Chapter 4 Applying Linear Functions Many situations in real life can be represented mathematically. You can write equations, create tables, or even construct graphs that display real-life data. Part of understanding mathematics is learning how it fits into our everyday world. Can you create a story describing how fast a speedboat is going? How about a table showing the boat s total distance traveled over time? During This Chapter You will solve problems involving direct variation. You will find the equation of the line that passes through a given point. You will use linear functions to model data. Application This chapter explores how lines model real-world situations. Linear equations describe how some relationships work, and can be used to approximate others. Knowing the linear rule behind a relationship makes it possible to solve for particular moments or predict certain events. The similar bone structure in the forelimbs of whale and humans indicates a common ancestor. The similar streamlined body shape of a shark and a dolphin, however, does not. What types of evidence can we use to learn about evolution and the relationships among species?
Section 4.1 Direct Variation Objectives Determine the rate of change from verbal representations Identify and solve problems of direct variation for an unknown New Vocabulary Direct variation Proportional Proportionality constant Inverse variation Did you know that it takes 2 lb of apples to make one 9 in. apple pie? In other words, every time a chef wants to make one apple pie, 2 lb of apples will go into the recipe. By knowing this piece of information, we can determine how many pounds of apples we will need to make any given number of pies. For example, if a chef wanted to bake 16 apple pies, he would need 32 lb of apples. How many apples would you need to create 15 apple pies? How many pies could you make with 21 lb of apples?
4.1 - Direct Variation What is Direct Variation? When two variables are related in such a way that the ratio of their values always remains the same, the two variables are said to be in direct variation. Direct variation is a term used to describe a proportional relationship between two variables. Two variables are said to be proportional if a constant change in one is always accompanied by a constant change in the other. Jump to Decimals, Fractions, & Percentages To make a batch of lemonade, the juice of one lemon is mixed with 5 cups of sweetened water. Likewise, two batches of lemonade require two lemons and 10 cups of sweetened water. Figure 4.1-1 visually shows this proportional relationship. Figure 4.1-1 The relationship between the amount of water and lemon juice needed to make lemonade is proportional. Notice that every 5 cups of water requires the juice of one lemon, as shown in Figure 4.1-2. In this example, the total amount of sweetened water and the number of lemons used vary directly because the ratio of cups of water to the number of lemons is always 5. Figure 4.1-2 The ratio of water to lemons always simplifies to 5. This constant ratio is called the proportionality constant (k). Direct variation situations can be modeled algebraically using the formula y = kx. Knowing the proportionality constant will help to set up an algebraic equation to solve future problems. For the lemonade scenario, the constant ratio is k = 5, so the direct variation formula would be y = 5x. Verbal representation y varies directly with x, and y = 10 when x = 2 Algebraic representation y = 5x Visual representation Direct variation relationships can be expressed verbally by stating that y varies directly with x. By finding the ratio y/x for given values of x and y, the proportionality constant (k) can be determined. The direct variation relationship between lemons and sweetened water used to make lemonade is represented verbally, algebraically, and visually in Table 4.1-1. Table 4.1-1 This table shows various representations of the direct variation between lemons and sweetened water used to make lemonade. Chapter 4 - Applying Linear Functions 231 Section continues
4.1 - Direct Variation Example problem"" If y varies directly with x, and x = 12 when y = 108, then what is " " " " the proportionality constant? Analyze" " " The problem mentions direct variation between two variables. It is " " " " asking for the proportionality constant (k) given values for x and y. Formulate" " " To find the proportionality constant, use the equation k = y x " " " " and substitute in the values given. Determine " " " k = y x " " " " k = 108 12 = 9"" Substitute given values and solve. Justify" " " Check to see if this is valid by creating an algebraic representation " " " " of the direct variation relationship. Since k = 9, the direct variation " " " " relationship is the equation y = 9x. Substitute in the values of x " " " " and y to check the proportionality constant. " " " " y = 9x " " " " 108 = 9(12) " " " " 108 = 108 Evaluate" " " The solution k = 9 is a reasonable solution. The ratio of y/x " " " " provides the proportionality constant as the ratio is derived from " " " " the direct variation formula y = kx. Chapter 4 - Applying Linear Functions 232 Section continues
4.1 - Direct Variation Slope & Proportionality Constant The proportionality constant can also be found when given a table of values. Table 4.1-2 displays the amounts of lemons and sweetened water needed for various amounts of lemonade. Dividing any y-value other than 0 by its corresponding x-value in the table will result in the proportionality constant. Number of lemons (x) Cups of sweetened water (y) Proportionality constant (k) 0 1 2 3 4 0 5 10 15 20 0 0 is undefined" 5 1 = 5 10 2 = 5 15 3 = 5 20 4 = 5 Table 4.1-2 This table displays various combinations of lemons and sweetened water that need to be mixed to make lemonade. Notice that the ratio y/x for each ordered pair is equal to 5. Notice that the value (0,0) is in Table 4.1-2. This value will always be present in a direct variation problem, as 0 multiplied by the proportionality constant will always be 0. In context of the lemonade scenario, this means that if there are no lemons, no water will be needed. Remember that dividing 0 by itself is undefined and will not yield the proportionality constant. Direct variation equation y = kx Linear function y = mx + b Jump to Number Properties The proportionality constant (k) found in direct variation equations and the slope of a linear equation (m) are very similar in definition. Both k and m are constant rates of change. Thus, a direct variation equation and a linear equation can have the same value. However, a direct variation equation will always pass through the origin while a linear equation can have a different y-intercept. Table 4.1-3 illustrates some of the similarities and differences in direct variation equations and linear equations. Proportionality constant k = 2 y-intercept at (0, 0) Slope m = 2 y-intercept at (0, 3) Jump to Slope Table 4.1-3 Notice that proportionality constants and slopes can be found in similar ways. Direct variation equations always contain the origin and cannot have any other y-intercept. Chapter 4 - Applying Linear Functions 233 Section continues
4.1 - Direct Variation Example problem"" Given the table below, find the proportionality constant. x 2 0 2 4 6 y 1.5 0 1.5 3 4.5 Analyze" " " The problem asks for the proportionality constant, which means " " " " the given table models direct variation. The table also includes the " " " " origin as a point. Formulate" " " To find the proportionality constant from a table, find the ratio " " " " " y/x for a set of x- and y-values. Do not use the point (0,0) because " " " " division by 0 is undefined. Determine" " " k = y x = 1.5 = 0.75" 2 Substitute a coordinate pair from " " " " " " " " table and simplify. Justify " " " Check to see if k is correct by using other values from the table. In " " " " each instance, the result is 0.75. " " " " 3 4 = 0.75, 4.5 6 = 0.75, 1.5 2 = 0.75 Evaluate " " " This method solves the problem quickly, and checking that the " " " " ratio is constant between terms is important to verify direct " " " " variation. Chapter 4 - Applying Linear Functions 234 Section continues
4.1 - Direct Variation Direct Variation Graphs Direct variation relationships have two key features that can be seen when graphed. First, direct variation graphs include the origin, which is (0,0). Second, direct variation graphs between x and y have a constant rate of change (the proportionality constant), which means their graphs are linear. Because a direct variation relationship is tightly defined, graphs of these relationships can be formed given very little information. Table 4.1-4 details how to create a graph given only the proportionality constant. Jump to Slope-Intercept Form Graphs can be used to predict solutions by extending their axes. The axes have been extended in Figure 4.1-3 to show more of the coordinate plane. With a larger viewing space, points such as (8,40) that share the proportionality constant can be seen. Creating a direct variation graph given only k = 5 Draw a coordinate plane and plot the point (0, 0) since all direct variation equations contain the origin. From the origin, use the proportionality constant as the slope to create a second point. Use these two points to create a line, forming the graph of the direct variation situation. Figure 4.1-3 By extending the axes of the graph, larger values of needed ingredients can be determined. Table 4.1-4 Graphs of direct variation situations can be created knowing only the proportionality constant. Chapter 4 - Applying Linear Functions 235 Section continues
4.1 - Direct Variation Example problem"" A car is traveling cross-country on Route 66 at a constant speed. " " " " In an hour, the car has driven 65 mi. Construct a graph that will " " " " predict how many miles the car will travel after 4 h on Route 66. Analyze" " " The problem gives the information that the car has traveled for " " " " 1 h and went 65 mi. If the car traveled 0 h, the car would go 0 mi. " " " " Since the car is driving at a constant speed and passes through " " " " the point (0,0), then the relationship between time and distance " " " " traveled is a direct variation. Formulate" " " In order to use a graph to estimate the distance the car will travel " " " " after 4 h, use the proportionality constant and one point. " " " " Since the proportionality constant is the ratio y/x, then " " " " k = 65 = 65. The point (0,0) is also on every direct variation 1 " " " " graph. " " " " Construct a graph with this information and estimate y when " " " " x = 4. Determine" " " Plot (0,0) and use the proportionality constant k = 65 1 " " " " second point. to create a Chapter 4 - Applying Linear Functions 236 Problem continues
4.1 - Direct Variation " " " " Draw a line through the points above and extend it until the value " " " " x = 4 is reached. " " " " " " " " " The graph shows that " " " " y = 260 when x = 4. Justify" " " You can use a list of values to determine if your solution is correct. " " " " If 1 h = 65 mi, then 2 h = 130 mi and 4 h = 260 mi. Evaluate" " " Number sense confirms the reasonableness of traveling 260 mi in " " " " 4 h. It can be observed on the graph that every hour adds 65 mi to " " " " the distance driven. It is important to be as accurate as possible " " " " when creating graphs to predict solutions. Chapter 4 - Applying Linear Functions 237 Problem complete
4.1 - Direct Variation Direct Variation Equations Creating equations from direct variation situations is a helpful way to find answers to problems that might be time consuming to graph. For example, if a person managed to procure 1200 lemons, how much sweetened water might be needed to create lemonade? Extending a graph or table of values to encompass an x-value of 1200 is not the most efficient means to answer this question. In this situation, it is far quicker to use a direct variation equation. Direct variation equations between x and y are always of the form y = kx, where k is the proportionality constant. Recall that k = 5 when making lemonade, where y is cups of water and x is lemons. Thus, we can use the formula y = 5x to find how much water is necessary for 1200 lemons. y = 5(1200) = 6000" It would take 6000 cups of water to make lemonade with 1200 lemons. Chapter 4 - Applying Linear Functions 238 Section continues
4.1 - Direct Variation Example problem"" The total cost of filling up a car with gasoline varies " " " " directly with the number of gallons of gas purchased. If a " " " " gallon of gas costs $3.28, how many gallons can be " " " " purchased with $18? Analyze" " " The problem is asking how much gas can be purchased " " " " with $18. The problem also states that gasoline is $3.28 " " " " per gallon, which provides the proportionality constant " " " " k = 3.28/1 = 3.28. Using k, a direct variation equation can " " " " be created and used to find x when y = 18. Formulate" " " Using k = 3.28, the direct variation equation is y = 3.28x. " " " " Substitute y = 18 into this equation to solve for x. Determine" " " y = 3.28x " " " " 18 = 3.28x" " Substitute 18 for y. " " " " 18 3.28 = x" " Solve for x and simplify. " " " " x = 5.49 gal Justify" " " To determine the validity of the answer, substitute 5.49 back into " " " " the original equation: 3.28(5.49) = 18. You can also check your " " " " answer by using estimation. Evaluate" " " The process was quick, provided a concrete answer, and could be " " " " repeated for other amounts of money. The answer is a reasonable " " " " quantity of gasoline. Chapter 4 - Applying Linear Functions 239 Section continues
4.1 - Direct Variation Example problem"" If y varies directly with x, and y = 102 when x = 17, find y when x = 12. Analyze" " " This is a direct variation problem based on the vocabulary used. One ordered pair is given, " " " " (17,102). The proportionality constant is not provided. Formulate" " " Use the ordered pair to find the proportionality constant, k = y. Create an equation of the form x " " " " y = kx and substitute in x = 12 to find the corresponding y-value. Determine" " " k = 102 17 = 6"" Find the constant of proportionality. " " " " y = 6x" " Create a direct variation equation. " " " " y = 6(12)" " Substitute x = 12 into the direct variation equation and simplify for y. " " " " y = 72 Justify" " " To determine the validity of the answer, test to see if the ratio of the new point is the same. " " " " 72 12 = 6 and 102 17 = 6 " " " " The answer can also be verified using a graphing calculator by plotting the points and graph on " " " " the same coordinate plane. Evaluate" " " The process used the direct variation information from the problem to establish a relationship that " " " " could be examined algebraically. The answer is reasonable because it should be less than 102 since " " " " the x-value (12) is less than 17. Chapter 4 - Applying Linear Functions 240 Section continues
4.1 - Direct Variation Inverse Variation Not all situations model direct variation where the constant (k) is the quotient of the y- and x-coordinates. Inverse variation exists when two variables are related in such a way that the product of their values is a nonzero constant. Any inverse variation between variables x and y can be represented by the equation xy = k or y = k. This resembles the form of an equation describing direct variation, but x it can now be seen that one variable varies directly with the reciprocal of the other variable. One example of a function exhibiting inverse variation is shown in Table 4.1-5. For each pair of x- and y-coordinates, the product xy is exactly 20. x 2 4 5 10 20 y 10 5 4 2 1 Table 4.1-5 The values of x and y in this table demonstrate inverse variation. Examining the values in Table 4.1-5 shows that when one variable increases, the other decreases. This behavior is a fundamental characteristic of inverse variation. Because the product xy must be constant, any increase in x must be compensated for by a decrease in y. The constant of proportionality (k) can be found for any relation exhibiting inverse variation as long as a pair of corresponding values is known. For example, suppose that variables a and b vary inversely, and it is known that a = 4 when b = 10. The proportionality constant can readily be found by substituting the known values into the equation ab = k. In this case, 4 10 = 40 = k. This constant can then be used in order to solve for other values in the relation with the equation xy = 40, or y = 40, as seen in Figure x 4.1-4. Inverse variation graphs differ from direct variation graphs in that they are not linear, do not pass through the origin, and are undefined for x = 0 and y = 0. Figure 4.1-4 Inverse variation graphs differ from direct variation graphs in the fact that they do not contain the point (0,0). Chapter 4 - Applying Linear Functions 241 Section continues
4.1 - Direct Variation Suppose that the variables x and y vary inversely. If the proportionality constant is known, then the value of y can be found for any corresponding value of x. For example, if k = 36 and x = 4, then y can be calculated by solving the inverse variation relationship xy = k for the unknown variable y, producing y = k/x. Substituting the known values of k and x into the equation results in the solution y = 36/4 = 9. In this way, the value corresponding to any known variable value can be determined as long as the proportionality constant is known. If the proportionality constant is not known, solutions to an inverse variation relation can be found as long as one corresponding pair of values is provided. This is done by substituting the pair of values into the inverse variation equation, given in Figure 4.1-5, and determining k. Once k is known, the corresponding value to any given value of one of the variables can be calculated. For example, suppose it takes 3 h for four men to assemble a structure. The principle of inverse variation can be used in order to determine how long it would take for six men to assemble the same structure, as pictured in Figure 4.1-6. Substituting the known corresponding values for the number of workers (w) and the time required (t) into the equation wt = k results in the constant of proportionality (k). In this problem, k = (4 workers)(3 h) = 12 man-hours. The inverse variation equation can then be used again, but this time be solved for the unknown quantity, t = 12/w. Substituting in the known values results in 12 man-hours t = = 2 h. Thus, six workers are needed to assemble the device in 2 h. 6 workers Figure 4.1-5 Once the constant of proportionality is known, the corresponding value to any x or y other than 0 can be calculated. Figure 4.1-6 The six men pictured here can finish the assembly in 2 h, which is faster than if there were only four men. Chapter 4 - Applying Linear Functions 242 Section complete
Section 4.2 Arithmetic Sequences Objectives Identify arithmetic sequences and express them with a linear function Use a recursively defined sequence to determine individual elements Convert between sequences defined functionally and recursively New Vocabulary Sequence Term (of a sequence) Subscript Arithmetic sequence Recursive rule Some groups of objects are arranged in a certain order. Look at this collection of stars. Is there a logical order that they should be arranged in? Do you think any stars are missing from this group? Can these patterns be described mathematically?
4.2 - Arithmetic Sequences Patterns in Sets of Numbers A sequence is a list of items, frequently numbers, that are arranged in a particular order. An element of a sequence is called a term. Usually, the terms of a sequence follow a pattern. For example, the sequence 2, 4, 6, 8, 10 has five terms, and each term can be found by adding 2 to the previous term. A sequence is not the same as a set, although they are both collections of items. The elements of a set are not arranged in any particular order, but the terms of a sequence are always arranged in a definite order. Sequences are sometimes written with parentheses and sometimes without, but not in braces. Jump to Set Look at the letters shown in Figure 4.2-1. Although this picture shows all the letters of the alphabet, the letters are out of order. Although all the letters of the alphabet are here, this is not a properly shown alphabet because the letters are in the wrong order. The alphabet is an example of a sequence. It is not only a collection of letters but also an ordering of those letters. Some sequences contain only a limited number of terms. The sequence 2, 4, 6, 8, 10 is an example of this. It is a finite sequence because it contains a limited number of terms. On the other hand, some sequences go on indefinitely. An example of this would be a sequence of all positive even numbers (2, 4, 6, 8, 10, ). This is an infinite sequence because it never comes to an end. Ellipses (three periods in a row) can be used when the pattern of the numbers is clear or described, to indicate that not all the terms are written. As a result, the sequence has an infinite number of terms. Figure 4.2-1 This figure shows the letters of the alphabet, but with some errors in the order of the letters. Because the terms of a sequence are arranged in a certain order, each term can be assigned a number representing its position in the sequence. These numbers are written as subscripts. A subscript is a symbol that is written smaller and lower than the surrounding characters. The terms of a sequence are sometimes referred to in the form a 1, a 2, a 3,, where the subscripts number the terms in the sequence. In this form, a 5 would represent the fifth term in the sequence. There are other numbering conventions for identifying the terms of a sequence. Often, the first term of the sequence is labeled a 1, but any integer could be used to start the indexing. For example, if a sequence s n was identified as 3, 6, 9, 12 and the first term of the sequence was identified as s 0 = 3, then s 3 would actually be the fourth term of the sequence because s 1 = 6, s 2 = 9, and s 3 = 12. Chapter 4 - Applying Linear Functions 244 Section continues
4.2 - Arithmetic Sequences Example problem"" Given the sequence 1, 2, 4, 7, 11, 16, 24, where a 1 = 1, identify the terms a 3 and a 6. Analyze" " " The problem asks for the terms a 3 and a 6 of the sequence 1, 2, 4, 7, 11, 16, 24. The term a 1 refers to " " " " the first value of the sequence. Formulate" " " Number the terms of the sequence to identify the third and sixth terms. Determine" " " Number the terms of the sequence. a 1 a 2 a 3 a 4 a 5 a 6 a 7 1 2 4 7 11 16 24 " " " " " " " " a 3 = 4"" " " " " Identify the third and sixth terms. " " " " " " " " a 6 = 16 Justify" " " Because the subscripts identify the order of the terms, a 3 and a 6 were identified by counting the " " " " terms from the beginning of the sequence. Evaluate" " " Numbering the terms of the sequence provided a straightforward way to identify the terms. The " " " " answer is reasonable because the numbers 4 and 16 are the third and sixth terms. Chapter 4 - Applying Linear Functions 245 Section continues
4.2 - Arithmetic Sequences Common Differences Consider the sequence 2, 4, 6, 8, 10. The terms of this sequence are evenly spaced on the number line. After the initial term, 2, each subsequent term is 2 greater than the previous term. Therefore, the difference between successive terms is always the same. Such a sequence is said to have a common difference. Many sequences follow an identifiable pattern, and some sequences can be classified according to the type of pattern they follow. One type of sequence is called an arithmetic (pronounced air-ith-met-ic ) sequence. An arithmetic sequence is a sequence whose successive terms have a common difference. In such a sequence, the difference between successive terms is always the same. That is, a n a n 1 = k. Consider the sequence 10, 13, 16, 19. Each term of this sequence can be found by adding 3 to the preceding term. Thus, there is a common difference, and this is an arithmetic sequence. Next, consider the sequence 8, 10, 13, 17. The difference between the first two terms is 2, the difference between the second and third terms is 3, and the difference between the last two terms is 4. There is no common difference, and so this sequence is not arithmetic. The differences in these sequences are illustrated in Figure 4.2-2. The common difference of an arithmetic sequence can be positive, negative, or even 0. If it is negative, then the numbers in the sequence will become progressively smaller. For example, the sequence 6, 4, 2, 0, 2, 4 has a common difference of 2. Each term of this sequence is 2 less than the preceding term, and so it is an arithmetic sequence. The sequence 5, 5, 5, 5, 5 can be considered an arithmetic sequence with a common difference of 0. Figure 4.2-2 The sequence 10, 13, 16, 19 is arithmetic because consecutive terms have a common difference, but the sequence 8,10,13,17 is not arithmetic because there is not a common difference. Chapter 4 - Applying Linear Functions 246 Section continues
4.2 - Arithmetic Sequences Example problem"" Find the seventh term of the arithmetic sequence 1, 4, 7, Analyze" " " The problem presents an infinite arithmetic sequence whose first three " " " " terms are 1, 4, and 7. It asks for the seventh term. Formulate" " " Determine the common difference (d) of the sequence. Repeatedly " " " " add this number to the last known term until the seventh term is " " " " determined. Determine" " " d = 7 4" " " Subtract consecutive terms. " " " " d = 3 " " " " a 4 = 7 + 3 = 10" " Add 3 to a 3 to find a 4. " " " " a 5 = 10 + 3 = 13" " Add 3 to a 4 to find a 5. " " " " a 6 = 13 + 3 = 16" " Add 3 to a 5 to find a 6. " " " " a 7 = 16 + 3 = 19" " Add 3 to a 6 to find a 7. Justify" " " Because the sequence is arithmetic, each term was calculated by " " " " adding the common difference to the previous term. Evaluate" " " Repeatedly adding the common difference provided a clear way " " " " to determine successive terms of the sequence. The answer is " " " " reasonable because it is the seventh term of the arithmetic " " " " sequence. Chapter 4 - Applying Linear Functions 247 Section continues
4.2 - Arithmetic Sequences Example problem"" Determine whether the sequence 4, 6, 9, 11, 13, 16 is an arithmetic " " " " sequence. Analyze" " " The problem presents the finite sequence 4, 6, 9, 11, 13, 16 and asks " " " " if it is arithmetic. Formulate" " " Subtract pairs of consecutive terms to determine whether there is " " " " a common difference. Determine" " " 6 4 = 2" " " Subtract consecutive terms. " " " " 9 6 = 3 " " " " 11 9 = 2 " " " " 13 11 = 2 " " " " 16 13 = 3 " " " " 2 3" " " " Compare the differences. Justify" " " Because the differences between consecutive terms are not the " " " " same, the sequence is not arithmetic. Evaluate" " " Comparing the differences between consecutive terms provided a " " " " quick way to determine whether the sequence was arithmetic. The " " " " answer is reasonable because the sequence does not follow the " " " " definition of an arithmetic sequence. Chapter 4 - Applying Linear Functions 248 Section continues
4.2 - Arithmetic Sequences Using a Recursive Rule (Arithmetic Sequences) A tournament bracket is shown in Figure 4.2-3. The number of teams in each round can be described by the sequence 16, 8, 4, 2, 1. Note that each term is half of the previous term. Because each term after the initial term can be calculated in a certain way from the previous term, it is possible to write a formula that can be used to calculate each term. For this sequence, the formula a n = a n 1 for a 1 = 16 and n = 1, 2, 3, 4, 5 shows that each term is equal 2 to the previous term divided by 2. The notation a n describes the term being calculated, and a n 1 describes the preceding term. A formula that describes the next term of a sequence using the previous term is called a recursive rule. When using a recursive rule, at least one term of the sequence must be specified. The sequence 16, 8, 4, 2, 1 is not an arithmetic sequence because there is not a common difference between consecutive terms. Although recursive rules can be used to describe many different kinds of sequences, the recursive rules for arithmetic sequences are all similar in form. A ruler with inch markings is shown in Figure 4.2-4. Each mark corresponds to a length of 1/32 of an inch greater than the previous mark. Because the difference between consecutive marks is the same across the entire ruler, these marks form an arithmetic sequence. This sequence can be described by the recursive rule a n = a n 1 + 1 32 for a 1 = 0. This formula states that any term can be found by adding 1/32 to the previous term. For example, if applied to the 25th term, the recursive rule takes the form a 25 = a 24 + 1 32. Note that the common difference of the sequence appears in the recursive rule. In the sequence of the ruler markings, the common difference is 1/32, and the recursive rule defines each term as equal to the previous term plus 1/32. Imagine a sequence that follows the recursive rule a n = a n 1 + 5. Because each term is equal to the previous term plus 5, it can be seen that the common difference of this sequence is 5. As a result, all arithmetic sequences follow the recursive rule a n = a n 1 + d where d is the common difference between consecutive terms. Again, at least one term must be specified. Figure 4.2-3 This is a figure of a tournament bracket. The number of teams in each round is described by the recursive rule a n = a n 1 2. Figure 4.2-4 This is a picture of a ruler with inch markings. The markings on this ruler follow an arithmetic sequence. Chapter 4 - Applying Linear Functions 249 Section continues
4.2 - Arithmetic Sequences Example problem"" Find the fifth term of the sequence that follows the recursive rule " " " " a n = a n 1 + 4 if a 1 = 2. Analyze" " " The problem presents an arithmetic sequence with a first term of 2 " " " " and common difference of 4, and it asks for the fifth term. Formulate" " " Beginning with the first term, repeatedly add 4 until the fifth term " " " " is determined. Determine" " " a 1 = 2 " " " " a 2 = 2 + 4 = 6" " " Add 4. " " " " a 3 = 6 + 4 = 10" " " Add 4. " " " " a 4 = 10 + 4 = 14" " " Add 4. " " " " a 5 = 14 + 4 = 18" " " Add 4. Justify" " " Because the sequence is arithmetic, the common difference of 4 " " " " was added to each term in order to find the value of the next term. Evaluate" " " This process required only one operation to be performed, but it " " " " was performed several times. The answer is reasonable because " " " " each term obeys the recursive rule a n = a n 1 + 4. Chapter 4 - Applying Linear Functions 250 Section continues
4.2 - Arithmetic Sequences Finding a Function Rule (Arithmetic Sequences) Consider the sequence 2, 6, 10, 14, This is an arithmetic sequence with a common difference of 4. The 11th term of this sequence could be found by writing out the first 11 terms. This process would involve adding four to the first term 10 times, as shown in Figure 4.2-5. However, because adding four 10 times is the same as adding 4(10), it is not necessary to find the intermediate terms. The 11th term is 2 + 4(10), or 42. It can be calculated directly from the first term and the common difference. The sequence in Figure 4.2-5 can be described by the recursive rule a n = a n 1 + 4, a 1 = 2. However, using this formula to find the 11th term of the sequence first requires that the 10th term be known, and then the 9th, and each preceding term as well. This process is simplified by writing a formula that enables any term of a sequence to be calculated without first needing to know the previous terms. Such a formula is known as an explicit formula because any term can be calculated directly, without needing preceding terms to be calculated. It is also called a function rule because each term is an output value that can be calculated from input values such as the first term and the common difference. Figure 4.2-5 This figure shows the first 11 terms of the arithmetic sequence a n = a n 1 + 4, where a 1 = 2. Look again at the sequence 2, 6, 10, 14, Beginning with the first term, the second term can be found by adding 4 one time. The third term can be found by adding 4 two times, and the fourth term can be found by adding 4 three times. By extension, the nth term can be found by adding 4 to the first term n 1 times. This concept is represented by the formula a n = 2 + 4(n 1). Notice that both the first term of the sequence, 2, and the common difference, 4, appear in this formula. As a result, this formula can be generalized to represent any arithmetic sequence, taking the form a n = a 1 + d(n 1), where a 1 is the first term and d is the common difference. Look at the Fahrenheit degree markings on the thermometer shown in Figure 4.2-6. The first mark is at 60 F, and each subsequent mark labels a temperature 2 F higher than the previous one. As a result, the marks form an arithmetic sequence that follows the recursive rule a n = a n 1 + 2, a 1 = 60. This sequence could instead be described by a function rule. Substituting the values for the first term and the common difference into the formula a n = a 1 + d(n 1) produces the rule a n = 60 + 2(n 1). This function can be used to find the value of any marking. For example, the temperature labeled by the 45th mark can be found by making the substitution n = 45, which produces a 45 = 60 + 2(45 1), or a 45 = 28. Thus, the 45th mark corresponds with the temperature of 28 F. Figure 4.2-6 The Fahrenheit degree markings on this thermometer form an arithmetic sequence that follows the rule a n = 60 + 2(n 1). Chapter 4 - Applying Linear Functions 251 Section continues
4.2 - Arithmetic Sequences Example problem"" Find the 21st term of the sequence a 1 = 20, a n = a n 1 + 5. Analyze" " " The problem presents an arithmetic sequence with a first term of 20 and a common " " " " difference of 5. It asks for the 21st term of this sequence. Formulate" " " Use the formula a n = a 1 + d(n 1) and the substitutions n = 21, d = 5, and a 1 = 20 to " " " " calculate a 21. Determine" " " a n = a 1 + d(n 1) " " " " a 21 = 20 + 5(21 1)" " " Substitute 21 for n, 5 for d, and 20 for a 1. " " " " a 21 = 20 + 5(20)" " " " Simplify. " " " " a 21 = 20 + 100 " " " " a 21 = 80 Justify" " " Because the sequence was arithmetic, the formula a n = a 1 + d(n 1) could be used to " " " " calculate the 21st term. Evaluate" " " The formula a n = a 1 + d(n 1) provided a quick way to solve directly for the 21st term. The " " " " answer is reasonable because it is obtained by adding the common difference to the first " " " " term 20 times. Solving the problem with a function rule was more efficient than solving the " " " " problem with a recursive rule because it did not take many steps. Chapter 4 - Applying Linear Functions 252 Section continues
4.2 - Arithmetic Sequences Example problem"" Find the function rule that describes the sequence 17, 13, 9, 5, Analyze" " " The problem presents an infinite sequence whose first four terms are 17, 13, 9, and 5, and " " " " asks for the explicit formula. Formulate" " " Find the common difference. Then substitute the common difference and the first term into " " " " the formula a n = a 1 + d(n 1). Determine" " " d = 13 17" " " Subtract consecutive terms to find the common difference. " " " " d = 4 " " " " a n = a 1 + d(n 1)" " Substitute 4 for d and 17 for a 1. " " " " a n = 17 + ( 4)(n 1) " " " " a n = 17 4(n 1)" " Simplify. Justify" " " Because the sequence is arithmetic, the explicit formula follows the form a n = a 1 + d(n 1), " " " " where a 1 is the first term and d is the common difference. Evaluate" " " Beginning with the generic form a n = a 1 + d(n 1) allowed the function rule to be found by " " " " making only two substitutions. The answer is reasonable because it follows the form of a " " " " function rule that describes an arithmetic sequence. Chapter 4 - Applying Linear Functions 253 Section complete
Section 4.3 Equation of a Line Objectives Find the equation of a line given two specific characteristics of the line Use the relationship of parallel and perpendicular lines to find a slope In the sunburst to the right, the multicolored rays of light depict a family of lines passing through a common point, the sun. What information can we use to distinguish a specific line from all the other lines through the same point?
4.3 - Equation of a Line Slope & Initial Condition Two pieces of information are needed in order to determine the equation of a specific line, but different types of information may work equally well. Knowing the coordinates of two different points on the line provides one way to determine its equation. Knowing one point on the line and the slope of the line also determines the equation of that line. The known point may be the point (a, 0) corresponding to the x-intercept a (the root, if it exists), or it may be the point (0, b) corresponding to the y-intercept b (the initial condition), or it may be an arbitrary point ( x 1, y 1). Any point will do as long as the slope m is known. Jump to Initial Conditions & y-intercepts The slope alone is not enough to determine the equation of a line. Specifying the slope by itself determines a family of parallel lines rather than a specific line. Once the slope is known, specifying a point on the line selects a particular line from the family of parallel lines. For example, the concrete ridges on the outside of the Figure 4.3-1 The concrete ridges on the outside of this building form a building shown in Figure 4.3-1 form a family of parallel lines. All of these lines have family of parallel lines with the same slope. the same slope. Suppose that one of the ridges is in need of repair. How will the maintenance crew know which ridge to fix? The damaged ridge can be identified by specifying a point on the line corresponding to that ridge. If the crew is asked to fix the fifth ridge up from the bottom left corner of the photo, that will select a specific ridge out of all the ridges shown. Let the bottom left corner of the photo in Figure 4.3-1 represent the origin of an xy-coordinate system. Assume that the lines are spaced approximately one unit apart along each axis. The fifth line up from the origin then has a y-intercept of 5. The rise of each line is roughly equal to its run, which means the slope of each line is approximately equal to 1. This provides enough information to determine the equation of the line in slope-intercept form. Letting m = 1 and b = 5 in the slope-intercept form y = mx + b produces the equation y = x + 5. This is the equation of the line corresponding to the particular ridge in need of repair. The more general equation y = x + b represents the entire family of parallel lines formed by all the ridges on the outside of the building. Jump to Slope-Intercept Form Chapter 4 - Applying Linear Functions 255 Section continues
4.3 - Equation of a Line Example problem"" A camera shop uses an online payment processing service to " " " " accept credit card payments for merchandise sold through its " " " " " online store. The online payment processing service costs the shop " " " " " a flat fee of 30 cents for each transaction plus 2.9% of the amount " " " " " of the sale. How much does it cost the shop to process a credit " " " " card payment of $1,000? Analyze" " " Let x represent the dollar amount of the sale and y represent the dollar " " " " cost of payment processing. The relationship between x and y is linear, " " " " with slope 0.029 and y-intercept 0.30. The problem asks for the value of y " " " " when x = 1000. Formulate" " " Substitute the slope and y-intercept into the slope-intercept form " " " " y = mx + b. Then let x = 1000 and calculate the value of y. Determine" " " y = mx + b " " " " y = 0.029x + 0.30" " " " Let m = 0.029 and b = 0.30. " " " " y = 0.029(1000) + 0.30" " " Now let x = 1000. " " " " y = 29.30" " " " " Simplify to calculate y. Justify" " " It costs the camera shop $29.30 to process a credit card payment of " " " " $1,000 using the online payment processing service. This follows from " " " " the calculations because x denotes the amount of the sale and y denotes " " " " the cost of payment processing. Evaluate" " " The process is effective because it leads step-by-step to a clear solution " " " " to the problem. The answer is reasonable because it is obtained by " " " " substituting correct values into the slope-intercept form of the equation " " " " of a line and calculating the numerical result. Chapter 4 - Applying Linear Functions 256 Section continues
4.3 - Equation of a Line Slope & Point Recall that the slope by itself describes a family of parallel lines. Specifying a point in addition to the slope selects a particular line from that family. Although the y-intercept b is sometimes used to determine the equation of a specific line with slope m, any arbitrary point ( x 1, y 1) can be used to determine the equation of a specific line when the slope is known. The photo of a grating pattern in Figure 4.3-2 depicts two families of parallel lines. Of the 10 lines shown, half of the lines slant upward and the other half slant downward. The upward-slanting lines have roughly equal rise and run, which means they all have slope approximately equal to 1. For the downward-slanting lines, the rise is roughly equal in size to the run but points down instead of up. All the downward-slanting lines therefore have slope approximately equal to 1. Specifying a slope of 1 selects one of the two families of parallel lines and reduces the number of possible lines to five. Specifying that the line also passes through the center of the photo selects a particular line out of those five. Thus, the center point and the slope of 1 together determine a unique line among the 10 lines shown. Let the bottom left corner of the photo in Figure 4.3-2 represent the origin of an xy-coordinate system. Notice that the lines are spaced approximately two units apart along each axis. This means that the center point has coordinates (3, 3). The point (3, 3) and the Figure 4.3-2 This photo of a grating pattern depicts two families of parallel lines. slope 1 provide enough information to determine the equation of the line in point-slope form. Letting x 1 = y 1 = 3 and m = 1 in the point-slope form y y 1 = m(x x 1 ) produces the equation y 3 = (x 3). This equation represents the downward-slanting line through the center of the photo. Solving the equation for y yields y = (x 3) + 3. Simplifying the right-hand side gives y = x + 6, which is the slope-intercept form of the equation. From the slope-intercept form, it can be determined by inspection that the y-intercept is 6. Counting upward from the origin in the bottom left corner of the photo confirms that the y-intercept of the downward-slanting center line is indeed 6. This shows that the point-slope form y 3 = (x 3) and the slope-intercept form y = x + 6 both describe the same line. Jump to Point-Slope Form Chapter 4 - Applying Linear Functions 257 Section continues
4.3 - Equation of a Line Example problem"" A local market takes grocery orders by phone and delivers the groceries to the customer for " " " " a fee. The market charges a fixed amount to prepare the order for delivery plus an " " " " additional $1.50 per mile to deliver the order. If a customer living eight miles away from the " " " " market pays $17.00 total for order preparation and delivery, how much does the market " " " " charge to prepare each order? Analyze" " " Let x represent the number of miles between the market and the customer. Let y represent the total " " " " dollar cost of preparing and delivering the order. The cost y is a linear function of the distance x. " " " " This line has slope 1.5 and passes through the point (8, 17). The problem asks for the cost of " " " " preparing each order. This cost is the value of y when x = 0, which is the y-intercept of the line. Formulate" " " Use the given information and the point-slope form y y 1 = m(x x 1 ) to determine the equation of " " " " the line. Then convert the equation into slope-intercept form y = mx + b to determine the " " " " y-intercept b. Determine" " " y y 1 = m(x x 1 ) " " " " y 17 = 1.5(x 8)"" " Let y 1 = 17, m = 1.5, and x 1 = 8. " " " " y = 1.5(x 8) + 17"" " Solve for y. " " " " y = 1.5x + 5"" " " Simplify the right side. " " " " y = 1.5(0) + 5" " " Let x = 0. " " " " y = 5" " " " " Simplify the right side. Justify" " " When x = 0, there are no miles to drive to deliver the order. The cost y comes entirely from " " " " preparing the order. Because y = 5 when x = 0, the market charges $5 to prepare each order. Evaluate" " " The process is effective because it leads step-by-step to a clear solution to the problem. The answer " " " " is reasonable because it is obtained by substituting the given information into the point-slope form " " " " of the equation of a line and converting the result into slope-intercept form. Chapter 4 - Applying Linear Functions 258 Section continues
4.3 - Equation of a Line Two Points In the same way that specifying a slope by itself describes a family of lines, specifying a point by itself also describes a family of lines. The clock face in Figure 4.3-3 illustrates what happens when only one point is given. The hands of the clock represent two of many possible lines passing through the center point. In addition, the big hand sweeps through all possible directions pointing outward from the center once every hour. Once every half hour, the big hand takes on all possible slopes. This shows that specifying one point alone does not determine a unique line. If a second point is specified, such as the two o clock tick mark on the outer rim of the clock face, that point and the center point together determine a unique line. Given two different points ( x 1, y 1) and ( x 2, y 2) on the same line, the slope of the line can be determined from the slope formula m = y y 2 1. Once the slope is known, either point can then be used x 2 x 1 to determine the equation of the line in point-slope form y y 1 = m(x x 1 ). If one of the points happens to be (0, b), it is quicker to use the y-intercept b to determine the equation of the line in slope-intercept form y = mx + b. Jump to Slope Formula Figure 4.3-3 This clock face illustrates the concept of a family of lines through a common point. Chapter 4 - Applying Linear Functions 259 Section continues
4.3 - Equation of a Line Example problem"" Find the equation of the line that passes through the points (2, 4) and (5, 6). Express the equation in standard form. Analyze" " " The problem describes the line that passes through the points (2,4) and (5,6). It asks for the equation of that line, and the equation " " " " must be given in standard form, Ax + By = C. Formulate" " " Use the slope formula to find the slope of the line. Then write an equation in point slope form using that slope and the " " " " coordinates of one of the given points. Simplify the resulting equation and rearrange it into standard form. Determine" " " m = 6 4 " " " Substitute the given coordinates into the slope formula. 5 2 " " " " m = 2 " " " Simplify to calculate the slope. 3 " " " " y 4 = 2 (x 2)" " Write the point slope form of the equation. 3 " " " " y 4 = 2 3 x 4 " " Use the distributive property. 3 " " " " 3y 12 = 2x 4" " Multiply each term by 3 to eliminate the denominators. " " " " 2x + 3y = 8" " Arrange the variable terms on the left and the constant on the right. " " " " 2x 3y = 8" " Multiply each term by 1 to make the coefficient of x positive. " " " " Justify" " " The two points given in the problem were sufficient for finding the slope, and knowing the slope made it possible to write the " " " " equation in point slope form. Because a linear equation can be converted from one form to another, the equation determined in " " " " point slope form was then converted to standard form. Evaluate" " " The problem did not give the information that would have been required to write the equation directly in standard form. " " " " However, it did give the necessary information for finding the slope and writing the point slope form of the equation, so the " " " " chosen strategy was effective with respect to the given information. The answer is reasonable because substituting the given " " " " points (2, 4) and (5, 6) into the equation both result in the true equation 8 = 8. Chapter 4 - Applying Linear 260 Section continues
4.3 - Equation of a Line For example, suppose that an American family vacations in Canada during the summer. America uses the Fahrenheit temperature scale, but Canada uses the Celsius temperature scale. In order to know how to dress comfortably for the weather conditions while on vacation, the family needs to convert the temperature from Celsius to Fahrenheit. If x represents the temperature in degrees Celsius and y represents the temperature in degrees Fahrenheit, then y is a linear function of x. The equation of this line can be determined from any two points on the line. Because water freezes at 0 C and 32 F, one point on the line is (0, 32). Because water boils at 100 C and 212 F, another point on the line is (100, 212). From these two points, the slope of the line is 212 32 m = 100 0 = 180 100 = 9. From the point (0, 32), the y-intercept is 32. The 5 equation of the line in slope-intercept form is y = 9 x + 32. Using this 5 equation, an early morning temperature of 20 C converts to 68 F, which the American family recognizes as room temperature. The thermometer in Figure 4.3-4 is marked with both temperature scales and visually confirms that 20 C converts to 68 F. If more than two points on a line are known, any two of those points can be used to determine the equation of the line. For example, suppose that a line passes through the points { (2, 7), (4, 13), (6, 19), (8, 25)}. Calculating Figure 4.3-4 This thermometer displays both Fahrenheit and Celsius the slope from the first two points yields m = 13 7 temperature scales. = 6/2 = 3. Using the 4 2 25 13 second and fourth points yields m = = 12/4 = 3, which is the same slope. Using the first point 8 4 to determine the point-slope form of the equation produces y 7 = 3(x 2), which simplifies to y = 3x + 1. Using the third point to determine the equation produces y 19 = 3(x 6), which also simplifies to y = 3x + 1. It makes no difference which points are selected from the list provided. The equation of the line is the same no matter which points are used to determine it. Chapter 4 - Applying Linear Functions 261 Section continues
4.3 - Equation of a Line Example problem"" The points in the table to the right all lie on the same line. Extend the table by adding one " " " " new point from the line at each end of the table. Analyze" " " The problem provides a table of points that lie on the same line. It asks for two new points " " " " (x, y) that also lie on the line. One point must satisfy x < 3 and the other point must satisfy " " " " x > 4. Formulate" " " Use the points (3, 12) and (4, 17) to calculate the slope of the line from the slope formula " " " " m = y 2 y 1 x 2 x 1. Then use the point (1, 2) to determine the point-slope form y y 1 = m(x x 1 ) of " " " " the equation of the line. To simplify calculations, convert the equation into slope-intercept " " " " form y = mx + b. Finally, compute the values of y when x = 4 and x = 5. x y 3 18 2 13 1 2 3 12 4 17 Determine" " " m = 17 12 4 3 = 5 1 = 5" " " Calculate the slope. " " " " y 2 = 5(x 1)" " " " Determine the point-slope form. " " " " y = 5x 3" " " " " Convert to slope-intercept form. " " " " y = 5( 4) 3 = 23" " " Let x = 4. " " " " y = 5(5) 3 = 22" " " " Let x = 5. Justify" " " The point ( 4, 23) extends the table at the beginning, and the point (5, 22) extends the table at the end. As a check, use these " " " " 22 + 23 points to find the equation of the original line. The slope is m = 5 + 4 = 45 = 5. The point-slope form of the equation is 9 " " " " y 22 = 5(x 5), which simplifies to y = 5x 3. This is the same slope-intercept form found above. Therefore, these two new " " " " points do lie on the same line as the other points in the table, and they extend the table at both ends. Evaluate" " " The process is effective because it leads step-by-step to a clear solution to the problem. The process is also flexible because " " " " different points can be used to get the same answer. The answer itself is open-ended. The answer is reasonable because the " " " " equation of the line determined by the two new points is the same as the equation of the line through the original points. Chapter 4 - Applying Linear Functions 262 Section continues
4.3 - Equation of a Line The equation of a line can also be determined solely by observing a graph, without explicitly being given the slope or any coordinate points. In Figure 4.3-5, a graphed function is displayed. Although no specific points are labeled in the figure, gathering coordinate points that fall on the line is still possible. For example, it can be observed that the line crosses through the points (2, 2) and (4, 3). Substituting the coordinates of these two points into the slope formula gives m = 1/2. Substituting this into the point-slope form along with one of the points gives the equation of the line, y = 1 2 x + 1. Another method to determine the equation of a line, without being given any information other than the graph, is to locate the y-intercept and use it to discover the slope. In Figure 4.3-6, the y-intercept has coordinates (0, 3). Using this as a starting point, the next coordinate point the line falls on is (1, 1), which involves rising two units up the y-axis and going to the right one unit on the x-axis. This shows that m = 2, and when the slope is combined with the previously located y-intercept, the equation of the line becomes y = 2x 3. Figure 4.3-5 The equation of y = f(x) can be determined by locating two points on the graph. Figure 4.3-6 Locating the y-intercept and using it as a starting point to find the slope can determine the equation of a line without any given information. Chapter 4 - Applying Linear Functions 263 Section continues
4.3 - Equation of a Line Parallel & Perpendicular Recall that parallel lines have the same slope. For example, the lines y = 2x 1, y = 2x, and y = 2x + 1 all have slope 2, which means these three lines are parallel. This relationship can be used to determine the equation of a line that is parallel to a given line. Jump to Parallel Lines Example problem"" Find the equation of the line that passes through the point ( 5, 4) and is parallel to the line 2x 3y = 12. Analyze" " " The problem gives one point on a line and the equation of another line that is parallel to the first line. It asks for the equation of " " " " the line through the given point. Formulate" " " Put the equation 2x 3y = 6 into slope-intercept form y = mx + b to determine the slope m. Because the two lines are parallel, the " " " " first line also has slope m. Use the given point ( 5, 4) and the point-slope form y y 1 = m(x x 1 ) to determine the equation of the " " " " first line. Determine" " " 2x 3y = 12 " " " " y = 2 x 4" " " " Solve for y. 3 " " " " m = 2 " " " " Find the slope. 3 " " " " y 4 = 2 3 ( x + 5)" " " Use the point-slope form. Justify" " " The equation of the line through ( 5, 4) and parallel to 2x 3y = 12 is y 4 = 2 3 ( x + 5). This follows from the point-slope form " " " " because the line 2x 3y = 12 has slope 2/3. Evaluate" " " The process is effective because it leads step-by-step to a clear solution to the problem. The answer is reasonable because parallel " " " " lines have the same slope. Chapter 4 - Applying Linear Functions 264 Section continues
4.3 - Equation of a Line Recall that perpendicular lines with defined, non-zero slopes have negative reciprocal slopes. For example, the lines y = 3x 5 and y = 1 x + 2 have slopes 3 and 1/3. These slopes are negative 3 reciprocals, which means the two lines are perpendicular. This relationship can be used to determine the equation of a line that is perpendicular to a given line. Jump to Perpendicular Lines Example problem"" Find the equation of the line that passes through the origin and is " " " " perpendicular to the line 2x + 4y = 1. Analyze" " " The problem gives one point on a line and the equation of another line " " " " that is perpendicular to the first line. It asks for the equation of the line " " " " through the given point. Formulate" " " Put the equation 2x + 4y = 1 into slope-intercept form y = m 2 x + b to " " " " determine the slope m 2. Because the two lines are perpendicular, the " " " " first line has slope m 1 = 1/m 2. Use the given point (0, 0) and the point- " " " " slope form y y 1 = m 1 (x x 1 ) to determine the equation of the first line. Determine" " " 2x + 4y = 1 " " " " y = 1 2 x + 1 " " " Solve for y. 4 " " " " m 2 = 1 " " " " Find the slope. 2 " " " " m 1 = 1 m 2 = 2" " " Compute the negative reciprocal. " " " " y 0 = 2(x 0)" " " Use the point-slope form. " " " " y = 2x" " " " Simplify the equation. Chapter 4 - Applying Linear Functions 265 Problem continues
4.3 - Equation of a Line Justify" " " The equation of the line through (0, 0) and perpendicular to 2x + 4y = 1 " " " " is y = 2x. This follows from the point-slope form because the " " " " perpendicular line 2x + 4y = 1 has slope 1/2. Evaluate" " " The process is effective because it leads step-by-step to a clear solution " " " " to the problem. The slope-intercept form of a line could have been used " " " " instead of the point-slope form because the given point was also the " " " " y-intercept. The answer is reasonable because perpendicular lines have " " " " negative reciprocal slopes. Chapter 4 - Applying Linear Functions 266 Problem complete
Section 4.4 Line of Reasonable Fit Objectives Analyze scatterplots to identify positive or negative correlation or no correlation Create and use lines of reasonable fit to model, estimate, and predict scatterplot data New Vocabulary Scatterplot Correlation Positive correlation Negative correlation Line of reasonable fit Interpolation Extrapolation Reliability As children grow older, a common way to record their growth is by marking their height and age on a wall or doorframe around the house. Every few months or so, the child s new height is marked in the same location and compared to the previous heights. Imagine what this data would look like if it were graphed. What type of relationship would the data exhibit? How could you summarize this information?
4.4 - Line of Reasonable Fit Scatterplots & Linear Approximations Graphical representations of data show relationships between variables. In most real-world situations, relationships between variables are not perfectly modeled by a straight line or a simple curve. For instance, the graph of the height of a growing child as a function of time would almost certainly not be a straight line or a smooth curve. Instead, the data might show a linear trend, with variations in the data on either side of the trend line. A scatterplot is a type of graph on the coordinate plane that is used to show the relationship between paired data. With scatterplots, the independent variable is graphed on the x-axis, and the dependent variable on the y-axis. Each pair of data represents one point on the graph. The more data points graphed, the clearer the relationship in the data usually becomes. Consider the ordered pairs in Table 4.4-1, which shows the height of a particular student in inches, measured each month for one year. By looking at the data in the table, it is clear that the student s height increases with time. What is more difficult to tell from the table is the mathematical relationship between time and height. Figure 4.4-1 is a scatterplot that was created by graphing the data from Table 4.4-1 on the coordinate plane. The independent variable is the number of the month. The dependent variable is the student s height in inches. Jump to Independent & Dependent Variables Month Height (in.) 1 61.5 2 61.6 3 62.1 4 62.7 5 63.3 6 63.4 7 63.8 8 64.2 9 64.5 10 65.1 11 65.2 12 65.5 Table 4.4-1 This table shows the height of a student measured each month for one year. Figure 4.4-1 This graph shows a scatterplot created from the data on the height of a student each month, as shown in Table 4.4-1. Chapter 4 - Applying Linear Functions 268 Section continues
4.4 - Line of Reasonable Fit The scatterplot in Figure 4.4-1 shows a linear relationship in the data. This means that the general trend is a line even though not all of the data points fall precisely on any single line. When data can be approximated with a line, the mathematical relationship between the variables can be determined. Figure 4.4-2 shows the data from Figure 4.4-1 with the addition of a line that approximates the data. Note that not all of the data points are actually on the line. Often, scatterplots that show linear relationships have no data points that fall on the line. What is important is that the line represents the trend of all of the data points taken together. The line in the example represents the average increase in the student s height over time. The h-intercept and slope of the line define the equation of the line and can be determined by examining the graph. In the example above, the equation of the line is h = 0.4n + 61. Jump to Slope Jump to Initial Conditions & y-intercepts Jump to Slope-Intercept Form of a Line Once the equation of the line approximating the data is known, it can be used to predict new ordered pairs of data. This is done by selecting a value for the independent variable and substituting it into the equation of the line to determine the corresponding value of the dependent variable. The accuracy of these predicted points depends largely on how well the line fits the data. If the data points are scattered and do not lie very near the line, then the predictions will be less accurate. Figure 4.4-2 This graph shows the scatterplot from Figure 4.4-1 with the addition of a line that approximates the data. Chapter 4 - Applying Linear Functions 269 Section continues
4.4 - Line of Reasonable Fit Correlation Scatterplots show the relationship between paired data. Sometimes the dependent variable increases as the independent variable increases. This type of trend is said to be increasing. An example of an increasing trend is the increasing height of a child as it changes with time. For other cases, the dependent variable decreases as the independent variable increases. This type of trend is said to be decreasing. An example of a decreasing trend is the value of a car as it depreciates with time. A third type of trend shows no change in the dependent variable as the independent variable increases. These trends are said to be constant. An example of a constant trend is the height of an adult as it remains the same over time. Figure 4.4-3 summarizes these three types of scatterplot trends. Figure 4.4-3 The scatterplot on the left shows an increasing trend. The scatterplot in the middle shows a constant trend, and the one on the right shows a decreasing trend. Chapter 4 - Applying Linear Functions 270 Section continues
4.4 - Line of Reasonable Fit There is another category of scatterplot that shows no apparent relationship between the dependent and independent variables. For these scatterplots, data pairs are scattered over the Cartesian plane in an apparently random pattern. Figure 4.4-4 shows an example of this type of scatterplot. Another term used to describe scatterplots is correlation. The correlation of a scatterplot describes both the trend and how strongly the data points follow it. The correlation of a scatterplot can be described by statistical analysis and has a value between 1 and 1. The calculated correlation for a scatterplot with an increasing trend will be a positive number, so that type of scatterplot is said to have positive correlation. Figure 4.4-5 shows an example of two scatterplots with positive correlation. The graph on the left has data points that fall very near the trend line, so it has a strong positive correlation. The graph on the right also has positive correlation, but the data points for this scatterplot lie farther from the trend line. This graph shows weaker positive correlation. Figure 4.4-4 This is an example of a scatterplot with no relationship between the independent and dependent variables. Figure 4.4-5 The graph on the left shows stronger positive correlation, and the one on the right shows weaker positive correlation. Chapter 4 - Applying Linear Functions 271 Section continues
4.4 - Line of Reasonable Fit A scatterplot with data that show a decreasing trend will have a correlation that is a negative number, so this type of scatterplot is said to have negative correlation. How tightly the data follow a linear trend defines whether it is a strong or weak negative correlation. Figure 4.4-6 shows examples of strong and weak negative correlation. When the graphed pairs of data in a scatterplot do not show a trend, then the data have no correlation. This means that changes in the independent variable lead to no predictable change in the dependent variable. Figure 4.4-7 shows this type of scatterplot. Three lines have been superimposed on the data to illustrate that a linear equation is not useful for modeling data with no correlation. Figure 4.4-6 The graph on the left shows stronger negative correlation, and the one on the right shows weaker negative correlation. Figure 4.4-7 This graph shows three identical scatterplots with no correlation. Chapter 4 - Applying Linear Functions 272 Section continues
4.4 - Line of Reasonable Fit Approximation The line drawn on a scatterplot to show the trend of the data is called a line of reasonable fit. When the data are strongly correlated, a line can be drawn in such a way that the points lie on or near the line. When the data are only weakly correlated, then the data points lie farther from the line. This is true for scatterplots with both positive and negative correlations. Example problem"" Describe the correlation of the data in the scatterplot shown. Analyze" " " The problem asks for a description of the correlation of the data " " " " in the scatterplot. Formulate" " " Observe the data points and decide whether there is a trend. If " " " " there is a trend, draw a line to show it. If the trend is decreasing, " " " " then the scatterplot has negative correlation. If the trend is " " " " increasing, then the scatterplot has positive correlation. If there is " " " " no trend, then the scatterplot has no correlation. If the data points " " " " fall very close to the line, then the correlation is strong; if they are " " " " more scattered, then the correlation is weak. Determine" " " Observe the data and decide whether there is a trend. Chapter 4 - Applying Linear Functions 273 Problem continues
4.4 - Line of Reasonable Fit " " " " Draw a line to show the trend. " " " " The slope of the line is negative and the trend is decreasing, " " " " meaning the correlationis negative. " " " " All of the data points lie very close to the trend line, so the " " " " correlation is strong. " " " " The scatterplot shows strong, negative correlation. Justify" " " The data was observed and a trend was noted. A line illustrating " " " " the trend was drawn. The trend in the data is decreasing because " " " " as the independent variable (the x-value) increases the dependent " " " " variable (the y-value) decreases. The line has a negative slope, so " " " " the correlation is negative. The data points all lie close to the line, " " " " so the correlation is strong. Evaluate" " " The process was clear. The data are strongly correlated, so seeing " " " " the trend in the data was quick. The process would be more " " " " difficult with weakly correlated data points. Chapter 4 - Applying Linear Functions 274 Problem complete
4.4 - Line of Reasonable Fit Linear Approximation & Predictions When looking at a scatterplot, the relationship between the two variables is shown only over the domain and range. When the linear relationship of a scatterplot is used to estimate a value that lies within the domain and range of the data, the process is called interpolation. The line showing a trend can also be used to predict values outside of the domain and range; however, there can be no certainty that an apparent trend would continue if more data were collected to extend the domain and range. When the linear relationship is used to generate a predicted value outside of the domain and range, the process is called extrapolation. Figure 4.4-8 illustrates this. Jump to Domain & Range The reliability of interpolated and extrapolated values is a measure of how close they are to actual data. In general, the more strongly correlated the data, the more reliable are any interpolated or extrapolated values. However, even for highly correlated data, the reliability of extrapolated values decreases as they move out farther from the domain and range of the data. This should be apparent in the example of the growing student. You could not use the equation of the line to extrapolate the student s height at age 30 the student will stop growing well before then. Figure 4.4-8 This graph shows the distinction between estimated data points (interpolation) and predicted data points (extrapolation) in scatterplots. Chapter 4 - Applying Linear Functions 275 Section continues
4.4 - Line of Reasonable Fit Example problem"" " " " " in months. The scatterplot to the right shows the heights of a group of children plotted against their ages " " " " Use the equation of the line to estimate the height of a child who is 28 months old, then use the " " " " equation to predict the height of a child who is 48 months old. Analyze" " " The problem asks to use the equation of the line from the provided scatterplot to estimate the " " " " height in centimeters of a child who is 28 months old and to predict the height in centimeters of " " " " a child who is 48 months old. Formulate" " " Interpolation is used to estimate a value that is within the domain and range of the data. " " " " Interpolate by substituting the value a = 28 into the equation of the line that follows the " " " " trend in the data, h = 61 + a. Extrapolation is used to predict a value that is outside the " " " " domain and range of the data. To extrapolate, use the same process, with a = 48. Determine" " " h = 61 + a = 61 + 28 = 89"" " Evaluate the equation for a = 28. The estimate is 89 cm. " " " " h = 61 + a = 61 + 48 = 109" " Evaluate the equation for a = 48. The prediction is 109 cm. Justify" " " The equation that represents the trend in the data on the scatterplot was used to estimate the " " " " height of a child who is 28 months old and to predict the height of one who is 48 months old. " " " " The age of the child, in months, was used in the equation for the line, and the equation was " " " " solved for h. Evaluate" " " The process of evaluating the linear equation for a = 28 and a = 48 was straightforward. The " " " " estimated and predicted heights lie on the line, but because the data have a relatively weak " " " " correlation, the values may not reliably match the heights of actual children at these ages. Chapter 4 - Applying Linear Functions 276 Section continues
4.4 - Line of Reasonable Fit Guiding Rules for Line Fitting To hand-draw a line of reasonable fit on a scatterplot, the first step is to observe the data and look for the trend. The line of reasonable fit will have the same slope as the trend in the data. Once the slope is chosen, the line is placed on top of the data points. If the data has a strong correlation, the line of reasonable fit may lie on top of many of the data points. If the data is more weakly correlated, place the line so that about half of the data points are above the line, and the other half are below it. It is not necessary that any data points fall exactly on the line, but it is fine if they do. Figure 4.4-9 shows three scatterplots with the same data. The blue line in the scatterplot on the left is the line of reasonable fit. The red line in the middle scatterplot has a slope that does not match the slope of the trend in the data, so it is not a reasonable fit. The green line in the scatterplot on the right has the same slope as the data, but is too high because more than half of the data points fall below it. For this reason, the green line is not a reasonable fit to the data. Figure 4.4-9 These three scatterplots with the same data are modeled by three very different lines. Chapter 4 - Applying Linear Functions 277 Section continues
4.4 - Line of Reasonable Fit Example problem"" Draw a line of reasonable fit for the data shown in the scatterplot " " " " below. Analyze" " " The problem asks for a line of reasonable fit to be drawn for the " " " " scatterplot shown. Formulate" " " To determine the line of best fit, first observe the data points to see " " " " whether there is a trend. If there is a trend, determine the slope " " " " that best fits it. Once the slope is determined, slide the line up and " " " " down on the graph until about half of the data points that are not " " " " actually on the line are above it, and the other half are below it. Determine" " " " " Observe the data and " " " " " " note that the data " " " " " " points show a " " " " " " negative correlation. " " " " " " Draw a line with a " " " " " " slope that matches the " " " " " " slope of the trend seen " " " " " " in the data. Chapter 4 - Applying Linear Functions 278 Problem continues
4.4 - Line of Reasonable Fit " " " " " Slide the line up and down " " " " " on the graph until about " " " " " half of the data points that " " " " " do not lie on the line are " " " " " above it, and about half are " " " " " below it. Justify" " " The scatterplot was observed to have a negative trend. The line of " " " " reasonable fit has the same slope as the trend seen in the data, and " " " " equal numbers of data points lie above and below the line. Evaluate" " " Determining the trend, the slope, and the position of the line of " " " " reasonable fit was quick and resulted in a line that matches the " " " " trend in the data. Chapter 4 - Applying Linear Functions 279 Problem complete
Virtual Manipulative - Line of Reasonable Fit Many different lines can be used to model a set of data. Some lines are better than others. The following virtual manipulative will allow you to generate a set of data and experiment with possible lines, changing the characteristics of the line to make it a better fit. As you went through the process of creating lines to model the data, you should have been able to determine if a line matched the correlation of the data. You also should have adjusted the slope and intercepts of the line to refine the model. Chapter 4 - Applying Linear Functions 280 Section complete
Section 4.5 Line of Best Fit Objectives Create a line of best fit for a scatterplot with positive or negative correlation Make predictions about real-world situations using a linear model Use the correlation coefficient to evaluate the fitness of a regression model New Vocabulary Line of best fit Linear regression Correlation coefficient Where is the best place to put a line? This is a strange question, but it has been asked before. When governments decide where to plan their public transit routes, they are looking for the best places to run them. Similarly, there is a best place to put a line on a scatterplot to model data. What makes one line a better model than another? How would you determine which line is the best?
4.5 - Line of Best Fit Scatterplots & Linear Regressions During childhood, height and weight are important measures of growth and health. The doctor can determine whether a child is developing proportionally by checking these two measurements together. Table 4.5-1 illustrates a data set of 25 height and weight measurements. Jump to Direct Variation Figure 4.5-1 models the data from the table along with two linear functions. Notice that both lines are reasonable fits to the data in that approximately half of the data points are above the line and half are below. While there are many more possible lines of reasonable fit, only one line best represents the data: the line of best fit. The line of best fit is the straight line that best summarizes the data in a scatterplot and it is determined by the correlation of the data. Jump to Guiding Rules for Line Fitting Index Height (in.) Weight (lb) 1 65.79 113.01 2 71.57 136.45 3 68.40 152.03 4 67.25 143.33 5 67.78 144.29 6 68.71 122.39 7 69.81 141.04 8 69.97 136.35 9 67.95 112.42 10 66.74 120.63 11 66.41 127.37 12 67.61 114.13 Index Height (in.) Weight (lb) 14 66.12 121.48 15 68.18 116.19 16 71.19 139.98 17 66.38 128.50 18 68.55 141.89 19 71.25 137.94 20 67.23 124.15 21 66.83 140.28 22 67.99 142.64 23 63.58 98.40 24 68.41 129.51 25 67.42 141.65 Figure 4.5-1 Height and weight are important indicators of health during childhood. 13 67.30 124.61 Table 4.5-1 This table shows the heights and weights of 25 individuals. Chapter 4 - Applying Linear Functions 282 Section continues
4.5 - Line of Best Fit Linear regression is the process used for finding the line of best fit; however, not all scatterplots are well-modeled by a line of best fit. Linear regression works better when there is a linear relationship between the points. Figure 4.5-2 displays data on the height of students and their academic grades. This data would not be a good candidate for linear regression because the data does not appear to be linear. If the number of hours a student spends studying is considered rather than height, then the scatterplot becomes a good candidate for linear regression, as displayed in Figure 4.5-3. Regression models are statistical processes that many technologies can perform. To find the line of best fit, the data points need to be entered into a list for the x-values and one for the y-values. Then, linear regression can be applied to the points, and an equation can be found. Different calculators, programs, and other technologies display the equation of the line differently, although most use the slope-intercept form. Figure 4.5-2 There does not appear to be a relationship between a person s height and academic grade. Figure 4.5-3 There does appear to be a relationship between hours studied and a student s academic grade. Chapter 4 - Applying Linear Functions 283 Section continues
4.5 - Line of Best Fit Example problem " Using technology, determine the line of best fit given the 10 data pairs in the table. Arm span (cm) Height (cm) Analyze" " " The problem asks to determine the line of best fit of the given data using " " " " technology. Formulate" " " The line of best fit comes from a linear regression. First, graph the scatterplot to " " " " make sure it has correlation. Many graphing calculators can perform linear " " " " regressions, as can computers with the right software. The data must be entered " " " " into whatever technology is being used. Check the work by graphing the " " " " scatterplot with the line of best fit. Determine" " " The scatterplot is shown below. Notice that there appears to be a linear " " " " relationship. " " " " Running a linear regression on the data gives y = 60.167 + 0.667x. The scatterplot " " " " and the line of best fit are shown below. 155 163 157 159 155 169 178 174 178 177 183 180 189 192 200 191 169 169 174 186 Justify" " " The line of best fit was determined using technology. The slope of the line is positive, like the correlation of the data. A graph of " " " " the line of best fit appears to summarize the data. Evaluate" " " Examining the scatterplot relationship first ensured that linear regression would be appropriate. Once the technology provided " " " " the line of best fit, graphing it with the scatterplot of the data verified that the line was an appropriate model. Chapter 4 - Applying Linear Functions 284 Section continues
4.5 - Line of Best Fit Line of Best Fit Predictions The line of best fit not only gives the best analytical linear representation of the data, but it also gives the ability to make predictions. With the equation for the line of best fit, predictions can be made about one variable given specific values of the other. Table 4.5-2 contains data on the arm span and height of 10 people. In this example, arm spans can predict height, and vice versa. Jump to Linear Approximations & Predictions Using technology, the regression model is y = 60.416 +.667x. The table shows that the arm span is the independent variable (x), and height is the dependent variable (y). Given an arm span of 179 cm, a prediction can be made about a person s height using the line of best fit. Substitute that value for x in the line of best fit equation. This yields y = 60.416 +.667(179), which is equal to 179.809, or about 180 cm. The same can be done to predict the independent variable s value by plugging in for y. Suppose the arm span (x) is needed of someone who is 184 cm tall (y). Using the line of best fit, plug in for y. This yields 184 = 60.416 +.667x. Solving this two-step equation for x gives 185.283, or about 185 cm. When making predictions, it is best to predict within the range of values given. The reason for staying within the range of values given is that it is not known whether the relationship maintains linearity beyond the points provided. For example, using the growth rates of teenagers to predict how tall they will be in middle age will not give a realistic prediction people do not typically grow taller during their 20s and 30s. Arm span (cm) Height (cm) 155 163 157 159 155 169 178 174 178 177 183 180 189 192 200 191 169 169 174 186 Table 4.5-2 The arm spans and heights of 10 people can be used to make predictions. Chapter 4 - Applying Linear Functions 285 Section continues
4.5 - Line of Best Fit Example problem"" The stride rate (steps per minute) of a runner is how quickly a runner " " " " takes steps. The pace (minutes per mile) measures the runner s speed. The " " " " pace and stride rate of five runners are given. Find the line of best fit, and " " " " use it to predict the stride rate for a pace of 7.15. Analyze" " " The problem asks to predict the stride rate for someone with a 7.15 pace " " " " based on the data. Formulate" " " Find the line of best fit using technology, and use the regression model to " " " " determine stride rate. Determine" " " A graph of these values is shown below. Running the linear regression " " " " gives the function y = 226.57 6.98x. Since x represents pace, substitute " " " " 7.15 for x, resulting in a stride rate of 176.66. Pace (x) Stride rate (y) 7.38 176.58 6.40 180.86 5.55 184.45 5.20 191.62 4.56 196.22 Justify" " " The slope of the regression model is negative, and the relationship between the points appears " " " " negative. This makes sense because the longer it takes to run a mile, the slower the stride rate. The " " " " pace that was plugged in fell between the paces of 7.38 and 6.40, and the stride rate fell in between " " " " the stride rates for those paces. Evaluate" " " Examining the scatterplot relationship first ensured that linear regression could be used. Once the " " " " technology provided the line of best fit, plugging in the value gave a predicted value that made " " " " sense compared to the data provided. Chapter 4 - Applying Linear Functions 286 Section continues
4.5 - Line of Best Fit Example problem"" The data in the table gives the number of Bob s Bicycle Shop ads run " " " " on television and the number of bikes sold. Use the data to find a " " " " line of best fit, and predict the number of ads run if 35 bikes are " " " " sold. Analyze" " " The problem asks to predict the number of ads if 35 bikes are sold. Formulate" " " Graph the data to ensure it is relatively linear. Find the line of best fit " " " " using technology, and use that value to determine pace. Determine" " " Running linear regression gives y = 1.30x + 2.44. Since y represents the " " " " number of bikes sold, substitute 35 for y. This gives 35 = 1.30x + 2.44. " " " " First, subtract 2.44 from both sides, resulting in 32.56 = 1.30x. Then, " " " " divide by 1.30 to get y = 25.04. Number of ads Number of run on TV (x) bikes sold (y) 6 14 19 30 0 5 15 17 9 11 15 19 35 50 17 24 Justify" " " The slope is positive, and the relationship between the points appears positive. This makes sense " " " " because the more ads that are run, the more bikes should be sold. The line of best fit can be plotted " " " " on the scatterplot to show that it is reasonable. Evaluate" " " Examining the scatterplot relationship first ensured that linear regression could be used. Once the " " " " technology provided the line of best fit, plugging in its value gave a predicted value that made " " " " sense when graphed compared to the data provided. Chapter 4 - Applying Linear Functions 287 Section continues
Virtual Manipulative - Linear Regression Modeling data with a line of best fit gives the best representation of the values, and even allows the opportunity to make predictions about data not provided. The following virtual manipulative will allow you to plot points on a coordinate plane and have the line of best fit calculated and graphed. You should have seen that as the graph was updated after each new point was plotted, the line of best fit adapted to the new data and adjusted to a function that was more fitting. You should have also seen that it is possible for the slope of the line of best fit can be changed from positive to negative (and vice versa) with the addition of only one point. Chapter 4 - Applying Linear Functions 288 Section continues
1.5 4.5 - Line of Best Fit Linear Correlation Coefficient The correlation coefficient, r, is a number between 1 and 1 that measures the linear relationship between two variables and is obtained using technology. The data values must be entered into separate lists to find the correlation coefficient, just as they were to create a scatterplot. Most calculators can be set to have the correlation coefficient display at the same time that the line of best fit is displayed. If r = 1, there is a perfect negative linear relationship; if r = + 1, there is a perfect positive linear relationship. Generally, if r 0.7, there is a strong linear relationship, and if 0 < r < 0.3, there is a weak linear relationship. A value of 0 indicates there is not a relationship at all. Finally, if 0.3 r < 0.7, there is a moderate correlation. These relationships are illustrated in Figure 4.5-4. It is important to note that simply looking at the scatterplot is not enough to determine the correlation. The correlation coefficient must be calculated to ensure that the strength of the relationship is correct. Jump to Correlation Figure 4.5-4 There are strong, weak, and no relationships shown in the given scatterplots. Chapter 4 - Applying Linear Functions 289 Section complete
Section 4.6 Linear Inequalities Objectives Create inequality statements in two variables to represent real-world situations Graph solutions to linear inequalities in two variables on a coordinate plane New Vocabulary Related equation Boundary Everyday life is full of algebraic inequalities. Credit cards have spending limits, roads have speed limits, and bridges have weight limits. Can you think of a real-life situation with two variables that are related by an inequality?
4.6 - Linear Inequalities Relationships Beyond Equality When y is a function of x, the two variables are related by the equation y = f(x). As x varies, this equation determines the value of y. Not all mathematical relationships work this way, however. When x and y are related by an inequality rather than an equation, both variables are free to vary. In that case, the value of one variable determines a range of values for the other variable rather than a single value. For example, imagine an upscale bakery that specializes in cakes and tarts, such as the ones shown in Figure 4.6-1. The bakery makes a profit of $14 on each cake and $18 on each tart. The bakery pays a total of $1,260 each month for rent and other expenses. In order to stay in business, the bakery must sell enough cakes and tarts each month to cover these operating expenses. The relationship between the number of cakes and tarts the bakery must sell to cover its monthly operating expenses can be described by a linear inequality. If c represents the number of cakes sold in one month, and t represents the number of tarts sold in that month, then the variables c and t can be related by an inequality. Since the bakery makes a profit of $14 on each cake, selling c cakes earns a profit of 14c dollars. Since the bakery makes a profit of $18 on each tart, selling t tarts earns a profit of 18t dollars. The total monthly profit is 14c + 18t dollars. In order for the bakery to stay in business, this amount must be at least as much as its $1,260 monthly operating expenses. The inequality 14c + 18t 1260 describes this relationship between the variables c and t. Figure 4.6-1 The number of cakes and tarts that a bakery must sell to cover its operating expenses is described by a linear inequality. Chapter 4 - Applying Linear Functions 291 Section continues
4.6 - Linear Inequalities A solution to the linear inequality 14c + 18t 1260 is an ordered pair of values (c, t) that makes the inequality true. A linear inequality has many possible solutions. For example, if the bakery sells only cakes and no tarts in a given month, then t = 0, and the inequality becomes 14c 1260. Solving for c yields c 1260/14, which simplifies to c 90. This means the bakery will meet or exceed its monthly operating expenses if it sells at least 90 cakes and no tarts. That makes the pair (90, 0) a solution to the linear inequality. The pairs (100, 0) and (110, 0) are solutions as well. On the other hand, if the bakery sells only tarts and no cakes in a given month, then c = 0, and the inequality becomes 18t 1260. Solving for t yields t 1260/18, which simplifies to t 70. This means the bakery will meet or exceed its monthly operating expenses if it sells no cakes and at least 70 tarts. The pair (0, 70) is a solution to the linear inequality, and so are the pairs (0, 80) and (0, 90). The graph in Figure 4.6-2 shows all six of these solutions to the linear inequality. Solutions that exactly meet the bakery s operating expenses of $1,260/ month lie on the line 14c + 18t = 1260. Solutions that exceed the monthly operating expenses lie above this line. Figure 4.6-2 Six of the pairs (c,t) that solve the linear inequality are depicted here as red points. Jump to One-Step Inequalities If the bakery sells both cakes and tarts in a given month, the pair (c, t) may or may not be a solution to the inequality 14c + 18t 1260. To test whether the pair is a solution, substitute the numerical values for c and t into the inequality and simplify. If the resulting numerical inequality is true, then (c, t) is a solution of the linear inequality. If the numerical inequality is false, then (c, t) is not a solution. For example, if the bakery sells 45 cakes and 35 tarts in the same month, the inequality becomes 14(45) + 18(35) 1260. This simplifies to 1260 1260, which is true. That means the pair (45, 35) solves the inequality 14c + 18t 1260. If the bakery sells only 44 cakes and 34 tarts, the inequality 14(44) + 18(34) 1260 simplifies to 1228 1260, which is false. That means the pair (44, 34) does not solve the inequality. Similarly, the pairs (44, 35) and (45, 34) produce the false numerical inequalities 1246 1260 and 1242 1260, so neither pair solves the linear inequality 14c + 18t 1260. The graph in Figure 4.6-3 shows one pair that solves the inequality and three pairs that do not. Figure 4.6-3 The pair (c,f) that satisfy the linear inequality is shown in light blue. The pairs that do not satisfy the inequality are shown in red. Chapter 4 - Applying Linear Functions 292 Section continues
4.6 - Linear Inequalities Linear Inequalities Algebraically Linear inequalities can be expressed as algebraic formulas, verbal descriptions, tables, and graphs. Translating linear inequalities from one form to another is a good way to solve problems and also helps explain what the solutions mean. For example, this is a verbal description of a linear inequality in two variables: A concession stand sells hot dogs at $2 each and cheeseburgers at $3 each to a customer who has $10 to spend on food. To translate this into algebraic form, let h denote the number of hot dogs and c denote the number of cheeseburgers. At $2 each, h hot dogs cost 2h dollars. At $3 each, c cheeseburgers cost 3c dollars. The total cost is 2h + 3c dollars, which cannot be more than the customer s $10 food budget. This relationship is expressed in algebraic form by the linear inequality 2h + 3c 10. When one of the variables takes on a specific value, the inequality restricts the values of the other variable to a range of possibilities. For example, to determine how many hot dogs the customer can afford after buying a specific number of cheeseburgers at a stand such as the one shown in Figure 4.6-4, solve the inequality for h. Subtracting 3c from both sides of the inequality 2h + 3c 10 yields 2h 10 3c. Dividing both sides of the inequality by 2 produces h 5 3 c. Suppose the 2 customer buys two cheeseburgers. Substituting c = 2 into the inequality results in h 5 3 (2). Simplifying yields 2 h 2, which shows that the customer has enough money left to buy up to two hot dogs. Jump to Two-Step Inequalities Figure 4.6-4 The number of hot dogs and cheeseburgers that a customer can buy from this concession stand is described by a linear inequality. Chapter 4 - Applying Linear Functions 293 Section continues
4.6 - Linear Inequalities When solving real-world problems, it is important to check that the variables take on reasonable values. For example, the customer cannot buy half a hot dog or a negative number of cheeseburgers. If cheeseburgers cost $3 each and the customer has $10 to spend, that customer can buy at most three cheeseburgers. This means the number of cheeseburgers (c) that the customer can buy must belong to the set {0, 1, 2, 3} in order to be reasonable for this problem. Letting c = 3 in the inequality h 5 3 2 c yields h 5 3 (3). Simplifying produces h 1/2. Since the customer cannot buy half a hot dog, this 2 means that buying three cheeseburgers does not leave enough money to buy any hot dogs on a $10 food budget. To put it another way, at $2 each on a $10 budget, the number of hot dogs (h) that the customer can buy must belong to the set {0, 1, 2, 3, 4, 5} to be reasonable for this problem. Since 1/2 does not belong to this set, h = 0 is the only reasonable solution to the inequality h 1/2. Any verbal statement about a maximum or minimum value of a variable is actually a statement about inequalities. Using only reasonable solutions to the linear inequality h 5 3 c, Table 4.6-1 shows the 2 maximum number of hot dogs the customer can afford after buying any reasonable number of cheeseburgers. This table provides another way to represent the inequality, in addition to the algebraic formulation 2h + 3c 10 and the verbal description presented earlier. If the customer buys this many cheeseburgers The maximum number of hot dogs the customer can afford to buy is 0 5 1 3 2 2 3 0 Table 4.6-1 This table describes reasonable solutions (h) of the linear inequality h 5 3 c for values of c in the set {0, 1, 2, 3}. 2 Chapter 4 - Applying Linear Functions 294 Section continues
4.6 - Linear Inequalities Linear equations provide valuable insight into linear inequalities. There are four varieties of linear inequalities in two variables, as shown in Table 4.6-2. An inequality s related equation is the equation obtained by replacing the inequality sign with the equals sign. Each of these four linear inequalities produces the related equation Ax + By = C. The graph of the related equation is a line in the xy-plane. This line divides the plane into two regions that lie on opposite sides of the line. All of the points in the same region share the same truth-value with respect to the inequality, so either every point on a particular side of the line satisfies the inequality or every point on that side does not. For example, consider the linear inequality 3x + 4y 24. Its related equation is 3x + 4y = 24. One of those regions satisfies 3x + 4y < 24, which means every point in that region solves the linear inequality 3x + 4y 24. The other region satisfies 3x + 4y > 24, which means none of the points in that region solve the linear inequality 3x + 4y 24. Thus, all of the points on one side of the line 3x + 4y = 24 solve the linear inequality 3x + 4y 24, but none of the points on the other side of the line solve it. Varieties of linear inequalities Ax + By < C Ax + By > C Ax + By C Ax + By C Table 4.6-2 This table shows the four varieties of linear inequalities. To determine which side of the line solves the linear inequality, it is enough to test any point that does not lie on the line. Since the line 3x + 4y = 24 does not pass through the origin, the point (0, 0) provides the simplest possible test. Substituting 0 for both x and y in the linear inequality 3x + 4y 24 yields the true inequality 0 24. This means (0, 0) solves the linear inequality. In addition, all points on the same side of the line 3x + 4y = 24 as the origin solve the inequality as well. The graph in Figure 4.6-5 shows the region in the xy-plane that solves the linear inequality 3x + 4y 24, which includes all points on the line 3x + 4y = 24. Figure 4.6-5 This graph shows the solution set of the linear inequality 3x + 4y 24, which includes the line 3x + 4y = 24 and the test point (0, 0). Chapter 4 - Applying Linear Functions 295 Section continues
4.6 - Linear Inequalities In general, the solution set of a linear inequality in two variables is a region that occupies half of the xy-plane. The solution set of the related equation Ax + By = C is a line called the boundary of the region. The boundary line is the border of the solution set of the linear inequality. The boundary line itself may or may not be included in the solution set. Whether it is included depends on the type of inequality. The linear inequalities Ax + By < C and Ax + By > C are strict inequalities, which means the left and right sides cannot be equal. In this case, the solution set of the linear inequality excludes the boundary line Ax + By = C. When the boundary is excluded, the graph of the solution set depicts the boundary as a dashed line. The linear inequalities Ax + By C and Ax + By C are nonstrict inequalities, which means the left and right sides are allowed to be equal. So in this case, the solution set of the linear inequality includes the boundary line Ax + By = C. When the boundary is included, the graph of the solution set depicts the boundary as a solid line. For example, consider the linear inequality 3x + 4y > 24, which has the related equation 3x + 4y = 24. As previously noted, the test point (0, 0) does not lie on the boundary line 3x + 4y = 24. Substituting this test point into the linear inequality yields the false inequality 0 > 24. This means the point (0, 0) does not solve the linear inequality 3x + 4y > 24. Therefore, solutions to the inequality lie on the side of the boundary line 3x + 4y = 24 opposite the origin (the side that lies above the line). Since 3x + 4y > 24 is a strict inequality, the boundary line 3x + 4y = 24 is excluded from the solution set of the linear inequality. That is why the boundary is depicted as a dashed line in the graph of the solution set in Figure 4.6-6. Figure 4.6-6 This graph shows the solution set of the strict linear inequality 3x + 4y > 24, which excludes the boundary line 3x + 4y = 24. Recall that the solution set of a linear inequality in one variable may or may not include the endpoint. Similarly, the solution set of a linear inequality in two variables may or may not include the boundary line. These two ideas are closely related. To see the connection, let y = 0 in the linear inequality 3x + 4y > 24. This yields the simpler inequality 3x > 24, which is solved by all real x with x > 8. Because the inequality is strict, the endpoint 8 is excluded from the solution set. This is depicted by an open circle around the point 8 on the number line that shows the solution set in Figure 4.6-7. Compare this to the dashed boundary line 3x + 4y = 24 in the graph of the solution set of 3x + 4y > 24 in Figure 4.6-6. Figure 4.6-7 This number line depicts the solution set consisting of all real x with x > 8. It excludes the endpoint 8. Jump to One-Step Inequalities Chapter 4 - Applying Linear Functions 296 Section continues
4.6 - Linear Inequalities Example problem"" A school district with a budget of $3,000 wants to buy softballs and bats for the athletic " " " " departments of all its schools. The best prices it could find for this equipment are given in " " " " the table to the right. " " " " Write a linear inequality that describes the number of softballs and bats the school district " " " " can afford. Graph a region that contains all of the reasonable solutions to this inequality. Equipment Price Softball $6 Bat $15 Analyze" " " The problem provides the amount of the budget and the cost of the equipment the school district wants to buy. It asks for a linear " " " " inequality that describes the number of softballs (s) and bats (b) the school district can afford. It also asks for the graph of a region " " " " that contains all of the reasonable solutions to the inequality. This means that s 0 and b 0. Formulate" " " Determine a formula for the total cost of the equipment and then write an inequality that puts the total cost within budget. Find " " " " the related equation of the inequality to determine the boundary of the solution set and note whether the boundary line is " " " " included or excluded. Test a point not on the boundary line to determine which side contains solutions to the inequality. Graph " " " " the solution set in the first quadrant of the sb-plane. Determine" " " The softballs cost 6s dollars, and the bats cost 15b dollars. The total cost for the equipment is 6s + 15b dollars, which cannot exceed " " " " the $3,000 budget. The inequality is 6s + 15b 3000. The related equation is 6s + 15b = 3000. The boundary line is included in the " " " " solution set because the inequality is nonstrict. The test point (0, 0) is not on the boundary and yields the true inequality 0 3000. " " " " Solutions to the inequality must lie on or below the boundary line in the first quadrant. Justify" " " The linear inequality 6s + 15b 3000 describes the number of softballs and bats the school district can " " " " afford. The graph of the region shown to the right contains all the reasonable solutions to the " " " " inequality. Evaluate" " " Representing the problem as an inequality and graphing it produced a clear solution to the problem " " " " in a systematic way. The answer is reasonable because the total cost of all the equipment is within the " " " " school district s budget, and the variables s and b take on values that are reasonable for this problem. Chapter 4 - Applying Linear Functions 297 Section continues
4.6 - Linear Inequalities Linear Inequalities Graphically Not only can graphs be produced from algebraic inequalities, but the reverse can be done as well. The graph of the solution set of a linear inequality can be used to determine the linear inequality in algebraic form. The key is to find the equation of the boundary line of the solution set. This equation can be determined using information provided by the graph, such as the coordinates of two points that lie on the boundary line. Jump to Slope-Intercept Form Once the equation of the boundary line is known, the type of inequality can also be determined from geometric information provided by the graph. If the boundary line is dashed, then it is excluded from the solution set, and the inequality is strict. If the boundary line is solid, then it is included in the solution set, and the inequality is nonstrict. The direction of the inequality can be determined by testing a point that does not lie on the boundary line. Chapter 4 - Applying Linear Functions 298 Section continues
4.6 - Linear Inequalities Example problem"" The graph to the right depicts the solution set of a linear inequality in two variables. Find the " " " " analytical representation of this linear inequality. Analyze" " " The problem provides a graphical representation of a linear inequality in the variables x and y. It asks " " " " for an analytical representation. Formulate" " " Use the intercepts and the definition of slope as rise over run to find the slope of the boundary line. " " " " Then, find the equation of the boundary line in slope-intercept form y = mx + b. The inequality is " " " " strict because the boundary line is dashed. Since the boundary line does not pass through the origin, " " " " use the test point (0, 0) to determine the direction of the inequality. Determine" " " The x- and y-intercepts of the boundary line are 1 and 2. The line has a rise of 2 and run of 1, which means the slope is 2. The " " " " slope-intercept form of the equation of the boundary line is y = 2x + 2. Substituting the test point (0, 0) into the linear inequality " " " " yields the false inequality 0 > 2. Since this test point lies outside the solution set, the direction of the strict inequality is y > 2x + 2. Justify" " " The analytical form of the linear inequality is y > 2x + 2. The related equation of this linear inequality is y = 2x + 2, which has the " " " " same intercepts as the boundary line of the solution set in the graph. The strict inequality of the analytical form is consistent with " " " " the dashed boundary line depicted in the graph. The test point ( 1, 2) lies in the graphical solution set. Substituting ( 1, 2) into the " " " " linear inequality y > 2x + 2 yields 2 > 2 + 2, which simplifies to the true inequality 2 > 0. This means the test point ( 1, 2) solves " " " " the analytical form of the inequality as well as the graphical form. Evaluate" " " The process is effective because it uses a combination of graphical information and algebraic methods to obtain the analytical form " " " " of the linear inequality from its graphical representation. The answer is reasonable because the related equation y = 2x + 2 of the " " " " analytical representation y > 2x + 2 is the equation of the boundary line of the solution set in the graphical representation. In " " " " addition, the test point ( 1, 2) solves both the analytical representation and the graphical representation of the linear inequality. Chapter 4 - Applying Linear Functions 299 Section complete