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