( ) multiply each term on the side by each term on the top (think of it as a multiplication table). Step 2: After we multiply, we get the following equation: x 2 + 3 x + 2 x + 6 Step 3: Combine the like terms to arrive at: x 2 + 5 x + 6 Multiplying More Complex Binomials Multiply ( 2 x + 7 ) ( 3 x − 5 ) . Step 1: Use the Distributive Property: ( 2 x ) ( 3 x ) − ( 2 x ) ( 5 ) + ( 7 ) ( 3 x ) − ( 7 ) ( 5 ) Step 2: After multiplying, get the following equation: 6 x 2 − 10 x + 21 x - 35 Step 3: Combine the like terms to arrive at: 6 x 2 + 11 x - 35 They Are Teaching Multiplication of Binomials in Elementary School Manipulatives are often used in elementary school for students to experience a hands-on way to experience the mathematics they are learning. Base Ten Blocks, or Dienes Blocks, are often used to introduce place value and the operation of whole numbers. When multiplying two-digit numbers, students can make an array to visualize the Distributive Property. shows the value of each Base Ten Block and shows how to multiply 17 and 23 using an area model and Base Ten Blocks. You can see how this helps students visualize the multiplication using the Distributive Property. Consider how ( 10 + 7 ) ( 20 + 3 ) can easily extend to ( 10 + x ) ( 20 + y ) in algebra! The Value of Each Base Ten Block How to Multiply 17 and 23 Using an Area Model and Base Ten Blocks Factoring Trinomials We’ve just covered how to multiply binomials. Now you will need to “undo” this multiplication—to start with the product and end up with the factors. Let us review an example of multiplying binomials to refresh your memory. ( x + 2 ) ( x + 3 ) = x 2 + 5 x + 6 To factor the trinomial means to start with the product, x 2 + 5 x + 6 , and end with the factors, ( x + 2 ) ( x + 3 ) . You need to think about where each of the terms in the trinomial came from. The first term came from multiplying the first term in each binomial. So, to get x 2 in the product, each binomial must start with an x . x 2 + 5 x + 6 ( x ) ( x ) The last term in the trinomial came from multiplying the last term in each binomial. So, the last terms must multiply to 6. What two numbers multiply to 6? The factors of 6 could be 1 and 6, or 2 and 3. How do you know which pair to use? Consider the middle term. It came from adding the outer and inner terms. So the numbers that must have a product of 6 will need a sum of 5. We’ll test both possibilities and summarize the results in the following table, which will be very helpful when you work with numbers that can be factored in many different ways. Factors of 6 Sum of Factors 1, 6 1 + 6 = 7 2, 3 2 + 3 = 5 We see that 2 and 3 are the numbers that multiply to 6 and add to 5. We have the factors of x 2 + 5 x + 6 . They are ( x + 2 ) ( x + 3 ) . x 2 + 5 x + 6 product ( x + 2 ) ( x + 3 ) factors You can check if the factors are correct by multiplying. Looking back, we started with x 2 + 5 x + 6 , which is of the form x 2 + b x + c , where b = 5 and c = 6 . We factored it into two binomials of the form ( x + m ) and ( x + n ) . x 2 + 5 x + 6 x 2 + b x + c ( x + 2 ) ( x + 3 ) ( x + m ) ( x + n ) To get the correct factors, we found two number m and n whose product is c and sum is b . With the area model ( ), start with an empty box and then put in the x 2 term and c . Continue by putting in two terms that add up to 5 x : 2 x and 3 x ( ): Then you find the terms of the binomials on the top and side ( ): Factoring Trinomials Factor x 2 + 7 x + 12 . The numbers that must have a product of 12 will need a sum of 7. We will summarize the results in a table below. Factors of 12 Sum of Factors 1, 12 1 + 12 = 13 2, 6 2 + 6 = 8 3, 4 3 + 4 = 7 We see that 3 and 4 are the numbers that multiply to 12 and add to 7. The factors of x 2 + 7 x + 12 are ( x + 3 ) ( x + 4 ) . Factoring More Complex Trinomials Factor x 2 - 11 x + 28 . The numbers that must have a product of 28 will need a sum of − 11 . We will summarize the results in a table. Factors of 28 Sum of Factors 1 , 28 1 + 28 = 29 2 , 14 2 + 14 = 16 4 , 7 4 + 7 = 11 We see that 4 and 7 are the numbers that multiply to 28 and add to 11. But we needed − 11 , so we will need to use − 4 and − 7 because ( − 4 ) ( − 7 ) = 28 and ( − 4 ) + ( − 7 ) = − 11 . The factors of x 2 − 11 x + 28 are ( x − 4 ) ( x − 7 ) . Factoring with the Box Method (Area Model) Solving Quadratic Equations by Graphing We have already solved and graphed linear equations in Graphing Linear Equations and Inequalities , equations of the form A x + B y = C . In linear equations, the variables have no exponents. Quadratic equations are equations in which the variable is squared. The following are some examples of quadratic equations: x 2 + 5 x + 6 = 0 3 y 2 + 4 y = 106 4 u 2 − 81 = 0 n ( n + 1 ) = 42 The last equation does not appear to have the variable squared, but when we simplify the expression on the left, we will get n 2 + n . The general form of a quadratic equation is a x 2 + b x + c = 0 , where a , b , and c are real numbers, with a ≠ 0 . Remember that a solution of an equation is a value of a variable that makes a true statement when substituted into the equation. The solutions of quadratic equations are the values of the variables that make the quadratic equation a x 2 + b x + c = 0 true. To solve quadratic equations, we need methods different than the ones we used in solving linear equations. We will start by solving a quadratic equation from its graph. Just like we started graphing linear equations by plotting points, we will do the same for quadratic equations. Let us look first at graphing the quadratic equation y = x 2 . We will choose integer values of x between − 2 and 2 and find their y values, as shown in the table below. y = x 2 x y 0 0 1 1 − 1 1 2 4 − 2 4 Notice when we let x = 1 and x = − 1 , we got the same value for y . y = x 2 y = 1 2 y = 1 y = x 2 y = ( − 1 ) 2 y = 1 The same thing happened when we let x = 2 and x = − 2 . Now, we will plot the points to show the graph of y = x 2 . See . The graph is not a line. This figure is called a parabola. Every quadratic equation has a graph that looks like this. When y = 0 the solution to the quadratic y = x 2 is 0 because x 2 = 0 at x = 0 . Graphing a Quadratic Equation Graph y = x 2 - 1 and list the solutions to the quadratic equation. We will graph the equation by plotting points. Step 1: Choose integer values for x , substitute them into the equation, and solve for y . Step 2: Record the values of the ordered pairs in the chart. y = x 2 - 1 x y 0 − 1 1 0 − 1 0 2 3 − 2 3 Step 3: Plot the points and then connect them with a smooth curve. The result will be the graph of the equation y = x 2 - 1 ( ). The solutions are x = 1 and x = - 1 . Solving a Quadratic Equation From Its Graph Find the solutions of y = x 2 + 5 x + 4 from its graph ( ). The solutions of a quadratic equations are the values of x that make the equation a true statement when set equal to zero (i.e. when y = 0 ). x 2 + 5 x + 4 = 0 at x = - 4 and x = - 1 . Solving Quadratic Equations by Factoring Another way of solving quadratic equations is by factoring. We will use the Zero Product Property that says that if the product of two quantities is zero, it must be that at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself. Solving a Quadratic Equation by Factoring Solve ( x + 1 ) ( x - 4 ) = 0 . Step 1. Set each factor equal to zero. The product equals zero, so at least one factor must equal zero. ( x + 1 ) ( x - 4 ) = 0 x + 1 = 0 or x - 4 = 0 Step 2. Solve the linear equations. Solve each equation. x = − 1 or x = 4 Step 3. Check. Substitute each solution separately into the original equation. x = - 1 ( x + 1 ) ( x - 4 ) = 0 ( − 1 + 1 ) ( − 1 − 4 ) = ? 0 ( 0 ) ( - 5 ) = ? 0 0 = 0 ✓ x = 4 ( x + 1 ) ( x - 4 ) = 0 ( 4 + 1 ) ( 4 − 4 ) = ? 0 ( 5 ) ( 0 ) = ? 0 0 = 0 ✓ Solve Another Quadratic Equation by Factoring Solve x 2 + 2 x − 8 = 0 . Step 1. Write the quadratic equation in standard form, ax 2 + bx + c = 0 . The equation is already in standard form. x 2 + 2 x - 8 = 0 Step 2. Factor the quadratic expression. Factor x 2 + 2 x - 8 ( x + 4 ) ( x - 2 ) ( x + 4 ) ( x - 2 ) = 0 Step 3. Use the Zero Product Property. Set each factor equal to zero. x + 4 = 0 or x - 2 = 0 Step 4. Solve the linear equations. We have two linear equations. x = - 4 or x = 2 Step 5. Check. Substitute each solution separately into the original equation. x 2 + 2 x - 8 = 0 x = - 4 ( − 4 ) 2 - 2 ( - 4 ) - 8 = ? 0 16 + ( - 8 ) - 8 = ? 0 0 = 0 ✓ x 2 + 2 x - 8 = 0 x = 2 2 2 - 2 ( 2 ) - 8 = ? 0 4 + 4 - 8 = ? 0 0 = 0 ✓ Solving Quadratics with the Zero Property Be careful to write the quadratic equation in standard form first. The equation must be set equal to zero in order for you to use the Zero Product Property! Often students start in Step 2 resulting in an incorrect solution. For example, x 2 + 2 x − 15 = - 7 cannot be factored to ( x − 3 ) ( x + 5 ) = − 7 and then solved by setting each factor equal to − 7 . The correct way to solve this quadratic equation is to set it equal to zero FIRST: x 2 + 2 x − 15 + 7 = − 7 + 7 which becomes x 2 + 2 x − 8 = 0 , then continue to factor. See the table below for the correct way to apply the Zero Product Property. x 2 + 2 x − 15 = − 7 x 2 + 2 x − 15 = − 7 Step 1 Skipped x 2 + 2 x − 15 + 7 = − 7 + 7 x 2 + 2 x − 8 = 0 Step 2 ( x − 3 ) ( x + 5 ) = − 7 ( x − 2 ) ( x + 4 ) = 0 Step 3 x − 3 = − 7 x + 5 = − 7 x − 2 = 0 x + 4 = 0 Step 4 x = − 4 x = − 12 x = 2 x = − 4 Step 5 ( − 4 ) 2 + 2 ( − 4 ) − 15 = 16 − 8 − 15 = − 7 ( − 12 ) 2 + 2 ( − 12 ) − 15 = 144 − 24 − 1 = 105 ≠ − 7 ( 2 ) 2 + 2 ( 2 ) − 8 = 4 + 4 − 8 = 0 ( − 4 ) 2 + 2 ( − 4 ) − 18 = 16 − 8 − 8 = 0 Solving Quadratic Equations Using the Square Root Property We just solved some quadratic equations by factoring. Let us use factoring to solve the quadratic equation x 2 = 9 . Step 1: Put the equation in standard form. x 2 − 9 = 0 Step 2: Factor the left side. ( x + 3 ) ( x - 3 ) = 0 Step 3: Use the Zero Product Property. x + 3 = 0 x − 3 = 0 Step 4: Solve each equation. x = − 3 x = 3 Step 5: Combine the two solutions into ± x = ± 3 The solution is read as “ x is equal to positive or negative 3.” What happens when we have an equation like x 2 = 7 ? Since 7 is not a perfect square, we cannot solve the equation by factoring. These equations are all of the form x 2 = k . We define the square root of a number in this way: If n 2 = m , then n is a square root of m . This leads to the Square Root Property . Using the Square Root Property to Solve a Quadratic Equation Solve using the square Root Property: x 2 = 169 . Step 1: Use the Square Root Property. x = ± 169 Step 2: Simplify the radical. x = ± 13 Step 3: Rewrite to show the two solutions. x = 13 x = - 13 Using the Square Root Property to Solve Another Quadratic Equation Solve using the Square Root Property: 144 q 2 = 25 . Step 1: Solve for q . q 2 = 25 144 Step 2: Use the Square Root Property. q = ± 25 144 Step 3: Simplify the radical. q = ± 5 12 Step 4: Rewrite to show the two solutions. q = 5 12 , q = - 5 12 Solving Quadratic Equations Using the Quadratic Formula This last method we will look at for solving quadratic equations is the quadratic formula . This method works for all quadratic equations, even the quadratic equations we could not factor! To use the quadratic formula, we substitute the values of a , b , and c into the expression on the right side of the formula. Then, we do all the math to simplify the expression. The result gives the solution(s) to the quadratic equation. Solving a Quadratic Equation Using the Quadratic Formula Solve using the quadratic formula: x 2 - 6 x + 5 = 0 . x 2 - 6 x + 5 = 0 This equation is in standard form. a x 2 + b x + c = 0 x 2 - 6 x + 5 = 0 Step 1: Identify the a , b , and c values. a = 1 , b = - 6 , c = 5 Step 2: Write the quadratic formula. x = - b ± b 2 - 4 a c 2 a Step 3: Substitute in the values of a , b , c . x = ( − 6 ) ± ( − 6 ) 2 − 4 ( 1 ) ( 5 ) 2.1 Step 4: Simplify. x = 6 ± 36 - 20 2 x = 6 ± 16 2 x = 6 ± 4 2 Step 5: Rewrite to show two solutions. x = 6 + 4 2 , x = 6 - 4 2 Step 6: Simplify. x = 10 2 , x = 2 2 x = 5 , x = 1 Step 7: Check. x 2 − 6 x + 5 = 0 x 2 − 6 x + 5 = 0 5 2 − 6 ⋅ 5 + 5 = ? 0 1 2 − 6 ⋅ 1 + 5 = ? 0 25 − 30 + 5 = ? 0 1 − 6 + 5 = ? 0 0 = 0 ✓ 0 = 0 ✓ Solving Another Quadratic Equation Using the Quadratic Formula Solve using the quadratic formula: 2 x 2 + 9 x − 5 = 0 . Step 1. Write the quadratic equation in standard form. Identify the a , b , c values. This equation is in standard form. ax 2 + bx + c = 0 2 x 2 + 9 x - 5 = 0 a = 2 , b = 9 , c = - 5 Step 2. Write the quadratic formula. Then substitute in the values of a , b , c . Substitute in a = 2 , b = 9 , c = - 5 x = − b ± b 2 − 4 a c 2 a x = − 9 ± 9 2 − 4 ( 2 ) ( − 5 ) 2 ⋅ 2 Step 3. Simplify the fraction, and solve for x . x = − 9 ± 81 − ( − 40 ) 4 x = − 9 ± 121 4 x = − 9 ± 11 4 x = − 9 + 11 4 x = − 9 − 11 4 x = 2 4 x = − 20 4 x = 1 2 x = − 5 Step 4. Check the solutions. Put each answer in the original equation to check. Substitute x = 1 2 . 2 x 2 + 9 x - 5 = 0 2 ( 1 2 ) 2 + 9 ⋅ 1 2 - 5 = ? 0 2 ⋅ 1 4 + 9 ⋅ 1 2 - 5 = ? 0 1 2 + 9 2 - 5 = ? 0 10 2 - 5 = ? 0 5 - 5 = ? 0 0 = 0 ✓ Substitute x = - 5 . 2 x 2 + 9 x - 5 = 0 2 ( - 5 ) 2 + 9 ( - 5 ) - 5 = ? 0 2 ⋅ 25 - 45 - 5 = ? 0 50 - 45 - 5 = ? 0 0 = 0 ✓ Solving Quadratics with the Quadratic Formula Solving Real-World Applications Modeled by Quadratic Equations There are problem solving strategies that will work well for applications that translate to quadratic equations. Here’s a problem-solving strategy to solve word problems: Step 1: Read the problem. Make sure all the words and ideas are understood. Step 2: Identify what we are looking for. Step 3: Name what we are looking for. Choose a variable to represent that quantity. Step 4: Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation. Step 5: Solve the equation using good algebra techniques. Step 6: Check the answer in the problem and make sure it makes sense. Step 7: Answer the question with a complete sentence. Finding Consecutive Integers The product of two consecutive integers is 132. Find the integers. Step 1: Read the problem. Step 2: Identify what we are looking for. We are looking for two consecutive integers. Step 3: Name what we are looking for. Let n = the first integer Let n + 1 = the next consecutive integer. Step 4: Translate into an equation. Restate the problem in a sentence. n ( n + 1 ) = 132 The product of the two consecutive integers is 132. The first integer times the next integer is 132. Step 5: Solve the equation. n 2 + n = 132 Bring all the terms to one side. n 2 + n - 132 = 0 Factor the trinomial. ( n - 11 ) ( n + 12 ) = 0 Use the zero product property. n - 11 = 0 or n + 12 = 0 Solve the equations. n = 11 , n = - 12 There are two values for n that are solutions to this problem. So, there are two sets of consecutive integers that will work. If the first integer is n = 11 , then the next integer is 12. If the first integer is n = - 12 , then the next integer is − 11 . Step 6: Check the answer. The consecutive integers are 11, 12 and − 11 , − 12 . The product of 11 and 12 = 132 and the product of - 11 ( - 12 ) = 132 . Both pairs of consecutive integers are solutions. Step 7: Answer the question. The consecutive integers are 11, 12, and − 11 , − 12 . Were you surprised by the pair of negative integers that is one of the solutions? In some applications, negative solutions will result from the algebra, but will not be realistic for the situation. Finding Length and Width of a Garden A rectangular garden has an area 15 square feet. The length of the garden is 2 feet more than the width. Find the length and width of the garden. Step 1: Read the problem. In problems involving geometric figures, a sketch can help you visualize the situation ( ). Step 2: Identify what you are looking for. We are looking for the length and width. Step 3: Name what you are looking for. The length is 2 feet more than width. Let W = the width of the garden. W + 2 = the length of the garden. Step 4: Translate into an equation. Restate the important information in a sentence. The area of the rectangular garden is 15 square feet. Use the formula for the area of a rectangle. A = L × W Substitute in the variables. 15 = ( W + 2 ) W Step 5: Solve the equation. Distribute first. 15 = W 2 + 2 W Get zero on one side. 0 = W 2 + 2 W - 15 Factor the trinomial. 0 = ( W + 5 ) ( W - 3 ) Use the Zero Product Property. 0 = W + 5 0 = W - 3 Solve each equation. - 5 = W 3 = W Since W is the width of the garden, it does not make sense for it to be negative. We eliminate that value for W . W = 3 Width is 3 feet. Find the value of the length. W + 2 = length. 3 + 2 5 Length is 5 feet. Step 6: Check the answer ( ). Does the answer make sense? Yes, this makes sense. Step 7: Answer the question. The width of the garden is 3 feet and the length is 5 feet. Completing the Square Recall the two methods used to solve quadratic equations of the form a x 2 + b x + c : by factoring and by using the quadratic formula. There are, however, many different methods for solving quadratic equations that were developed throughout history. Egyptian, Mesopotamian, Chinese, Indian, and Greek mathematicians all solved various types of quadratic equations, as did Arab mathematicians of the ninth through the twelfth centuries. It is one of these Arab mathematicians' methods that we wish to investigate with this activity. Muhammad ibn Musa al-Khwarizmi was employed as a scholar at the House of Wisdom in Baghdad, located in present day Iraq. One of the many accomplishments of Al-Khwarizmi was his book on the topic of algebra. In that book, he asks, “What must be the square which, when increased by ten of its own roots, amounts to 39?” Al-Khwarizmi, like many Arab mathematicians of his time, was well versed in Euclid's Elements. Like Euclid, he viewed algebra very geometrically, and thus had a geometric approach to solving a problem like the one above. In his approach, he used a method which today we refer to as completing the square . His description of the solution method for the above problem is: halve the number of roots, which in the present instance yields 5. This you multiply by itself; the product is 25. Add this to 39; the sum is 64. Now take the root of this which is 8, and subtract from it half the number of the roots, which is 5; the remainder is 3. This is the root of the square which you sought for. Thus the square is 9. So, what does all of this mean? Al-Khwarizmi would start with a square of unknown length of side (we will label the side length x ). See So, this square has area x 2 x He would then proceed to halve the number of roots (i.e., there are 10 roots by which the square is increasing) to get 5; this he would add to the first square. See The area of the two new pieces added into the original square are both 5 x . At this point, we have x 2 + 10 x = 39 . Now Al-Khwarizmi needed to “complete the square” by adding into the drawing a small square. See This square has an area of 25. x 2 + 10 x + 25 = 39 + 25 , or x 2 + 10 x + 25 = 64 . Notice that the completed square has side length x + 5 , so the large square has area ( x + 5 ) 2 . (Notice algebraically that the left half of the equation x 2 + 10 x + 25 = 64 factors to ( x + 5 ) 2 = 64. ) This means the area of large square equals 64. If ( x + 5 ) 2 = 64 , then x + 5 = 8 ; so x must be equal to 3 or − 13 to make this true. Note that Al-Kwarimi would not have considered the possibility of a negative solution, since he approached the solution geometrically, and negative distances do not exist. Check Your Understanding Key Terms monomial polynomial binomial trinomial quadratic equation Zero Product Property Key Concepts A quadratic equation is an algebraic equation where the highest power (degree) of the equation is two. To solve a quadratic equation is to find the value(s) that when substituted in for the variables, will make the equation equal to zero. There can be two, one, or no solutions to any quadratic equation. There are several methods to solve a quadratic equation. These methods include factoring quadratic equations, graphic quadratic equations, using the square root method, and using the quadratic formula. Videos Factoring with the Box Method (Area Model) Solving Quadratics with the Zero Property Solving Quadratics with the Quadratic Formula

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Functions A small group of elementary students learning from their teacher. (credit: modification of work "Our school" by Woodleywonderworks/Flickr, CC BY 2.0 ) Learning Objectives After completing this section, you should be able to: Use function notation. Determine if a relation is a function with different representations. Apply the vertical line test. Determine the domain and range of a function. In this section, we will learn about relations and functions. As we go about our daily lives, we have many data items or quantities that are paired to our names. Our social security number, student ID number, email address, phone number, and our birthday are matched to our name. There is a relationship between our name and each of those items. When your teacher gets their class roster, the names of all the students in the class are listed in one column and then the student ID number is likely to be in the next column. If we think of the correspondence as a set of ordered pairs, where the first element is a student name and the second element is that student’s ID number, we call this a relation. (Student name, Student ID #) The set of all the names of the students in the class is called the domain of the relation and the set of all student ID numbers paired with these students is the range of the relation. In general terms, a relation is any set of ordered pairs, ( x , y ). All the x -values in the ordered pairs together make up the domain . All the y -values in the ordered pairs together make up the range. There are many situations similar to the student's name and student ID # where one variable is paired or matched with another. The set of ordered pairs that records this matching is a relation. A special type of relation, called a function , occurs extensively in mathematics. A function is a relation that assigns to each element in its domain exactly one element in the range. For each ordered pair in the relation, each x -value is matched with only one y -value. Let us look at the relation between your friends and their birthdays in . Every friend has a birthday, but no one has two birthdays. It is okay for two people to share a birthday. It is okay that Danny and Stephen share July 24 as their birthday and that June and Liz share August 2. Since each person has exactly one birthday, the relation is a function. Birthday Mapping Use Function Notation It is very convenient to name a function; most often functions are named f , g , h , F , G , or H . In any function, for each x -value from the domain, we get a corresponding y -value in the range. In the function f , we write this range value y as f ( x ). This notation f ( x ) is called function notation and is read "f of x " or "the value of f at x ." In this case the parentheses do not indicate multiplication. We call x the independent variable as it can be any value in the domain. We call y the dependent variable as its value depends on x . Much like when you first encountered the variable x , function notation may be rather unsettling. But the more you use the notation, the more familiar you become with the notation, and the more comfortable you will be with it. Let’s review the equation y = 4 x − 5 . To find the value of y when x = 2 , we know to substitute x = 2 into the equation and then simplify. y = 4 x − 5 Let x = 2 . y = 4 ⋅ 2 − 5 y = 3 The value of the function at x = 2 is 3. We do the same thing using function notation, the equation y = 4 x − 5 can be written as f ( x ) = 4 x − 5 . To find the value when x = 2 , we write: f ( x ) = 4 x − 5 Let x = 2 . f ( 2 ) = 4 ⋅ 2 − 5 f ( 2 ) = 3 The value of the function at x = 2 is 3. This process of finding the value of f ( x ) for a given value of x is called evaluating the function . Evaluating the Function For the function f ( x ) = 2 x 2 + 3 x − 1 , evaluate the function. f ( 3 ) f ( − 2 ) f ( a ) To evaluate f ( 3 ) , substitute 3, for x . Simplify. f ( x ) = 2 x 2 + 3 x − 1 f ( 3 ) = 2 ( 3 ) 2 + 3 ⋅ 3 − 1 f ( 3 ) = 2 ⋅ 9 + 3 ⋅ 3 − 1 f ( 3 ) = 18 + 9 − 1 f ( 3 ) = 26 To evaluate f ( − 2 ) , substitute − 2 for x . Simplify. f ( x ) = 2 x 2 + 3 x − 1 f ( − 2 ) = 2 ( − 2 ) 2 + 3 ( − 2 ) − 1 f ( − 2 ) = 2 ⋅ 4 + ( − 6 ) − 1 f ( − 2 ) = 8 + ( − 6 ) − 1 f ( − 2 ) = 1 To evaluate f ( a ) , substitute a for x . Simplify. f ( x ) = 2 x 2 + 3 x − 1 f ( a ) = 2 ( a ) 2 + 3 ⋅ a − 1 f ( a ) = 2 a 2 + 3 a − 1 Evaluating the Function in an Application The number of unread emails in Sylvia’s inbox is 75. This number grows by 10 unread emails a day. The function N ( t ) = 75 + 10 t represents the relation between the number of emails, N , and the time, t , measured in days. Find N (5). Explain what this result means. Find N (5). Explain what this result means. Substitute in t = 5 . Simplify. N ( 5 ) = 75 + 10 ⋅ 5 N ( t ) = 75 + 10 t N ( 5 ) = 75 + 50 N ( 5 ) = 125 If 5 is the number of days, N ( 5 ) is the number of unread emails after 5 days. After 5 days, there are 125 unread emails in Sylvia’s inbox. Determining If a Relation Is a Function with Different Representations We can determine whether a relation is a function by identifying the input and the output values. If each input value leads to only one output value, classify the relation as a function. If any input value leads to two or more outputs, do not classify the relation as a function. We will review three different representations of relations and determine if they are functions: ordered pairs, mapping, and equations. Determining If a Relation Is a Function with a Set of Ordered Pairs Use the set of ordered pairs to determine whether the relation is a function. { ( − 3 , 27 ) , ( − 2 , 8 ) , ( − 1 , 1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) } { ( 9 , − 3 ) , ( 4 , − 2 ) , ( 1 , − 1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 4 , 2 ) , ( 9 , 3 ) } Each x -value is matched with only one y -value. This relation is a function. The x -value 9 is matched with two y -values, both 3 and − 3 . This relation is not a function. A mapping is sometimes used to show a relation. The arrows show the pairing of the elements of the domain with the elements of the range. Consider the example of the relation between your friends and their birthdays used in . In this particular example, the domain is the set of people’s names, and the range is the set of their birthdays. This mapping was a function because everybody’s name maps to exactly one birthday. Determining If a Relation Is a Function with Mapping Use the mapping in to determine whether the relation is a function. Both Lydia and Marty have two phone numbers. Each x -value is not matched with only one y -value. This relation is not a function. In algebra functions will usually be represented by an equation. It is easiest to see if the equation is a function when it is solved for y . If each value of x results in only one value of y , then the equation defines a function. Determining If a Relation Is a Function with an Equation Determine whether each equation is a function. Assume x is the independent variable. 2 x + y = 7 y = x 2 + 1 x + y 2 = 3 2 x + y = 7 For each value of x , we multiply it by − 2 and then add 7 to get the y -value. For example, if x = 3 : y = − 2 x + 7 y = − 2 ⋅ 3 + 7 y = 1 We have that when x = 3 , then y = 1 . It would work similarly for any value of x . Since each value of x , corresponds to only one value of y the equation defines a function. y = x 2 + 1 For each value of x , we square it and then add 1 to get the y -value. For example, if x = 2 y = x 2 + 1 y = 2 2 + 1 y = 5 We have that when x = 2 , then y = 5 . It would work similarly for any value of x . Since each value of x corresponds to only one value of y , the equation defines a function. x + y 2 = 3 x + y 2 = 3 Isolate the y term. y 2 = − x + 3 Let us substitute x = 2 . y 2 = − 2 + 3 y 2 = 1 This gives us two values for y . y = 1 , y = − 1 We have shown that when x = 2 , then y = 1 and y = − 1 . It would work similarly for any value of x . Since each value of x does not corresponds to only one value of y the equation does not define a function. Relations and Functions Applying the Vertical Line Test We reviewed how to determine if a relation is a function. The relations we looked at were expressed as a set of ordered pairs, a mapping, or an equation. We will now cover how to tell if a graph is that of a function. An ordered pair ( x , y ) is a solution of a linear equation, if the equation is a true statement when the x -values and y -values of the ordered pair are substituted into the equation. The graph of a linear equation is a straight line where every point on the line is a solution of the equation, and every solution of this equation is a point on this line. we can see that in the graph of the equation y = 2 x − 3 , for every x -value there is only one y -value, as shown in the accompanying table. Graph of the Equation y = 2 x − 3 A relation is a function if every element of the domain has exactly one value in the range. The relation defined by the equation y = 2 x − 3 is a function. If we look at the graph, each vertical dashed line only intersects the solid line at one point. This makes sense as in a function, for every x -value there is only one y -value. If the vertical line hit the graph twice, the x -value would be mapped to two y -values, and so the graph would not represent a function. This leads us a graphical method of determining functions called the vertical line test , which states that a set of points in a rectangular coordinate system is the graph of a function if every vertical line intersects the graph in at most one point. If any vertical line intersects the graph in more than one point, the graph does not represent a function. Applying the Vertical Line Test Determine whether the graph ( ) is the graph of a function applying the vertical line test. On the graph ( ), only three vertical dashed lines are drawn. However, it can be determined that any vertical dashed line that is drawn will intersect the solid line at exactly one point. It is the graph of a function. Applying the Vertical Line Test to a Parabola Determine whether the graph is the graph of a function ( ). does not represent a function since the vertical dashed lines shown on the graph below intersect the solid line at two points. Determining the Domain and Range of a Function For the function y = f ( x ) , x is the independent variable as it can be any value in the domain, and y is the dependent variable since its value depends on x . For the function y = f ( x ) , the values of x make up the domain and the values of y make up the range. Finding the Domain and Range of Ordered Pairs For { ( 1 , 1 ) , ( 2 , 4 ) , ( 3 , 9 ) , ( 4 , 16 ) , ( 5 , 25 ) } : Find the domain of the relation. Find the range of the relation. The domain is the set of all x -values of the relation: { 1 , 2 , 3 , 4 , 5 } The range is the set of all y -values of the relation: { 1 , 4 , 9 , 16 , 25 } Finding the Domain and Range on a Graph Use to: List the ordered pairs of the relation. Find the domain of the relation. Find the range of the relation. The ordered pairs of the relation are: { ( 1 , 5 ) , ( − 3 , − 1 ) , ( 4 , − 2 ) , ( 0 , 3 ) , ( 2 , − 2 ) , ( − 3 , 4 ) } . The domain is the set of all x -values of the relation: { − 3 , 0 , 1 , 2 , 4 } . Notice that while − 3 repeats, it is only listed once. The range is the set of all y -values of the relation: { − 2 , − 1 , 3 , 4 , 5 } . Notice that while − 2 repeats, it is only listed once. Domain and Range on Graphs Function and Function Notation In 1673, Gottfried Leibniz, the German mathematician who co-invented calculus, seems to be the first person to use the word function in a mathematical sense, although his use of it does not exactly fit with the modern use and definition. The person who is credited with the modern definition of function is Swiss mathematician Johann Bernoulli, who wrote about it in a letter to Leibniz in 1698. Supposedly, Leibniz wrote Bernoulli back, approving of this use of the word. In 1734, the use of the notation f ( x ) for a function was first used by Swiss mathematician Leonhard Euler (pronounced “Oiler”). Euler had a knack for inventing notation. He also introduced the notation e for the base of natural logs (1727), i for the square root of − 1 (1777), ∑ for summation (1755), and many others. Euler also introduced many other ideas associated with functions. Euler defined exponential functions and defined logarithmic functions as their inverse; he also introduced the beta and gamma functions, and was the first person to consider the trigonometric identities (sine, cosine, etc.) as functions. Check Your Understanding Key Terms relation domain function mapping vertical line test Key Concepts A relation is any set of ordered pairs ( x , y ) . All of the x -values of the set are the domain, and all of the y -values of the set are the range. A relation is a function if each x -value in the domain is assigned to exactly one element in the range. A y -value in the range can have more than one x -value assigned to it; but each x -value can only be assigned to one y -value. For the function y = f ( x ) , f is the name of the function, x is the domain value variable, and y = f ( x ) is the range value variable. The vertical line test is a test that can be done on the graph of a relation to determine if it is a function. Videos Relations and Functions Domain and Range on Graphs

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