Split (1)

train · 15k rows

text stringlengths 235 3.08k
an equation is also useful if we know any two points through which a line passes. Suppose, for example, we know that a line passes through the points (0, 1) and (3, 2). We can use the coordinates of the two points to find the slope. SECTION 2.1 linear Functions 133 m = y2 − y1 _ x2 − x1 2 − 1 _ 3 − 0 1 __ = 3 = Now we can use the slope we found and the coordinates of one of the points to find the equation for the line. Let use (0, 1) for our point. y − y1 = m(x − x1) 1 __ (x − 0) y − 1 = 3 As before, we can use algebra to rewrite the equation in the slope-intercept form. 1 __ y − 1 = (x − 0) 3 1 __ y − 1 = x 3 1 __ x + 1 y = 3 1 __. Distribute the 3 Add 1 to each side. Both equations describe the line shown in Figure 8. y 6 5 4 3 2 1 –6 –5 –4 –3 –2 3 4 5 6 x 21 –1 –1 Figure 8 –2 –3 –4 –5 –6 Example 6 Writing Linear Equations Using Two Points Write the point-slope form of an equation of a line that passes through the points (5, 1) and (8, 7). Then rewrite it in the slope-intercept form. Solution Let’s begin by finding the slope. m = = y2 − y1 _ x2 − x1 7 − 1 _ 8 − 5 6 __ = 3 = 2 So m = 2. Next, we substitute the slope and the coordinates for one of the points into the general point-slope equation. We can choose either point, but we will use (5, 1). y − y1 = m(x − x1) y − 1 = 2(x − 5) The point-slope equation of the line is y2 − 1 = 2(x2 − 5). To rewrite the equation in slope-intercept form, we use algebra. y − 1 = 2(x − 5) y − 1 = 2x − 10 y = 2x − 9 The slope-intercept equation of the line is y = 2x − 9. 134 CHAPTER 2 linear Functions Try It #4 Write the point-slope form of an equation of a line that passes
through the points (−1, 3) and (0, 0). Then rewrite it in the slope-intercept form. Writing and Interpreting an equation for a linear Function Now that we have written equations for linear functions in both the slope-intercept form and the point-slope form, we can choose which method to use based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function f in Figure 9. y f 10 8 6 4 2 –2 –2 –4 (0, 7) (4, 4) 42 6 8 10 x Figure 9 –10 –8 –6 –4 We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let’s choose (0, 7) and (4, 4). We can use these points to calculate the slope. m = y2 − y1 _ x2 − x1 = __ 4 Now we can substitute the slope and the coordinates of one of the points into the point-slope form. If we want to rewrite the equation in the slope-intercept form, we would find y − y1 = m(x − x1) y − 4 = − 3 __ (x − 4) 4 y − 4 = − 3 __ (x − 4) 4 y − 4 = − 3 __ x + 3 4 y = − 3 __ x + 7 4 If we wanted to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y-axis when the output value is 7. Therefore, b = 7. We now have the initial value b and the slope m so we can substitute m and b into the slope-intercept form of a line. f (x) = mx + b ↑ ↑ − 3 __ 7 4 f (x) = − 3 __ x + 7 4 So the function is f (x) = − 3 x + 7, and the linear equation would be y = − 3 __ __ x + 7. 4 4 SECTION 2.1 linear Functions 135 How To… Given the graph of a linear function, write an equation to represent the function. 1. Identify two points on the line. 2. Use the two points to calculate the slope. 3. Determine where the line crosses the y
-axis to identify the y-intercept by visual inspection. 4. Substitute the slope and y-intercept into the slope-intercept form of a line equation. Example 7 Writing an Equation for a Linear Function Write an equation for a linear function given a graph of f shown in Figure 10. y f 642 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 Figure 10 Solution Identify two points on the line, such as (0, 2) and (−2, −4). Use the points to calculate the slope. m = y2 − y1 _ x2 − x1 −4 − 2 _ −2 − 0 −6 _ −2 = = Substitute the slope and the coordinates of one of the points into the point-slope form. = 3 y − y1 = m(x − x1) y − (−4) = 3(x − (−2)) y + 4 = 3(x + 2) We can use algebra to rewrite the equation in the slope-intercept form. y + 4 = 3(x + 2) y + 4 = 3x + 6 y = 3x + 2 Analysis This makes sense because we can see from Figure 11 that the line crosses the y-axis at the point (0, 2), which is the y-intercept, so b = 2. y 10 8 6 4 2 –10 –2 –4 –6 –8 (−2, −4) –2 –4 –6 –8 –10 (0, 2) 642 8 10 x Figure 11 136 CHAPTER 2 linear Functions Example 8 Writing an Equation for a Linear Cost Function Suppose Ben starts a company in which he incurs a fixed cost of $1,250 per month for the overhead, which includes his office rent. His production costs are $37.50 per item. Write a linear function C where C(x) is the cost for x items produced in a given month. Solution The fixed cost is present every month, $1,250. The costs that can vary include the cost to produce each item, which is $37.50 for Ben. The variable cost, called the marginal cost, is represented by 37.5. The cost Ben incurs is the sum of these two costs, represented by C(x) = 1250 + 37.5x. Analysis If Ben produces 100 items in a month, his monthly cost
is represented by So his monthly cost would be $5,000. C(100) = 1250 + 37.5(100) = 5000 Example 9 Writing an Equation for a Linear Function Given Two Points If f is a linear function, with f (3) = −2, and f (8) = 1, find an equation for the function in slope-intercept form. Solution We can write the given points using coordinates. We can then use the points to calculate the slope. f (3) = −2 → (3, −2) f (8) = 1 → (8, 1) m = y2 − y1 _ x2 − x1 1 − (−2) ________ 8 − 3 = Substitute the slope and the coordinates of one of the points into the point-slope form. = 3 __ 5 y − y1 = m(x − x1) y − (−2) = 3 __ (x − 3) 5 We can use algebra to rewrite the equation in the slope-intercept form. y + 2 = 3 __ (x − 3 __ __ 5 5 19 y = 3 __ __ x − 5 5 Try It #5 If f (x) is a linear function, with f (2) = −11, and f (4) = −25, find an equation for the function in slope-intercept form. Modeling Real-World Problems with linear Functions In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems. SECTION 2.1 linear Functions 137 How To… Given a linear function f and the initial value and rate of change, evaluate f (c). 1. Determine the initial value and the rate of change (slope). 2. Substitute the values into f (x) = mx + b. 3. Evaluate the function at x = c. Example 10 Using a Linear Function to Determine the Number of Songs in a Music Collection Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, N, in his collection as a function of time, t, the number of
months. How many songs will he own in a year? Solution The initial value for this function is 200 because he currently owns 200 songs, so N(0) = 200, which means that b = 200. The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that m = 15. We can substitute the initial value and the rate of change into the slope-intercept form of a line. f (x) = mx + b ↑ ↑ 15 200 N(t) = 15t + 200 Figure 12 We can write the formula N(t) = 15t + 200. With this formula, we can then predict how many songs Marcus will have in 1 year (12 months). In other words, we can evaluate the function at t = 12. N(12) = 15(12) + 200 = 180 + 200 = 380 Marcus will have 380 songs in 12 months. Analysis Notice that N is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well. Example 11 Using a Linear Function to Calculate Salary Plus Commission Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income, I, depends on the number of new policies, n, he sells during the week. Last week he sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies and earned $920. Find an equation for I(n), and interpret the meaning of the components of the equation. Solution The given information gives us two input-output pairs: (3,760) and (5,920). We start by finding the rate of change. m = 920 − 760 ________ 5 − 3 $160 _______ 2 policies = $80 per policy = Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies increased by 2, so the rate of change is $80 per policy. Therefore, Ilya earns a commission of $80 for each policy sold during the week. 138 CHAPTER 2 linear Functions We can then solve for the initial value. I(n) = 80n + b 760 = 80(3) + b when n = 3, I(3) = 760 760 − 80(3) = b 520 = b The value of b is the starting value for the function and
represents Ilya’s income when n = 0, or when no new policies are sold. We can interpret this as Ilya’s base salary for the week, which does not depend upon the number of policies sold. We can now write the final equation. I(n) = 80n + 520 Our final interpretation is that Ilya’s base salary is $520 per week and he earns an additional $80 commission for each policy sold. Example 12 Using Tabular Form to Write an Equation for a Linear Function Table 1 relates the number of rats in a population to time, in weeks. Use the table to write a linear equation. w, number of weeks 0 2 4 6 P(w), number of rats 1,000 1,080 1,160 1,240 Table 1 Solution We can see from the table that the initial value for the number of rats is 1000, so b = 1000. Rather than solving for m, we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week. If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using (2, 1080) and (6, 1240) P(w) = 40w + 1000 m = 1240 − 1080 __________ 6 − 2 = 160 ___ 4 = 40 Q & A… Is the initial value always provided in a table of values like Table 1? No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into f (x) = mx + b, and solve for b. Try It #6 A new plant food was introduced to a young tree to test its effect on the height of the tree. Table 2 shows the height of the tree, in feet, x months since the measurements began. Write a linear function, H(x), where x is the number of months since the start of the experiment. x 0 2 4 8 12 H(x) 12.5 13.5 14.5 16.5 18.5 Table 2 Access this online
resource for additional instruction and practice with linear functions. • linear Functions (http://openstaxcollege.org/l/linearfunctions) SECTION 2.1 section exercises 139 2.1 SeCTIOn exeRCISeS VeRBAl 1. Terry is skiing down a steep hill. Terry’s elevation, E(t), in feet after t seconds is given by E(t) = 3000 − 70t. Write a complete sentence describing Terry’s starting elevation and how it is changing over time. 2. Maria is climbing a mountain. Maria’s elevation, E(t), in feet after t minutes is given by E(t) = 1200 + 40t. Write a complete sentence describing Maria’s starting elevation and how it is changing over time. 3. Jessica is walking home from a friend’s house. After 2 minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate in miles per hour? 4. Sonya is currently 10 miles from home and is walking farther away at 2 miles per hour. Write an equation for her distance from home t hours from now. 5. A boat is 100 miles away from the marina, sailing directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after t hours. 6. Timmy goes to the fair with $40. Each ride costs $2. How much money will he have left after riding n rides? AlGeBRAIC For the following exercises, determine whether the equation of the curve can be written as a linear function. 7. y = 1 __ x + 6 4 8. y = 3x − 5 9. y = 3x 2 − 2 10. 3x + 5y = 15 11. 3x 2 + 5y = 15 12. 3x + 5y 2 = 15 13. −2x 2 + 3y2 = 6 14. − x − 3 ______ 5 = 2y For the following exercises, determine whether each function is increasing or decreasing. 15. f (x) = 4x + 3 16. g(x) = 5x + 6 17. a(x) = 5 − 2x 18. b(x) = 8 − 3x 19. h(x) = −2x + 4 20. k(x) = −4x + 1 21. j(x) =
1 __ x − 3 2 24. m(x) = − 3 __ x + 3 8 22. p(x) = 1 __ x − 5 4 23. n(x) = − 1 __ x − 2 3 For the following exercises, find the slope of the line that passes through the two given points. 25. (2, 4) and (4, 10) 26. (1, 5) and (4, 11) 27. (−1, 4) and (5, 2) 28. (8, −2) and (4, 6) 29. (6, 11) and (−4, 3) 140 CHAPTER 2 linear Functions For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible. 30. f (−5) = −4, and f (5) = 2 31. f (−1) = 4 and f (5) = 1 32. (2, 4) and (4, 10) 33. Passes through (1, 5) and (4, 11) 34. Passes through (−1, 4) and (5, 2) 35. Passes through (−2, 8) and (4, 6) 36. x-intercept at (−2, 0) and y-intercept at (0, −3) 37. x-intercept at (−5, 0) and y-intercept at (0, 4) GRAPHICAl For the following exercises, find the slope of the lines graphed. 39. 321 4 5 6 x –6 –5 –4 –3 –2 38. –6 –5 –4 –3 –2 y 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 40. 321 4 5 6 x –6 –5 –4 –3 –2 For the following exercises, write an equation for the lines graphed. 41. 44. –6 –5 –4 –3 –2 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 42. 321 4 5 6 x –6 –5 –4 –3 –2 45. 321 4 5 x –6 –
5 –4 –3 –2 43. 321 4 5 6 x –5 –4 –3 –2 46. 321 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 y 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 y 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 321 4 5 6 x 321 4 5 x 321 4 5 6 x SECTION 2.1 section exercises 141 nUMeRIC For the following exercises, which of the tables could represent a linear function? For each that could be linear, find a linear equation that models the data. 47. x g(x) 5 15 0 5 −10 −25 −40 10 48. x h(x) 0 5 5 30 10 105 15 230 49. x 0 f (x) −5 10 28 4 16 20 58 6 36 25 73 8 56 50. x k(x) 5 13 53. x 2 f (x) −4 TeCHnOlOGY 51. x g(x) 2 6 0 6 −19 −44 −69 4 52. x f (x) 2 13 54. x k(x) 0 6 2 31 6 106 8 231 5 20 4 23 10 45 6 43 15 70 8 53 55. If f is a linear function, f (0.1) = 11.5, and f (0.4) = −5.9, find an equation for the function. 56. Graph the function f on a domain of [−10, 10] : f (x) = 0.02x − 0.01. Enter the function in a graphing utility. For the viewing window, set the minimum value of x to be −10 and the maximum value of x to be 10. 57. Graph the function f on a domain of [−10, 10] : f (x) = 2,500x + 4,000 58. Table 3 shows the input, w, and output, k, for a linear function k. a. Fill in the missing values of the table. b. Write the linear function k, round to 3
decimal places. w −10 5.5 67.5 b k 30 −26 a −44 Table 3 59. Table 4 shows the input, p, and output, q, for a linear function q. a. Fill in the missing values of the table. b. Write the linear function k. p q 0.5 400 0.8 700 12 a b 1,000,000 Table 4 1 __ 60. Graph the linear function f on a domain of [−10, 10] for the function whose slope is and y-intercept is 8 31 __. 16 Label the points for the input values of −10 and 10. 61. Graph the linear function f on a domain of [−0.1, 0.1] for the function whose slope is 75 and y-intercept is −22.5. Label the points for the input values of −0.1 and 0.1. 62. Graph the linear function f where f (x) = ax + b on the same set of axes on a domain of [−4, 4] for the following values of a and b. a. a = 2; b = 3 b. a = 2; b = 4 c. a = 2; b = −4 d. a = 2; b = −5 exTenSIOnS 63. Find the value of x if a linear function goes through the following points and has the following slope: (x, 2), (−4, 6), m = 3 64. Find the value of y if a linear function goes through the following points and has the following slope: (10, y), (25, 100), m = −5 65. Find the equation of the line that passes through the 66. Find the equation of the line that passes through the following points: (a, b) and (a, b + 1) following points: (2a, b) and (a, b + 1) 67. Find the equation of the line that passes through the following points: (a, 0) and (c, d) 142 CHAPTER 2 linear Functions ReAl-WORlD APPlICATIOnS 68. At noon, a barista notices that she has $20 in her 69. A gym membership with two personal training tip jar. If she makes an average of $0.50 from each customer, how much will she have in her tip jar if she serves n more customers during her shift? sessions costs $125, while
gym membership with five personal training sessions costs $260. What is cost per session? 70. A clothing business finds there is a linear relationship between the number of shirts, n, it can sell and the price, p, it can charge per shirt. In particular, historical data shows that 1,000 shirts can be sold at a price of $30, while 3,000 shirts can be sold at a price of $22. Find a linear equation in the form p(n) = mn + b that gives the price p they can charge for n shirts. 71. A phone company charges for service according to the formula: C(n) = 24 + 0.1n, where n is the number of minutes talked, and C(n) is the monthly charge, in dollars. Find and interpret the rate of change and initial value. 72. A farmer finds there is a linear relationship between the number of bean stalks, n, she plants and the yield, y, each plant produces. When she plants 30 stalks, each plant yields 30 oz of beans. When she plants 34 stalks, each plant produces 28 oz of beans. Find a linear relationship in the form y = mn + b that gives the yield when n stalks are planted. 73. A city’s population in the year 1960 was 287,500. In 1989 the population was 275,900. Compute the rate of growth of the population and make a statement about the population rate of change in people per year. 74. A town’s population has been growing linearly. In 2003, the population was 45,000, and the population has been growing by 1,700 people each year. Write an equation, P(t), for the population t years after 2003. 75. Suppose that average annual income (in dollars) for the years 1990 through 1999 is given by the linear function: I(x) = 1,054x + 23,286, where x is the number of years after 1990. Which of the following interprets the slope in the context of the problem? a. As of 1990, average annual income was $23,286. b. In the ten-year period from 1990–1999, average annual income increased by a total of $1,054. c. Each year in the decade of the 1990s, average annual income increased by $1,054. d. Average annual income rose to a level of $23,286 by the end of 1999
. 76. When temperature is 0 degrees Celsius, the Fahrenheit temperature is 32. When the Celsius temperature is 100, the corresponding Fahrenheit temperature is 212. Express the Fahrenheit temperature as a linear function of C, the Celsius temperature, F (C). a. Find the rate of change of Fahrenheit temperature for each unit change temperature of Celsius. b. Find and interpret F (28). c. Find and interpret F (−40). SECTION 2.2 graphs oF linear Functions 143 leARnInG OBjeCTIVeS In this section, you will: • • • • • Graph linear functions. Write the equation for a linear function from the graph of a line. Given the equations of two lines, determine whether their graphs are parallel or perpendicular. Write the equation of a line parallel or perpendicular to a given line. Solve a system of linear equations. 2.2 GRAPHS OF lIneAR FUnCTIOnS Two competing telephone companies offer different payment plans. The two plans charge the same rate per long distance minute, but charge a different monthly flat fee. A consumer wants to determine whether the two plans will ever cost the same amount for a given number of long distance minutes used. The total cost of each payment plan can be represented by a linear function. To solve the problem, we will need to compare the functions. In this section, we will consider methods of comparing functions using graphs. Graphing linear Functions In Linear Functions, we saw that the graph of a linear function is a straight line. We were also able to see the points of the function as well as the initial value from a graph. By graphing two functions, then, we can more easily compare their characteristics. There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the y-intercept and slope. And the third is by using transformations of the identity function f (x) = x. Graphing a Function by Plotting Points To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. The input values and corresponding output values form coordinate pairs. We then plot the coordinate pairs on a grid. In general, we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph. For example, given the function, f (x) = 2x, we might use the input values 1 and 2.
Evaluating the function for an input value of 1 yields an output value of 2, which is represented by the point (1, 2). Evaluating the function for an input value of 2 yields an output value of 4, which is represented by the point (2, 4). Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error. How To… Given a linear function, graph by plotting points. 1. Choose a minimum of two input values. 2. Evaluate the function at each input value. 3. Use the resulting output values to identify coordinate pairs. 4. Plot the coordinate pairs on a grid. 5. Draw a line through the points. Graphing by Plotting Points Example 1 Graph f (x) = − 2 __ x + 5 by plotting points. 3 Solution Begin by choosing input values. This function includes a fraction with a denominator of 3, so let’s choose multiples of 3 as input values. We will choose 0, 3, and 6. 144 CHAPTER 2 linear Functions Evaluate the function at each input value, and use the output value to identify coordinate pairs0) = − 2 __ (0) + 5 = 5 3 f (3) = − 2 __ (3) + 5 = 3 3 f (6) = − 2 __ (6) + 5 = 1 3 (0, 5) (3, 3) (6, 1) Plot the coordinate pairs and draw a line through the points. Figure 1 represents the graph of the function 2 __ x + 5. f (x) = − 3 f(x) (0, 5) f 6 5 4 3 2 1 (3, 3) (6, 1) – – __ x + 5. Figure 1 The graph of the linear function f (x) = − 3 Analysis The graph of the function is a line as expected for a linear function. In addition, the graph has a downward slant, which indicates a negative slope. This is also expected from the negative constant rate of change in the equation for the function. Try It #1 Graph f (x) = − 3 __ x + 6 by plotting points. 4 Graphing a Function Using y-intercept and Slope Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. The first characteristic is its y-intercept, which is the point at which the input value is zero. To find
the y-intercept, we can set x = 0 in the equation. The other characteristic of the linear function is its slope m, which is a measure of its steepness. Recall that the slope is the rate of change of the function. The slope of a function is equal to the ratio of the change in outputs to the change in inputs. Another way to think about the slope is by dividing the vertical difference, or rise, by the horizontal difference, or run. We encountered both the y-intercept and the slope in Linear Functions. Let’s consider the following function. f (x) = 1 __ x + 1 2 1 __ The slope is. Because the slope is positive, we know the graph will slant upward from left to right. The y-intercept is 2 the point on the graph when x = 0. The graph crosses the y-axis at (0, 1). Now we know the slope and the y-intercept. We can begin graphing by plotting the point (0, 1). We know that the slope is rise over run, m = From our example, we have m = 1 __, which means that the rise is 1 and the run is 2. So starting from our y-intercept 2 (0, 1), we can rise 1 and then run 2, or run 2 and then rise 1. We repeat until we have a few points, and then we draw a line through the points as shown in Figure 2. rise ___. run y 5 4 3 2 (0, 1) –2 –1 f y-intercept ←Rise = 1 ↑Run = 2 1 3 4 2 Figure 2 x 5 6 7 SECTION 2.2 graphs oF linear Functions 145 graphical interpretation of a linear function In the equation f (x) = mx + b • b is the y-intercept of the graph and indicates the point (0, b) at which the graph crosses the y-axis. • m is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope: m = change in output (rise) __ change in input (run) = Δy _ Δx = y2 − y1 _ x2 − x1 Q & A… Do all linear functions have y-intercepts? Yes. All linear functions cross the y-axis and therefore have y-intercepts. (Note: A
vertical line parallel to the y-axis does not have a y-intercept, but it is not a function.) How To… Given the equation for a linear function, graph the function using the y-intercept and slope. 1. Evaluate the function at an input value of zero to find the y-intercept. 2. Identify the slope as the rate of change of the input value. 3. Plot the point represented by the y-intercept. 4. Use to determine at least two more points on the line. rise ___ run 5. Sketch the line that passes through the points. Graphing by Using the y-intercept and Slope Example 2 Graph f (x) = − 2 __ x + 5 using the y-intercept and slope. 3 Solution Evaluate the function at x = 0 to find the y-intercept. The output value when x = 0 is 5, so the graph will cross the y-axis at (0, 5). According to the equation for the function, the slope of the line is − 2 __. This tells us that for each vertical decrease in 3 the “rise” of −2 units, the “run” increases by 3 units in the horizontal direction. We can now graph the function by first plotting the y-intercept on the graph in Figure 3. From the initial value (0, 5) we move down 2 units and to the right 3 units. We can extend the line to the left and right by repeating, and then draw a line through the points. f f(x) 6 5 4 3 2 1 – –1 1 2 3 4 5 6 7 x Figure 3 Analysis The graph slants downward from left to right, which means it has a negative slope as expected. Try It #2 Find a point on the graph we drew in Example 2 that has a negative x-value. 146 CHAPTER 2 linear Functions Graphing a Function Using Transformations Another option for graphing is to use transformations of the identity function f (x) = x. A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression. Vertical Stretch or Compression In the equation f (x) = mx, the m is acting as the vertical stretch or compression of the identity function. When m is negative, there is also a vertical reflection of the graph. Notice in Figure 4 that multiplying the equation of f
(x) = x by m stretches the graph of f by a factor of m units if m > 1 and compresses the graph of f by a factor of m units if 0 < m < 1. This means the larger the absolute value of m, the steeper the slope. y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 1 2 3 4 5 6 x Figure 4 Vertical stretches and compressions and reflections on the function f ( x ) = x. Vertical Shift In f (x) = mx + b, the b acts as the vertical shift, moving the graph up and down without affecting the slope of the line. Notice in Figure 5 that adding a value of b to the equation of f (x) = x shifts the graph of f a total of b units up if b is positive and ∣b∣ units down if b is negative. f(x) = x + 4 f(x) = x + 2 f(x) = x f(x) = x – 2 f(x) = x – 4 x 42 6 8 10 y 10 8 6 4 2 –2 –2 –4 –6 –8 –10 –10 –8 –6 –4 Figure 5 This graph illustrates vertical shifts of the function f ( x ) = x. Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method. How To… Given the equation of a linear function, use transformations to graph the linear function in the form f (x) = mx + b. 1. Graph f (x) = x. 2. Vertically stretch or compress the graph by a factor m. 3. Shift the graph up or down b units. SECTION 2.2 graphs oF linear Functions 147 Graphing by Using Transformations Example 3 Graph f (x) = 1 __ x − 3 using transformations. 2 Solution The equation for the function shows that m = 1 1 __ __. The so the identity function is vertically compressed by 2 2 equation for the function also shows that b = −3 so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression as in Figure 6. Then show the vertical shift as in Figure 7. y y = x –
7 –6 –5 –4 –3 –2 4 3 2 1 –1 –1 –2 –3 –4 y = x1 2 x 5 6 7 21 3 4 y 5 4 3 2 1 –1 –1 –2 –3 –4 –7 –6 –5 –4 –3 –2 y = x1 2 1 y = x − 3 2 x 21 3 4 5 6 7 Figure 6 The function, y =x compressed by a factor of 2. 1__ Figure 7 The function y = __1 2 x, shifted down 3 units. –5 Try It #3 Graph f (x) = 4 + 2x, using t ransformations. Q & A… In Example 3, could we have sketched the graph by reversing the order of the transformations? No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following the order: Let the input be 2. f (2) = 1 __ (22 Writing the equation for a Function from the Graph of a line Recall that in Linear Functions, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 8. We can see right away that the graph crosses the y-axis at the point (0, 4) so this is the y-intercept. y f 42 6 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 Figure 8 Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point (−2, 0). To get from this point to the y-intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be m = rise ___ run = 4 __ = 2 2 Substituting the slope and y-intercept into the slope-intercept form of a line gives y = 2x + 4 148 CHAPTER 2 linear Functions How To… Given a graph of linear function, find the equation to describe the function. 1. Identify the y-intercept of an equation. 2. Choose two points to determine the slope. 3. Substitute
the y-intercept and slope into the slope-intercept form of a line. Example 4 Matching Linear Functions to Their Graphs Match each equation of the linear functions with one of the lines in Figure 9. a. f (x) = 2x + 3 b. g(x) = 2x − 3 c. h(x) = −2x + 3 d. j(x) = 1 __ x + 3 2 –7 –6 –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 I II III 21 3 4 5 6 7 x IV Figure 9 Solution Analyze the information for each function. a. This function has a slope of 2 and a y-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function g has the same slope, but a different y-intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through (0, 3) so f must be represented by line I. b. This function also has a slope of 2, but a y-intercept of −3. It must pass through the point (0, −3) and slant upward from left to right. It must be represented by line III. c. This function has a slope of −2 and a y-intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right. 1 __ and a y-intercept of 3. It must pass through the point (0, 3) and slant upward d. This function has a slope of 2 from left to right. Lines I and II pass through (0, 3), but the slope of j is less than the slope of f so the line for j must be flatter. This function is represented by Line II. Now we can re-label the lines as in Figure 10. 1 j(x) = x + 3 2 –7 –6 –5 –4 –3 –2 g(x) = 2x − 3 21 1 –1 –2 –3 –4 –5 f (x) = 2x + 3 h(x) = −2x + 3 SECTION 2.
2 graphs oF linear Functions 149 Finding the x-intercept of a Line Figure 10 So far, we have been finding the y-intercepts of a function: the point at which the graph of the function crosses the y-axis. A function may also have an x-intercept, which is the x-coordinate of the point where the graph of the function crosses the x-axis. In other words, it is the input value when the output value is zero. To find the x-intercept, set a function f (x) equal to zero and solve for the value of x. For example, consider the function shown. Set the function equal to 0 and solve for x. f (x) = 3x − 6 0 = 3x − 6 6 = 3x 2 = x x = 2 The graph of the function crosses the x-axis at the point (2, 0). Q & A… Do all linear functions have x-intercepts? No. However, linear functions of the form y = c, where c is a nonzero real number are the only examples of linear functions with no x-intercept. For example, y = 5 is a horizontal line 5 units above the x-axis. This function has no x-intercepts, as shown in Figure 11. y = 5 21 1–1 –2 –3 –4 –5 –5 –4 –3 –2 Figure 11 x-intercept The x-intercept of the function is value of x when f (x) = 0. It can be solved by the equation 0 = mx + b. Finding an x-intercept Example 5 Find the x-intercept of f (x) = 1 __ x − 3. 2 Solution Set the function equal to zero to solve for x. 0 = 1 __ x − 3 2 3 = 1 __ x 2 6 = x x = 6 150 CHAPTER 2 linear Functions The graph crosses the x-axis at the point (6, 0). Analysis A graph of the function is shown in Figure 12. We can see that the x-intercept is (6, 0) as we expected. –10 –8 –6 –4 y 4 2 –2 –2 –4 –6 –8 –10 642 8 10 x 1 __ x − 3. Figure 12 The graph of the linear function f (x) = 2 Try It #4 Find the x-intercept of f (x) = 1 __
x − 4. 4 Describing Horizontal and Vertical Lines There are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line indicates a constant output, or y-value. In Figure 13, we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use m = 0 in the equation f (x) = mx + b, the equation simplifies to f (x) = b. In other words, the value of the function is a constant. This graph represents the function f (x) = 24 −5 –4 –3 –2 –1 21 3 4 5 x Figure 13 A horizontal line representing the function f (x) = 2. A vertical line indicates a constant input, or x-value. We can see that the input value for every point on the line is 2, but the output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator will be zero, so the slope of a vertical line is undefined. m = change of output __ change of input ← Non-zero real number ← 0 Notice that a vertical line, such as the one in Figure 14, has an x-intercept, but no y-intercept unless it’s the line x = 0. This graph represents the line x = 2. y –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x 2 2 x y −4 −2 2 0 2 2 2 4 SECTION 2.2 graphs oF linear Functions 151 Figure 14 The vertical line, x = 2, which does not represent a function. horizontal and vertical lines Lines can be horizontal or vertical. A horizontal line is a line defined by an equation in the form f (x) = b. A vertical line is a line defined by an equation in the form x = a. Example 6 Writing the Equation of a Horizontal Line Write the equation of the line graphed in Figure 15. –10 –8 –6 –4 y –2 –2 –4 –6 –8 –10 642 8 10 x f Figure 15 Solution
For any x-value, the y-value is −4, so the equation is y = −4. Example 7 Writing the Equation of a Vertical Line Write the equation of the line graphed in Figure 16. y 10 8 6 4 2 –2 –2 –4 –6 –8 –10 –10 –8 –6 –4 f 42 6 8 10 x Figure 16 Solution The constant x-value is 7, so the equation is x = 7. Determining Whether lines are Parallel or Perpendicular The two lines in Figure 17 are parallel lines: they will never intersect. Notice that they have exactly the same steepness, which means their slopes are identical. The only difference between the two lines is the y-intercept. If we shifted one line vertically toward the y-intercept of the other, they would become the same line. 152 CHAPTER 2 linear Functions 1 –1 –2 –3 –6 –5 –4 –3 –2 21 3 4 5 6 x Figure 17 Parallel lines We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel. f (x) = −2x + 6 } f (x) = −2x − 4 parallel f (x) = 3x + 2 } f (x) = 2x + 2 not parallel Unlike parallel lines, perpendicular lines do intersect. Their intersection forms a right, or 90-degree, angle. The two lines in Figure 18 are perpendicular. –6 –5 –4 –3 –2 y 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –7 –8 21 3 4 5 6 x Figure 18 Perpendicular lines Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is 1. So, if m1 and m2 are negative reciprocals of one another, they can be multiplied together to yield −1. m1m2 = −1 1 1 __ __ is 8. To, and the reciprocal of To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is 8 8 find the negative reciprocal, first find the reciprocal and then change
the sign. As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes, assuming that the lines are neither horizontal nor perpendicular. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular. The product of the slopes is −1. f (x) = 1 1 __ __ is −4 x + 2 negative reciprocal of 4 4 1 __ f (x) = −4x + 3 negative reciprocal of −4 is 4 1 __  = −1 −4  4 parallel and perpendicular lines Two lines are parallel lines if they do not intersect. The slopes of the lines are the same. f (x) = m1x + b1 and g(x) = m2x + b2 are parallel if m1 = m2. SECTION 2.2 graphs oF linear Functions 153 If and only if b1 = b2 and m1 = m2, we say the lines coincide. Coincident lines are the same line. Two lines are perpendicular lines if they intersect at right angles. f (x) = m1x + b1 and g(x) = m2x + b2 are perpendicular if m1m2 = −1, and so m2 = − 1 _ m1 Example 8 Identifying Parallel and Perpendicular Lines Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines. f (x) = 2x + 3 g(x) = 1 __ x − 4 2 h(x) = −2x + 2 j(x) = 2x − 6 Solution Parallel lines have the same slope. Because the functions f (x) = 2x + 3 and j(x) = 2x − 6 each have a slope 1 __ of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and are negative 2 reciprocals, the equations, g(x) = 1 __ x − 4 and h(x) = −2x + 2 represent perpendicular lines. 2 Analysis A graph of the lines is shown in Figure 19. h(x) = −2x + 2 y f (x) = 2x + 3 j(x) = 2x − 6 1 g(x) = x − 4 2 42 6 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 The
graph shows that the lines f (x) = 2x + 3 and j(x) = 2x − 6 are parallel, and the lines g(x) = 1 __ x − 4 and 2 h(x) = −2x + 2 are perpendicular. Figure 19 Writing the equation of a line Parallel or Perpendicular to a Given line If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line. Writing Equations of Parallel Lines Suppose for example, we are given the following equation. f (x) = 3x + 1 We know that the slope of the line formed by the function is 3. We also know that the y-intercept is (0, 1). Any other line with a slope of 3 will be parallel to f (x). So the lines formed by all of the following functions will be parallel to f (x). g(x) = 3x + 6 h(x) = 3x + 1 p(x) = 3x + 2 __ 3 Suppose then we want to write the equation of a line that is parallel to f and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for b will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form. 154 CHAPTER 2 linear Functions y − y1 = m(x − x1) y − 7 = 3(x − 1) y − 7 = 3x − 3 y = 3x + 4 So g(x) = 3x + 4 is parallel to f (x) = 3x + 1 and passes through the point (1, 7). How To… Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point. 1. Find the slope of the function. 2. Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line. 3. Simplify. Example 9 Finding a Line Parallel to a Given Line Find a line parallel to the graph of f (x) = 3x + 6 that passes through the point (3, 0). Solution The slope of the given line is 3. If we choose the slope-intercept
form, we can substitute m = 3, x = 3, and f (x) = 0 into the slope-intercept form to find the y-intercept. g(x) = 3x + b 0 = 3(3) + b b = −9 The line parallel to f (x) that passes through (3, 0) is g(x) = 3x − 9. Analysis We can confirm that the two lines are parallel by graphing them. Figure 20 shows that the two lines will never intersect. y 5 4 3 2 1 Right 1 –4 –6 –5 –3 y = 3x + 6 –2 –1 –1 –2 –3 –4 –5 Up 3 Up 3 1 2 4 5 6 3 Right 1 x y = 3x − 9 Writing Equations of Perpendicular Lines We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function: f (x) = 2x + 4. Any function with a slope of − 1 The slope of the line is 2, and its negative reciprocal is − 1 __ __ will be perpendicular to 2 2 f (x). So the lines formed by all of the following functions will be perpendicular to f (x). Figure 20 g(x) = − 1 __ x + 4 2 h(x) = − 1 __ x + 2 2 x − 1 p(x) = − 1 __ __ 2 2 As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to f (x) and passes through the point (4, 0). We already know that the slope is − 1 __. Now we can use the point to find the y-intercept by substituting 2 the given values into the slope-intercept form of a line and solving for b. SECTION 2.2 graphs oF linear Functions 155 g (x) = mx + b 0 = − 1 __ (4) + b 2 0 = −2 + b 2 = b The equation for the function with a slope of − 1 __ and a y-intercept of 2 is 2 b = 2 g(x) = − 1 __ x + 2. 2 So g(x) = − 1 __ x +
2 is perpendicular to f (x) = 2x + 4 and passes through the point (4, 0). Be aware that perpendicular 2 lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature. Q & A… A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not −1. Doesn’t this fact contradict the definition of perpendicular lines? No. For two perpendicular linear functions, the product of their slopes is −1. However, a vertical line is not a function so the definition is not contradicted. How To… Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line. 1. Find the slope of the function. 2. Determine the negative reciprocal of the slope. 3. Substitute the new slope and the values for x and y from the coordinate pair provided into g(x) = mx + b. 4. Solve for b. 5. Write the equation for the line. Example 10 Finding the Equation of a Perpendicular Line Find the equation of a line perpendicular to f (x) = 3x + 3 that passes through the point (3, 0). Solution The original line has slope m = 3, so the slope of the perpendicular line will be its negative reciprocal, or − 1 __. Using this slope and the given point, we can find the equation for the line. 3 g(x) = − 1 __ x + b 3 0 = − 1 __ (3 The line perpendicular to f (x) that passes through (3, 0) is g(x) = − 1 __ x + 1. 3 Analysis A graph of the two lines is shown in Figure 211 –1 –2 –3 –6 –5 –4 –3 –2 y f (x) = 3x + 6 1 g(x) = − x + 1 3 x 21 3 4 5 6 Figure 21 156 CHAPTER 2 linear Functions Try It #5 Given the function h(x) = 2x − 4, write an equation for the line passing through (0, 0) that is a. parallel to h(x) b. perpendicular to h(x) How To… Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point. 1. Determine the slope of the line passing through the
points. 2. Find the negative reciprocal of the slope. 3. Use the slope-intercept form or point-slope form to write the equation by substituting the known values. 4. Simplify. Example 11 Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point A line passes through the points (−2, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5). Solution From the two points of the given line, we can calculate the slope of that line. Find the negative reciprocal of the slope. m1 = 5 − 6 _______ 4 − (−2) = −1 ___ 6 = − 1 __ 6 m2 = −1 _ − 1 __ 6 = −1  − 6 __  1 = 6 We can then solve for the y-intercept of the line passing through the point (4, 5). g (x) = 6x + b 5 = 6(4) + b 5 = 24 + b −19 = b b = −19 The equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4, 5) is y = 6x − 19 Try It #6 A line passes through the points, (−2, −15) and (2,−3). Find the equation of a perpendicular line that passes through the point, (6, 4). SECTION 2.2 graphs oF linear Functions 157 Solving a System of linear equations Using a Graph A system of linear equations includes two or more linear equations. The graphs of two lines will intersect at a single point if they are not parallel. Two parallel lines can also intersect if they are coincident, which means they are the same line and they intersect at every point. For two lines that are not parallel, the single point of intersection will satisfy both equations and therefore represent the solution to the system. To find this point when the equations are given as functions, we can solve for an input value so that f (x) = g(x). In other words, we can set the formulas for the lines equal to one another, and solve for the input that satisfies the equation. Example 12 Finding a Point of Intersection Algebraically Find the point of intersection of the lines h(t) = 3t − 4 and j(t) = 5 − t. Solution Set h(t) = j(t). 3t − 4
= 5 − t 4t = 9 t = 9 __ 4 9 __ This tells us the lines intersect when the input is. 4 We can then find the output value of the intersection point by evaluating either function at this input.  = 5 − 9 9 __ __ j  4 4 11 __ 4 = 9 __ These lines intersect at the point , 4 11 __ . 4 Analysis Looking at Figure 22, this result seems reasonable. –5 –4 –3 –2 h(t), 9 4 11 4 x 21 3 4 5 j(t) y 5 4 3 2 1 –1–1 –2 –3 –4 –5 Figure 22 Q & A… If we were asked to find the point of intersection of two distinct parallel lines, should something in the solution process alert us to the fact that there are no solutions? Yes. After setting the two equations equal to one another, the result would be the contradiction “0 = non-zero real number.” 158 CHAPTER 2 linear Functions Try It #7 Look at the graph in Figure 22 and identify the following for the function j(t): a. y-intercept d. Is j(t) parallel or perpendicular to h(t) (or neither)? e. Is j(t) an increasing or decreasing function (or neither)? f. Write a transformation description for j(t) from the identity toolkit function f (x) = x. b. x-intercept(s) c. slope Example 13 Finding a Break-Even Point A company sells sports helmets. The company incurs a one-time fixed cost for $250,000. Each helmet costs $120 to produce, and sells for $140. a. Find the cost function, C, to produce x helmets, in dollars. b. Find the revenue function, R, from the sales of x helmets, in dollars. c. Find the break-even point, the point of intersection of the two graphs C and R. Solution a. The cost function in the sum of the fixed cost, $250,000, and the variable cost, $120 per helmet. C(x) = 120x + 250,000 b. The revenue function is the total revenue from the sale of x helmets, R(x) = 140x. c. The break-even point is the point of intersection of the graph of the cost and revenue functions. To find the x-coordinate of the
coordinate pair of the point of intersection, set the two equations equal, and solve for x. C(x) = R(x) 250,000 + 120x = 140x 250,000 = 20x 12,500 = x x = 12,500 To find y, evaluate either the revenue or the cost function at 12,500. R(x) = 140(12,500) = $1,750,000 The break-even point is (12,500, 1,750,000). Analysis This means if the company sells 12,500 helmets, they break even; both the sales and cost incurred equaled 1.75 million dollars. See Figure 23. y 2,000,000 Cost < Sales Company loses money 1,000,000 C R Sales > Cost Company makes a profit (12,500, 1,750,000) 0 0 x 20,000 Figure 23 Access these online resources for additional instruction and practice with graphs of linear functions. • Finding Input of Function from the Output and Graph (http://openstaxcollege.org/l/findinginput) • Graphing Functions Using Tables (http://openstaxcollege.org/l/graphwithtable) SECTION 2.2 section exercises 159 2.2 SeCTIOn exeRCISeS VeRBAl 1. If the graphs of two linear functions are parallel, 2. If the graphs of two linear functions are describe the relationship between the slopes and the y-intercepts. perpendicular, describe the relationship between the slopes and the y-intercepts. 4. Explain how to find a line parallel to a linear function that passes through a given point. 3. If a horizontal line has the equation f (x) = a and a vertical line has the equation x = a, what is the point of intersection? Explain why what you found is the point of intersection. 5. Explain how to find a line perpendicular to a linear function that passes through a given point. AlGeBRAIC For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular: 6. 4x − 7y = 10 7x + 4y = 1 9. 6x − 9y = 10 3x + 2y = 1 7. 3y + x = 12 −y = 8x + 1 10. y = 2 __ x + 1 3 3x + 2y = 1 For the following exercises, find the x- and y
-intercepts of each equation 12. f (x) = −x + 2 15. k(x) = −5x + 1 13. g(x) = 2x + 4 16. −2x + 5y = 20 8. 3y + 4x = 12 −6y = 8x + 1 11. y = 3 __ x + 1 4 −3x + 4y = 1 14. h(x) = 3x − 5 17. 7x + 2y = 56 For the following exercises, use the descriptions of each pair of lines given below to find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular, or neither? 18. Line 1: Passes through (0, 6) and (3, −24) Line 2: Passes through (−1, 19) and (8, −71) 19. Line 1: Passes through (−8, −55) and (10, 89) Line 2: Passes through (9, −44) and (4, −14) 20. Line 1: Passes through (2, 3) and (4, −1) Line 2: Passes through (6, 3) and (8, 5) 21. Line 1: Passes through (1, 7) and (5, 5) Line 2: Passes through (−1, −3) and (1, 1) 22. Line 1: Passes through (0, 5) and (3, 3) Line 2: Passes through (1, −5) and (3, −2) 23. Line 1: Passes through (2, 5) and (5, −1) Line 2: Passes through (−3, 7) and (3, −5) 24. Write an equation for a line parallel to f (x) = −5x − 3 25. Write an equation for a line parallel to and passing through the point (2, −12). g(x) = 3x − 1 and passing through the point (4, 9). 26. Write an equation for a line perpendicular to h(t) = −2t + 4 and passing through the point (−4, −1). 27. Write an equation for a line perpendicular to p(t) = 3t + 4 and passing through the point (3, 1). 160 CHAPTER 2 linear Functions 28. Find the point at which the line f (x
) = −2x − 1 29. Find the point at which the line f (x) = 2x + 5 intersects the line g(x) = −x. intersects the line g(x) = −3x − 5. 30. Use algebra to find the point at which the line 73 __. 10 intersects h(x) = 9 __ x + 4 f (x) = − 4 __ x + 5 274 ___ 25 31. Use algebra to find the point at which the line f (x) = 7 __ x + 4 457 ___ 60 intersects g(x) = 4 __ x + 3 31 __. 5 GRAPHICAl For the following exercises, match the given linear equation with its graph in Figure 24. B 5 A 4 3 2 1 –5 –4 –3 –2 –1–1 –2 –3 –4 –5 21 3 4 5 E F Figure 24 C D 32. f (x) = −x − 1 33. f (x) = −2x − 1 34. f (x) = − 1 __ x − 1 2 35. f (x) = 2 36. f (x) = 2 + x 37. f (x) = 3x + 2 For the following exercises, sketch a line with the given features. 38. An x-intercept of (−4, 0) and y-intercept of (0, −2) 39. An x-intercept of (−2, 0) and y-intercept of (0, 4) 40. A y-intercept of (0, 7) and slope − 3 __ 2 2 __ 41. A y-intercept of (0, 3) and slope 5 42. Passing through the points (−6, −2) and (6, −6) 43. Passing through the points (−3, −4) and (3, 0) For the following exercises, sketch the graph of each equation. 44. f (x) = −2x − 1 45. g(x) = −3x + 2 47. k(x) = 2 __ x − 3 3 50. x = 3 53. q(x) = 3 48. f (t) = 3 + 2t 51. x = −2 54. 4x = −9y + 36 56. 3x − 5y = 15 57. 3x = 15 46. h(x) = 1 __ x + 2
3 49. p(t) = −2 + 3t 52. r(x) = 4 y x _ __ = 1 − 55. 4 3 58. 3y = 12 59. If g(x) is the transformation of f (x) = x after a 3 __, a shift right by 2, and vertical compression by 4 a shift down by 4 a. Write an equation for g(x). b. What is the slope of this line? c. Find the y-intercept of this line. 60. If g(x) is the transformation of f (x) = x after a 1 __, a shift left by 1, and a vertical compression by 3 shift up by 3 a. Write an equation for g(x). b. What is the slope of this line? c. Find the y-intercept of this line. SECTION 2.2 section exercises 161 For the following exercises, write the equation of the line shown in the graph. 62. 21 3 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1–1 –2 –3 –4 –5 –6 63. 21 3 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1–1 –2 –3 –4 –5 –6 21 3 4 5 6 x 21 3 4 5 x 61. 64. –6 –5 –4 –3 –2 –5 –4 –3 –2 y 6 5 4 3 2 1 –1–1 –2 –3 –4 –5 –6 y 5 4 3 2 1 –1–1 –2 –3 –4 –5 For the following exercises, find the point of intersection of each pair of lines if it exists. If it does not exist, indicate that there is no point of intersection. 65. y = 3 __ x + 1 4 −3x + 4y = 12 68. x − 2y + 2 = 3 x − y = 3 exTenSIOnS 66. 2x − 3y = 12 5y + x = 30 69. 5x + 3y = − 65 x − y = − 5 67. 2x = y − 3 y + 4x = 15 70. Find the equation of the line parallel to the line g(x) = −0.01x + 2.01 through the point (1, 2). 71. Find the equation of the line perpendicular
to the line g(x) = −0.01x + 2.01 through the point (1, 2). For the following exercises, use the functions f (x) = −0.1x + 200 and g(x) = 20x + 0.1. 72. Find the point of intersection of the lines f and g. 73. Where is f (x) greater than g(x)? Where is g(x) greater than f (x)? ReAl-WORlD APPlICATIOnS 74. A car rental company offers two plans for renting a car. Plan A: $30 per day and $0.18 per mile Plan B: $50 per day with free unlimited mileage How many miles would you need to drive for plan B to save you money? 75. A cell phone company offers two plans for minutes. Plan A: $20 per month and $1 for every one hundred texts. Plan B: $50 per month with free unlimited texts. How many texts would you need to send per month for plan B to save you money? 76. A cell phone company offers two plans for minutes. Plan A: $15 per month and $2 for every 300 texts. Plan B: $25 per month and $0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money? 162 CHAPTER 2 linear Functions leARnInG OBjeCTIVeS In this section, you will: • • • Identify steps for modeling and solving. Build linear models from verbal descriptions. Build systems of linear models. 2.3 MODelInG WITH lIneAR FUnCTIOnS Figure 1 (credit: eeK Photography/Flickr) Emily is a college student who plans to spend a summer in Seattle. She has saved $3,500 for her trip and anticipates spending $400 each week on rent, food, and activities. How can we write a linear model to represent her situation? What would be the x-intercept, and what can she learn from it? To answer these and related questions, we can create a model using a linear function. Models such as this one can be extremely useful for analyzing relationships and making predictions based on those relationships. In this section, we will explore examples of linear function models. Identifying Steps to Model and Solve Problems When modeling scenarios with linear functions and solving problems involving quantities with a constant rate of change, we
typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them: 1. Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system. 2. Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value. 3. Carefully read the problem to determine what we are trying to find, identify, solve, or interpret. 4. Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem. 5. When needed, write a formula for the function. 6. Solve or evaluate the function using the formula. 7. Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically. 8. Clearly convey your result using appropriate units, and answer in full sentences when necessary. SECTION 2.3 modeling with linear Functions 163 Building linear Models Now let’s take a look at the student in Seattle. In her situation, there are two changing quantities: time and money. The amount of money she has remaining while on vacation depends on how long she stays. We can use this information to define our variables, including units. • Output: M, money remaining, in dollars • Input: t, time, in weeks So, the amount of money remaining depends on the number of weeks: M(t) We can also identify the initial value and the rate of change. • Initial Value: She saved $3,500, so $3,500 is the initial value for M. • Rate of Change: She anticipates spending $400 each week, so −$400 per week is the rate of change, or slope. Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week. The rate of change is constant, so we can start with the linear model M(t) = mt + b. Then we can substitute the intercept and slope provided. M(t) = mt + b ↑ ↑ −400 3500 M(t) = − 400t + 3500 To find the x-intercept, we set
the output to zero, and solve for the input. 0 = −400t + 3500 t = 3500 ____ 400 = 8.75 The x-intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Emily will have no money left after 8.75 weeks. When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be valid—almost no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make sense to talk about input values less than zero. A negative input value could refer to a number of weeks before she saved $3,500, but the scenario discussed poses the question once she saved $3,500 because this is when her trip and subsequent spending starts. It is also likely that this model is not valid after the x-intercept, unless Emily will use a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is 0 ≤ t ≤ 8.75. In the above example, we were given a written description of the situation. We followed the steps of modeling a problem to analyze the information. However, the information provided may not always be the same. Sometimes we might be provided with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information we are given, and use it appropriately to build a linear model. Using a Given Intercept to Build a Model Some real-world problems provide the y-intercept, which is the constant or initial value. Once the y-intercept is known, the x-intercept can be calculated. Suppose, for example, that Hannah plans to pay off a no-interest loan from her parents. Her loan balance is $1,000. She plans to pay $250 per month until her balance is $0. The y-intercept is the initial amount of her debt, or $1,000. The rate of change, or slope, is −$250 per month. We can then use the slopeintercept form and the given information to develop a linear model. f (x) = mx + b = −250x + 1000 164 CHAPTER 2 linear Functions Now we can set the function equal to 0, and solve for x to find the x-intercept. 0 = −250x + 1000 1000 = 250x 4 = x x
= 4 The x-intercept is the number of months it takes her to reach a balance of $0. The x-intercept is 4 months, so it will take Hannah four months to pay off her loan. Using a Given Input and Output to Build a Model Many real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes instead be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output. How To… Given a word problem that includes two pairs of input and output values, use the linear function to solve a problem. 1. Identify the input and output values. 2. Convert the data to two coordinate pairs. 3. Find the slope. 4. Write the linear model. 5. Use the model to make a prediction by evaluating the function at a given x-value. 6. Use the model to identify an x-value that results in a given y-value. 7. Answer the question posed. Example 1 Using a Linear Model to Investigate a Town’s Population A town’s population has been growing linearly. In 2004 the population was 6,200. By 2009 the population had grown to 8,100. Assume this trend continues. a. Predict the population in 2013. b. Identify the year in which the population will reach 15,000. Solution The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y-intercept would correspond to the year 0, more than 2,000 years ago! To make computation a little nicer, we will define our input as the number of years since 2004: • Input: t, years since 2004 • Output: P(t), the town’s population To predict the population in 2013 (t = 9), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope. To determine the rate of change, we will use the change in output per change in input. m = change in output __ change in input The problem gives us two input-output pairs. Converting them to match our defined variables,
the year 2004 would correspond to t = 0, giving the point (0, 6200). Notice that through our clever choice of variable definition, we have “given” ourselves the y-intercept of the function. The year 2009 would correspond to t = 5, giving the point (5, 8100). The two coordinate pairs are (0, 6200) and (5, 8100). Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope. SECTION 2.3 modeling with linear Functions 165 m = 8100 − 6200 __________ 5 − 0 = 1900 ____ 5 = 380 people per year We already know the y-intercept of the line, so we can immediately write the equation: P(t) = 380t + 6200 To predict the population in 2013, we evaluate our function at t = 9. P(9) = 380(9) + 6,200 = 9,620 If the trend continues, our model predicts a population of 9,620 in 2013. To find when the population will reach 15,000, we can set P(t) = 15000 and solve for t. 15000 = 380t + 6200 8800 = 380t t ≈ 23.158 Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027. Try It #1 A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It costs $0.25 to produce each doughnut. a. Write a linear model to repre sent the cost C of the company as a function of x, the number of doughnuts produced. b. Find and interpret the y-intercept. Try It #2 A city’s population has been growing linearly. In 2008, the population was 28,200. By 2012, the population was 36,800. Assume this trend continues. a. Predict the population in 2014. b. Identify the year in which the population will reach 54,000. Using a Diagram to Model a Problem It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate
the variables. Often, geometrical shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful. Example 2 Using a Diagram to Model Distance Walked Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after they start walking will they fall out of radio contact? Solution In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question: “How long will it take them to be 2 miles apart?” 166 CHAPTER 2 linear Functions In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables. • Input: t, time in hours. • Output: A(t), distance in miles, and E(t), distance in miles Because it is not obvious how to define our output variable, we’ll start by drawing a picture such as Figure 2. Anna walking east, 4 miles/hour Distance between them Emanuel walking south, 3 miles/hour Figure 2 Initial Value: They both start at the same intersection so when t = 0, the distance traveled by each person should also be 0. Thus the initial value for each is 0. Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for A is 4 and the slope for E is 3. Using those values, we can write formulas for the distance each person has walked. A(t) = 4t E(t) = 3t For this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the “starting point” at the intersection where they both started. Then we can use the variable, A, which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, E, to represent Emanuel’s position
, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure. We can then define a third variable, D, to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful, as we can see from Figure 3. Recall that we need to know how long it takes for D, the distance between them, to equal 2 miles. Notice that for any given input t, the outputs A(t), E(t), and D(t) represent distances. E A D Figure 3 Figure 3 shows us that we can use the Pythagorean Theorem because we have drawn a right angle. Using the Pythagorean Theorem, we get: SECTION 2.3 modeling with linear Functions 167 D(t)2 = A(t)2 + E(t)2 = (4t)2 + (3t)2 = 16t2 + 9t2 = 25t2 D(t) = ± √ 25t2 — Solve for D(t) using the square root. In this scenario we are considering only positive values of t, so our distance D(t) will always be positive. We can simplify this answer to D(t) = 5t. This means that the distance between Anna and Emanuel is also a linear function. Because D is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output D(t) = 2 and solve for t. = ±5 \| t \| D(t) = 2 5t = 2 t = 2 __ = 0.4 5 They will fall out of radio contact in 0.4 hours, or 24 minutes. Q & A… Should I draw diagrams when given information based on a geometric shape? Yes. Sketch the figure and label the quantities and unknowns on the sketch. Example 3 Using a Diagram to Model Distance Between Cities There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough? Solution It might help here to draw a picture of the situation. See Figure
4. It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates (30, 10), and Eastborough at (20, 0). Agritown (30, 10) (0, 0) Westborough (20, 0) Eastborough 20 miles Figure 4 Using this point along with the origin, we can find the slope of the line from Westborough to Agritown: m = 10 − 0 = 1 __ ______ 30 − 0 3 The equation of the road from Westborough to Agritown would be W(x) = 1 __ x 3 From this, we can determine the perpendicular road to Eastborough will have slope m = −3. Because the town of Eastborough is at the point (20, 0), we can find the equation: E(x) = −3x + b 0 = −3(20) + b b = 60 E(x) = −3x + 60 Substitute in (20, 0). 168 CHAPTER 2 linear Functions We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal, 1 __ x = −3x + 60 3 10 __ x = 60 3 10x = 180 x = 18 y = W(18) = 1 __ (18) 3 = 6 Substitute this back into W(x). The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from Westborough to the junction. distance = √ — (x2 − x1)2 + (y2 − y1)2 = √ —— (18 − 0)2 + (6 − 0)2 ≈ 18.974 miles Analysis One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points. Try It #3 There is a straight road leading from the town of Timpson to Ashburn 60 miles east and 12 miles north. Partway down the road, it junctions with a second road, perpendicular to the first, leading to the town of Garrison. If the town of Garrison is located 22 miles directly east of the town of Timpson, how far is the road junction from Timpson? Building Systems of linear Models Real-world situations including two or more linear functions may be modeled
with a system of linear equations. Remember, when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are three types of answers possible, as shown in Figure 5. Exactly one solution x g Infinitely many solutions y f (a) y y x g f x No solutions (c) (b) Figure 5 How To… Given a situation that represents a system of linear equations, write the system of equations and identify the solution. 1. Identify the input and output of each linear model. 2. Identify the slope and y-intercept of each linear model. 3. Find the solution by setting the two linear functions equal to one another and solving for x, or find the point of intersection on a graph. SECTION 2.3 modeling with linear Functions 169 Example 4 Building a System of Linear Models to Choose a Truck Rental Company Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile[9]. When will Keep on Trucking, Inc. be the better choice for Jamal? Solution The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions. Input Outputs Initial Value Rate of Change d, distance driven in miles K(d): cost, in dollars, for renting from Keep on Trucking M(d) cost, in dollars, for renting from Move It Your Way Up-front fee: K(0) = 20 and M(0) = 16 K(d) = $0.59/mile and P(d) = $0.63/mile Table 1 A linear function is of the form f (x) = mx + b. Using the rates of change and initial charges, we can write the equations K(d) = 0.59d + 20 M(d) = 0.63d + 16 Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when K(d) < M(d). The solution pathway will lead us to find the equations for the two functions, find
the intersection, and then see where the K(d) function is smaller. These graphs are sketched in Figure 6, with K(d) in red. $ 120 110 100 90 80 70 60 50 40 30 20 10 0 K(d) = 0.59d + 20 (100, 80) M(d) = 0.63d + 16 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 d Figure 6 To find the intersection, we set the equations equal and solve: K(d) = M(d) 0.59d + 20 = 0.63d + 16 4 = 0.04d 100 = d d = 100 This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that K(d) is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is d > 100. Access this online resources for additional instruction and practice with linear function models. • Interpreting a linear Function (http://openstaxcollege.org/l/interpretlinear) 9 Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/ 170 CHAPTER 2 linear Functions 2.3 SeCTIOn exeRCISeS VeRBAl 1. Explain how to find the input variable in a word 2. Explain how to find the output variable in a word problem that uses a linear function. problem that uses a linear function. 3. Explain how to interpret the initial value in a word 4. Explain how to determine the slope in a word problem that uses a linear function. problem that uses a linear function. AlGeBRAIC 5. Find the area of a parallelogram bounded by the y-axis, the line x = 3, the line f (x) = 1 + 2x, and the line parallel to f (x) passing through (2, 7). 6. Find the area of a triangle bounded by the x-axis, 1 __ the line f (x) = 12 – x, and the line perpendicular 3 to f (x) that passes through the origin. 7. Find the area of a triangle bounded by the y-axis, 8. Find the area of a parallelogram bounded by the 6 __ the line f (x) = 9 –
x, and the line perpendicular to 7 f (x) that passes through the origin. x-axis, the line g (x) = 2, the line f (x) = 3x, and the line parallel to f (x) passing through (6, 1). For the following exercises, consider this scenario: A town’s population has been decreasing at a constant rate. In 2010 the population was 5,900. By 2012 the population had dropped 4,700. Assume this trend continues. 9. Predict the population in 2016. 10. Identify the year in which the population will reach 0. For the following exercises, consider this scenario: A town’s population has been increased at a constant rate. In 2010 the population was 46,020. By 2012 the population had increased to 52,070. Assume this trend continues. 11. Predict the population in 2016. 12. Identify the year in which the population will reach 75,000. For the following exercises, consider this scenario: A town has an initial population of 75,000. It grows at a constant rate of 2,500 per year for 5 years. 13. Find the linear function that models the town’s 14. Find a reasonable domain and range for the population P as a function of the year, t, where t is the number of years since the model began. function P. 15. If the function P is graphed, find and interpret the 16. If the function P is graphed, find and interpret the x-and y-intercepts. slope of the function. 17. When will the output reached 100,000? 18. What is the output in the year 12 years from the onset of the model? SECTION 2.3 section exercises 171 For the following exercises, consider this scenario: The weight of a newborn is 7.5 pounds. The baby gained one-half pound a month for its first year. 19. Find the linear function that models the baby’s 20. Find a reasonable domain and range for the weight, W, as a function of the age of the baby, in months, t. function W. 21. If the function W is graphed, find and interpret the 22. If the function W is graphed, find and interpret the x- and y-intercepts. slope of the function. 23. When did the baby weight 10.4 pounds? 24. What is the output when the input is 6.2? Interpret
your answer. For the following exercises, consider this scenario: The number of people afflicted with the common cold in the winter months steadily decreased by 205 each year from 2005 until 2010. In 2005, 12,025 people were afflicted. 25. Find the linear function that models the number of people inflicted with the common cold, C, as a function of the year, t. 26. Find a reasonable domain and range for the function C. 27. If the function C is graphed, find and interpret the 28. If the function C is graphed, find and interpret the x-and y-intercepts. slope of the function. 29. When will the output reach 0? 30. In what year will the number of people be 9,700? GRAPHICAl For the following exercises, use the graph in Figure 7, which shows the profit, y, in thousands of dollars, of a company in a given year, t, where t represents the number of years since 1980. y 155 150 145 140 135 130 0 0 10 15 Figure 7 20 25 t 31. Find the linear function y, where y depends on t, the number of years since 1980. 32. Find and interpret the y-intercept. 33. Find and interpret the x-intercept. 34. Find and interpret the slope. 172 CHAPTER 2 linear Functions For the following exercises, use the graph in Figure 8, which shows the profit, y, in thousands of dollars, of a company in a given year, t, where t represents the number of years since 1980. y 500 450 400 350 300 250 200 150 100 50 0 t 0 10 15 20 25 Figure 8 35. Find the linear function y, where y depends on t, the number of years since 1980. 36. Find and interpret the y-intercept. 37. Find and interpret the x-intercept. 38. Find and interpret the slope. nUMeRIC For the following exercises, use the median home values in Mississippi and Hawaii (adjusted for inflation) shown in Table 2. Assume that the house values are changing linearly. Year 1950 2000 Mississippi $25,200 $71,400 Table 2 Hawaii $74,400 $272,700 39. In which state have home values increased at a higher rate? 40. If these trends were to continue, what would be the median home value in Mississippi in 2010? 41. If we assume the linear trend existed before 1950 and continues after 2000, the two states’ median house values
will be (or were) equal in what year? (The answer might be absurd.) For the following exercises, use the median home values in Indiana and Alabama (adjusted for inflation) shown in Table 3. Assume that the house values are changing linearly. Year 1950 2000 Indiana $37,700 $94,300 Table 3 Alabama $27,100 $85,100 42. In which state have home values increased at a higher rate? 43. If these trends were to continue, what would be the median home value in Indiana in 2010? 44. If we assume the linear trend existed before 1950 and continues after 2000, the two states’ median house values will be (or were) equal in what year? (The answer might be absurd.) SECTION 2.3 section exercises 173 ReAl-WORlD APPlICATIOnS 45. In 2004, a school population was 1,001. By 2008 the population had grown to 1,697. Assume the population is changing linearly. a. How much did the population grow between the 46. In 2003, a town’s population was 1,431. By 2007 the population had grown to 2,134. Assume the population is changing linearly. a. How much did the population grow between the year 2004 and 2008? year 2003 and 2007? b. How long did it take the population to grow from b. How long did it take the population to grow from 1,001 students to 1,697 students? 1,431 people to 2,134 people? c. What is the average population growth per year? d. What was the population in the year 2000? e. Find an equation for the population, P, of the c. What is the average population growth per year? d. What was the population in the year 2000? e. Find an equation for the population, P of the school t years after 2000. town t years after 2000. f. Using your equation, predict the population of f. Using your equation, predict the population of the school in 2011. the town in 2014. 47. A phone company has a monthly cellular plan where a customer pays a flat monthly fee and then a certain amount of money per minute used on the phone. If a customer uses 410 minutes, the monthly cost will be $71.50. If the customer uses 720 minutes, the monthly cost will be $118. a. Find a linear equation for the monthly cost of the cell plan as a function of x,
the number of monthly minutes used. 48. A phone company has a monthly cellular data plan where a customer pays a flat monthly fee of $10 and then a certain amount of money per megabyte (MB) of data used on the phone. If a customer uses 20 MB, the monthly cost will be $11.20. If the customer uses 130 MB, the monthly cost will be $17.80. a. Find a linear equation for the monthly cost of the data plan as a function of x, the number of MB used. b. Interpret the slope and y-intercept of the b. Interpret the slope and y-intercept of the equation. equation. c. Use your equation to find the total monthly cost c. Use your equation to find the total monthly cost if 687 minutes are used. if 250 MB are used. 49. In 1991, the moose population in a park was measured to be 4,360. By 1999, the population was measured again to be 5,880. Assume the population continues to change linearly. a. Find a formula for the moose population, 50. In 2003, the owl population in a park was measured to be 340. By 2007, the population was measured again to be 285. The population changes linearly. Let the input be years since 1990. a. Find a formula for the owl population, P. Let the P since 1990. input be years since 2003. b. What does your model predict the moose b. What does your model predict the owl population to be in 2003? population to be in 2012? 51. The Federal Helium Reserve held about 16 billion cubic feet of helium in 2010 and is being depleted by about 2.1 billion cubic feet each year. a. Give a linear equation for the remaining federal helium reserves, R, in terms of t, the number of years since 2010. b. In 2015, what will the helium reserves be? c. If the rate of depletion doesn’t change, in what 52. Suppose the world’s oil reserves in 2014 are 1,820 billion barrels. If, on average, the total reserves are decreasing by 25 billion barrels of oil each year: a. Give a linear equation for the remaining oil reserves, R, in terms of t, the number of years since now. b. Seven years from now, what will the oil reserves be? year will the Federal Helium Reserve be depleted? c. If the rate at which the reserves are decreasing
is constant, when will the world’s oil reserves be depleted? 174 CHAPTER 2 linear Functions 53. You are choosing between two different prepaid cell phone plans. The first plan charges a rate of 26 cents per minute. The second plan charges a monthly fee of $19.95 plus 11 cents per minute. How many minutes would you have to use in a month in order for the second plan to be preferable? 54. You are choosing between two different window washing companies. The first charges $5 per window. The second charges a base fee of $40 plus $3 per window. How many windows would you need to have for the second company to be preferable? 55. When hired at a new job selling jewelry, you are 56. When hired at a new job selling electronics, you are given two pay options: • Option A: Base salary of $17,000 a year with a given two pay options: • Option A: Base salary of $14,000 a year with a commission of 12% of your sales commission of 10% of your sales • Option B: Base salary of $20,000 a year with a • Option B: Base salary of $19,000 a year with a commission of 5% of your sales commission of 4% of your sales How much jewelry would you need to sell for option A to produce a larger income? How much electronics would you need to sell for option A to produce a larger income? 57. When hired at a new job selling electronics, you are 58. When hired at a new job selling electronics, you are given two pay options: • Option A: Base salary of $20,000 a year with a given two pay options: • Option A: Base salary of $10,000 a year with a commission of 12% of your sales commission of 9% of your sales • Option B: Base salary of $26,000 a year with a • Option B: Base salary of $20,000 a year with a commission of 3% of your sales commission of 4% of your sales How much electronics would you need to sell for option A to produce a larger income? How much electronics would you need to sell for option A to produce a larger income? SECTION 2.4 Fitting linear models to data 175 leARnInG OBjeCTIVeS In this section, you will: • Draw and interpret scatter plots. • Find the line of best fit. • • Distinguish between linear and nonlinear relations. Use a linear
model to make predictions. 2.4 FITTInG lIneAR MODelS TO DATA A professor is attempting to identify trends among final exam scores. His class has a mixture of students, so he wonders if there is any relationship between age and final exam scores. One way for him to analyze the scores is by creating a diagram that relates the age of each student to the exam score received. In this section, we will examine one such diagram known as a scatter plot. Drawing and Interpreting Scatter Plots A scatter plot is a graph of plotted points that may show a relationship between two sets of data. If the relationship is from a linear model, or a model that is nearly linear, the professor can draw conclusions using his knowledge of linear functions. Figure 1 shows a sample scatter plot. Final Exam Score vs. Age 100 90 80 70 60 50 40 30 20 10 10 20 30 40 50 Age Figure 1 A scatter plot of age and final exam score variables. Notice this scatter plot does not indicate a linear relationship. The points do not appear to follow a trend. In other words, there does not appear to be a relationship between the age of the student and the score on the final exam. Example 1 Using a Scatter Plot to Investigate Cricket Chirps Table 1 shows the number of cricket chirps in 15 seconds, for several different air temperatures, in degrees Fahrenheit[10]. Plot this data, and determine whether the data appears to be linearly related. Chirps Temperature 44 80.5 35 70.5 20.4 57 33 66 31 68 35 72 18.5 52 37 73.5 26 53 Table 1 Solution Plotting this data, as depicted in Figure 2 suggests that there may be a trend. We can see from the trend in the data that the number of chirps increases as the temperature increases. The trend appears to be roughly linear, though certainly not perfectly so. 10 Selected data from http://classic.globe.gov/fsl/scientistsblog/2007/10/. Retrieved Aug 3, 2010. 176 CHAPTER 2 linear Functions Cricket Chirps vs. Temperature ) 90 80 70 60 50 40 0 0 10 20 30 40 50 Cricket Chirps in 15 Seconds Figure 2 Finding the line of Best Fit Once we recognize a need for a linear function to model that data, the natural follow-up question is “what is that linear function?” One way to approximate our linear function is to sketch the line that seems to best fit
the data. Then we can extend the line until we can verify the y-intercept. We can approximate the slope of the line by extending it until we can estimate the rise ___. run Example 2 Finding a Line of Best Fit Find a linear function that fits the data in Table 1 by “eyeballing” a line that seems to fit. Solution On a graph, we could try sketching a line. Using the starting and ending points of our hand drawn line, points (0, 30) and (50, 90), this graph has a slope of m = = 1.2 60 __ 50 and a y-intercept at 30. This gives an equation of T(c) = 1.2c + 30 where c is the number of chirps in 15 seconds, and T(c) is the temperature in degrees Fahrenheit. The resulting equation is represented in Figure 3. 90 80 70 60 50 40 30 ) Cricket Chirps vs. Temperature 10 20 30 40 50 c, Number of Chirps Figure 3 Analysis This linear equation can then be used to approximate answers to various questions we might ask about the trend. SECTION 2.4 Fitting linear models to data 177 Recognizing Interpolation or Extrapolation While the data for most examples does not fall perfectly on the line, the equation is our best guess as to how the relationship will behave outside of the values for which we have data. We use a process known as interpolation when we predict a value inside the domain and range of the data. The process of extrapolation is used when we predict a value outside the domain and range of the data. Figure 4 compares the two processes for the cricket-chirp data addressed in Example 2. We can see that interpolation would occur if we used our model to predict temperature when the values for chirps are between 18.5 and 44. Extrapolation would occur if we used our model to predict temperature when the values for chirps are less than 18.5 or greater than 44. 90 80 70 60 50 40 30 ) Cricket Chirps vs. Temperature Extrapolation Interpolation 10 20 30 40 50 c, Number of Chirps Figure 4 Interpolation occurs within the domain and range of the provided data whereas extrapolation occurs outside. There is a difference between making predictions inside the domain and range of values for which we have data and outside that domain and range. Predicting a value outside of the domain and range has its limitations. When our
model no longer applies after a certain point, it is sometimes called model breakdown. For example, predicting a cost function for a period of two years may involve examining the data where the input is the time in years and the output is the cost. But if we try to extrapolate a cost when x = 50, that is in 50 years, the model would not apply because we could not account for factors fifty years in the future. interpolation and extrapolation Different methods of making predictions are used to analyze data. • The method of interpolation involves predicting a value inside the domain and/or range of the data. • The method of extrapolation involves predicting a value outside the domain and/or range of the data. • Model breakdown occurs at the point when the model no longer applies. Example 3 Understanding Interpolation and Extrapolation Use the cricket data from Table 1 to answer the following questions: a. Would predicting the temperature when crickets are chirping 30 times in 15 seconds be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable. b. Would predicting the number of chirps crickets will make at 40 degrees be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable. 178 Solution CHAPTER 2 linear Functions a. The number of chirps in the data provided varied from 18.5 to 44. A prediction at 30 chirps per 15 seconds is inside the domain of our data, so would be interpolation. Using our model: T(30) = 30 + 1.2(30) = 66 degrees Based on the data we have, this value seems reasonable. b. The temperature values varied from 52 to 80.5. Predicting the number of chirps at 40 degrees is extrapolation because 40 is outside the range of our data. Using our model: 40 = 30 + 1.2c 10 = 1.2c c ≈ 8.33 We can compare the regions of interpolation and extrapolation using Figure 5. 90 80 70 60 50 40 30 ) Cricket Chirps vs. Temperature Interpolation Extrapolation 10 20 30 40 50 c, Number of Chirps Figure 5 Analysis Our model predicts the crickets would chirp 8.33 times in 15 seconds. While this might be possible, we have no reason to believe our model is valid outside the domain and range. In fact, generally crickets stop chirping altogether below around 50 degrees. Try It #1 According to the
data from Table 1, what temperature can we predict it is if we counted 20 chirps in 15 seconds? Finding the Line of Best Fit Using a Graphing Utility While eyeballing a line works reasonably well, there are statistical techniques for fitting a line to data that minimize the differences between the line and data values[11]. One such technique is called least squares regression and can be computed by many graphing calculators, spreadsheet software, statistical software, and many web-based calculators[12]. Least squares regression is one means to determine the line that best fits the data, and here we will refer to this method as linear regression. 11 Technically, the method minimizes the sum of the squared differences in the vertical direction between the line and the data values. 12 For example, http://www.shodor.org/unchem/math/lls/leastsq.html SECTION 2.4 Fitting linear models to data 179 How To… Given data of input and corresponding outputs from a linear function, find the best fit line using linear regression. 1. Enter the input in List 1 (L1). 2. Enter the output in List 2 (L2). 3. On a graphing utility, select Linear Regression (LinReg). Example 4 Finding a Least Squares Regression Line Find the least squares regression line using the cricket-chirp data in Table 1. Solution 1. Enter the input (chirps) in List 1 (L1). 2. Enter the output (temperature) in List 2 (L2). See Table 2. L1 L2 44 80.5 35 70.5 20.4 57 33 66 31 68 35 72 18.5 52 37 73.5 26 53 Table 2 3. On a graphing utility, select Linear Regression (LinReg). Using the cricket chirp data from earlier, with technology we obtain the equation: T(c) = 30.281 + 1.143c Analysis Notice that this line is quite similar to the equation we “eyeballed” but should fit the data better. Notice also that using this equation would change our prediction for the temperature when hearing 30 chirps in 15 seconds from 66 degrees to: T(30) = 30.281 + 1.143(30) = 64.571 ≈ 64.6 degrees The graph of the scatter plot with the least squares regression line is shown in Figure 6 90 80 70 60 50 40 30 Number of Cricket Ch
irps vs. Temperature 0 10 20 30 40 50 c, Number of Chirps Figure 6 Q & A… Will there ever be a case where two different lines will serve as the best fit for the data? No. There is only one best fit line. 180 CHAPTER 2 linear Functions Distinguishing Between linear and non-linear Models As we saw above with the cricket-chirp model, some data exhibit strong linear trends, but other data, like the final exam scores plotted by age, are clearly nonlinear. Most calculators and computer software can also provide us with the correlation coefficient, which is a measure of how closely the line fits the data. Many graphing calculators require the user to turn a ”diagnostic on” selection to find the correlation coefficient, which mathematicians label as r. The correlation coefficient provides an easy way to get an idea of how close to a line the data falls. We should compute the correlation coefficient only for data that follows a linear pattern or to determine the degree to which a data set is linear. If the data exhibits a nonlinear pattern, the correlation coefficient for a linear regression is meaningless. To get a sense for the relationship between the value of r and the graph of the data, Figure 7 shows some large data sets with their correlation coefficients. Remember, for all plots, the horizontal axis shows the input and the vertical axis shows the output. 1.0 0.8 0.4 0.0 −0.4 −0.8 −1.0 1.0 1.0 1.0 −1.0 −1.0 −1.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 Figure 7 Plotted data and related correlation coefficients. (credit: “DenisBoigelot,” Wikimedia Commons) correlation coefficient The correlation coefficient is a value, r, between −1 and 1. • r > 0 suggests a positive (increasing) relationship • r < 0 suggests a negative (decreasing) relationship • The closer the value is to 0, the more scattered the data. • The closer the value is to 1 or −1, the less scattered the data is. Example 5 Finding a Correlation Coefficient Calculate the correlation coefficient for cricket-chirp data in Table 1. Solution Because the data appear to follow a linear pattern, we can use technology to calculate r. Enter the inputs and corresponding outputs and select the Linear Regression. The calculator
will also provide you with the correlation coefficient, r = 0.9509. This value is very close to 1, which suggests a strong increasing linear relationship. Note: For some calculators, the Diagnostics must be turned “on” in order to get the correlation coefficient when linear regression is performed: [2nd]>[0]>[alpha][x − 1], then scroll to DIAGNOSTICSON. Predicting with a Regression line Once we determine that a set of data is linear using the correlation coefficient, we can use the regression line to make predictions. As we learned above, a regression line is a line that is closest to the data in the scatter plot, which means that only one such line is a best fit for the data. SECTION 2.4 Fitting linear models to data 181 Example 6 Using a Regression Line to Make Predictions Gasoline consumption in the United States has been steadily increasing. Consumption data from 1994 to 2004 is shown in Table 3[13]. Determine whether the trend is linear, and if so, find a model for the data. Use the model to predict the consumption in 2008. Year ‘94 ‘95 ‘96 ‘97 ‘98 ‘99 ‘00 ‘01 ‘02 ‘03 ‘04 Consumption (billions of gallons) 113 116 118 119 123 125 126 128 131 133 136 The scatter plot of the data, including the least squares regression line, is shown in Figure 8. Table ( 150 140 130 120 110 100 0 Gas Consumption vs. Year 0 1 2 3 4 6 5 7 Years After 1994 8 9 10 11 12 13 14 Figure 8 Solution We can introduce new input variable, t, representing years since 1994. The least squares regression equation is: C(t) = 113.318 + 2.209t Using technology, the correlation coefficient was calculated to be 0.9965, suggesting a very strong increasing linear trend. Using this to predict consumption in 2008 (t = 14), The model predicts 144.244 billion gallons of gasoline consumption in 2008. C(14) = 113.318 + 2.209(14) = 144.244 Try It #2 Use the model we created using technology in Example 6 to predict the gas consumption in 2011. Is this an interpolation or an extrapolation? Access these online resources for additional instruction and practice with fitting linear models to data. • Introduction to Regression Analysis (http://openstaxcollege.org/l/introreg
ress) • linear Regression (http://openstaxcollege.org/l/linearregress) 13 http://www.bts.gov/publications/national_transportation_statistics/2005/html/table_04_10.html 182 CHAPTER 2 linear Functions 2.4 SeCTIOn exeRCISeS VeRBAl 1. Describe what it means if there is a model breakdown when using a linear model. 2. What is interpolation when using a linear model? 3. What is extrapolation when using a linear model? 4. Explain the difference between a positive and a negative correlation coefficient. 5. Explain how to interpret the absolute value of a correlation coefficient. AlGeBRAIC 6. A regression was run to determine whether there is 7. A regression was run to determine whether there is a a relationship between hours of TV watched per day (x) and number of sit-ups a person can do (y). The results of the regression are given below. Use this to predict the number of situps a person who watches 11 hours of TV can do. y = ax + b a = −1.341 b = 32.234 r = −0.896 relationship between the diameter of a tree (x, in inches) and the tree’s age (y, in years). The results of the regression are given below. Use this to predict the age of a tree with diameter 10 inches. y = ax + b a = 6.301 b = −1.044 r = −0.970 For the following exercises, draw a scatter plot for the data provided. Does the data appear to be linearly related? 8. 10. 2 0 8 4 −22 −19 −15 −11 −6 6 10 −2 100 12 250 12.6 300 13.1 450 14 600 14.5 750 15.2 9. 11. 1 46 1 1 2 50 3 9 3 59 5 28 4 75 7 65 5 100 6 136 9 125 11 216 12. For the following data, draw a scatter plot. If we 13. For the following data, draw a scatter plot. If we wanted to know when the population would reach 15,000, would the answer involve interpolation or extrapolation? Eyeball the line, and estimate the answer. wanted to know when the temperature would reach 28°F, would the answer involve interpolation or extrapolation? Eyeball the line and estimate the answer. Year 2010
Population 11,500 12,100 12,700 13,000 13,750 2000 2005 1990 1995 Temperature, °F Time, seconds 16 46 18 50 20 54 25 55 30 62 SECTION 2.4 section exercises 183 GRAPHICAl For the following exercises, match each scatterplot with one of the four specified correlations in Figure 9 and Figure 10. (a) (b) Figure 9 14. r = 0.95 16. r = 0.26 (c) (d) Figure 10 15. r = −0.89 17. r = −0.39 For the following exercises, draw a best-fit line for the plotted data. 18. 12 10 8 6 4 2 0 19. 10 10 0 2 4 6 8 10 10 10 184 CHAPTER 2 linear Functions 20. 10 10 nUMeRIC 21. 10 10 22. The U.S. Census tracks the percentage of persons 25 years or older who are college graduates. That data for several years is given in Table 4[14]. Determine whether the trend appears linear. If so, and assuming the trend continues, in what year will the percentage exceed 35%? Year Percent Graduates 1990 21.3 1992 21.4 1994 22.2 1996 23.6 1998 24.4 2000 25.6 2002 26.7 2004 27.7 2006 28 2008 29.4 Table 4 23. The U.S. import of wine (in hectoliters) for several years is given in Table 5. Determine whether the trend appears linear. If so, and assuming the trend continues, in what year will imports exceed 12,000 hectoliters? Year Imports 1992 2665 1994 2688 1996 3565 1998 4129 2000 4584 2002 5655 2004 6549 2006 7950 2008 8487 2009 9462 Table 5 24. Table 6 shows the year and the number of people unemployed in a particular city for several years. Determine whether the trend appears linear. If so, and assuming the trend continues, in what year will the number of unemployed reach 5 people? Year Number Unemployed 1990 750 1992 670 1994 650 1996 605 Table 6 1998 550 2000 510 2002 460 2004 420 2006 380 2008 320 TeCHnOlOGY For the following exercises, use each set of data to calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to 3 decimal places of accuracy. 25. 26. x y x y 8 23 5 4 15 41 7 12 26 53 10 17 31 72 12 22 56
103 15 24 14 http://www.census.gov/hhes/socdemo/education/data/cps/historical/index.html. Accessed 5/1/2014. SECTION 2.4 section exercises 185 3 21.9 11 15.76 4 22.22 12 13.68 5 22.74 13 14.1 6 22.26 14 14.02 7 20.78 15 11.94 8 17.6 16 12.76 9 16.52 17 11.28 10 18.54 18 9.1 4 44.8 5 43.1 6 38.8 7 39 8 38 9 32.7 10 30.1 11 29.3 12 27 13 25.8 21 17 100 2000 900 70 25 11 80 1798 988 80 30 2 60 1589 1000 82 31 −1 55 1580 1010 84 40 −18 40 1390 1200 105 50 −40 20 1202 1205 108 27. 28. 29. 30. 31 exTenSIOnS 32. Graph f (x) = 0.5x + 10. Pick a set of 5 ordered 33. Graph f (x) = −2x − 10. Pick a set of 5 ordered pairs using inputs x = −2, 1, 5, 6, 9 and use linear regression to verify that the function is a good fit for the data. pairs using inputs x = −2, 1, 5, 6, 9 and use linear regression to verify the function. For the following exercises, consider this scenario: The profit of a company decreased steadily over a ten-year span. The following ordered pairs shows dollars and the number of units sold in hundreds and the profit in thousands of over the ten-year span, (number of units sold, profit) for specific recorded years: (46, 600), (48, 550), (50, 505), (52, 540), (54, 495). 34. Use linear regression to determine a function P 35. Find to the nearest tenth and interpret the where the profit in thousands of dollars depends on the number of units sold in hundreds. x-intercept. 36. Find to the nearest tenth and interpret the y-intercept. 186 CHAPTER 2 linear Functions ReAl-WORlD APPlICATIOnS For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs shows the population and the year over the ten-year span, (population
, year) for specific recorded years: (2500, 2000), (2650, 2001), (3000, 2003), (3500, 2006), (4200, 2010) 37. Use linear regression to determine a function y, 38. Predict when the population will hit 8,000. where the year depends on the population. Round to three decimal places of accuracy. For the following exercises, consider this scenario: The profit of a company increased steadily over a ten-year span. The following ordered pairs show the number of units sold in hundreds and the profit in thousands of over the ten-year span, (number of units sold, profit) for specific recorded years: (46, 250), (48, 305), (50, 350), (52, 390), (54, 410). 39. Use linear regression to determine a function y, 40. Predict when the profit will exceed one million where the profit in thousands of dollars depends on the number of units sold in hundreds. dollars. For the following exercises, consider this scenario: The profit of a company decreased steadily over a ten-year span. The following ordered pairs show dollars and the number of units sold in hundreds and the profit in thousands of over the ten-year span (number of units sold, profit) for specific recorded years: (46, 250), (48, 225), (50, 205), (52, 180), (54, 165). 41. Use linear regression to determine a function y, 42. Predict when the profit will dip below the $25,000 where the profit in thousands of dollars depends on the number of units sold in hundreds. threshold. CHAPTER 2 review 187 CHAPTeR 2 ReVIeW Key Terms correlation coefficient a value, r, between −1 and 1 that indicates the degree of linear correlation of variables, or how closely a regression line fits a data set. decreasing linear function a function with a negative slope: If f (x) = mx + b, then m < 0. extrapolation predicting a value outside the domain and range of the data horizontal line a line defined by f (x) = b, where b is a real number. The slope of a horizontal line is 0. increasing linear function a function with a positive slope: If f (x) = mx + b, then m > 0. interpolation predicting a value inside the domain and range of the data least squares regression a statistical technique for fitting a line to data in a way that minimizes the differences between the line
and data values linear function a function with a constant rate of change that is a polynomial of degree 1, and whose graph is a straight line model breakdown when a model no longer applies after a certain point parallel lines two or more lines with the same slope perpendicular lines two lines that intersect at right angles and have slopes that are negative reciprocals of each other point-slope form the equation for a line that represents a linear function of the form y − y1 = m(x − x1) slope the ratio of the change in output values to the change in input values; a measure of the steepness of a line slope-intercept form the equation for a line that represents a linear function in the form f (x) = mx + b vertical line a line defined by x = a, where a is a real number. The slope of a vertical line is undefined. x-intercept the point on the graph of a linear function when the output value is 0; the point at which the graph crosses the horizontal axis y-intercept the value of a function when the input value is zero; also known as initial value Key equations slope-intercept form of a line f (x) = mx + b slope m = change in output (rise) __ = change in input (run) ∆y _ ∆x = y2 − y1 _ x2 − x1 point-slope form of a line y − y1 = m(x − x1) Key Concepts 2.1 Linear Functions • The ordered pairs given by a linear function represent points on a line. • Linear functions can be represented in words, function notation, tabular form, and graphical form. See Example 1. • The rate of change of a linear function is also known as the slope. • An equation in the slope-intercept form of a line includes the slope and the initial value of the function. • The initial value, or y-intercept, is the output value when the input of a linear function is zero. It is the y-value of the point at which the line crosses the y-axis. • An increasing linear function results in a graph that slants upward from left to right and has a positive slope. • A decreasing linear function results in a graph that slants downward from left to right and has a negative slope. • A constant linear function results in a graph that is a horizontal line. 188 CHAPTER 2 linear Functions • Analyzing the slope within the context
of a problem indicates whether a linear function is increasing, decreasing, or constant. See Example 2. • The slope of a linear function can be calculated by dividing the difference between y-values by the difference in corresponding x-values of any two points on the line. See Example 3 and Example 4. • The slope and initial value can be determined given a graph or any two points on the line. • One type of function notation is the slope-intercept form of an equation. • The point-slope form is useful for finding a linear equation when given the slope of a line and one point. See Example 5. • The point-slope form is also convenient for finding a linear equation when given two points through which a line passes. See Example 6. • The equation for a linear function can be written if the slope m and initial value b are known. See Example 7, Example 8, and Example 9. • A linear function can be used to solve real-world problems. See Example 10 and Example 11. • A linear function can be written from tabular form. See Example 12. 2.2 Graphs of Linear Functions • Linear functions may be graphed by plotting points or by using the y-intercept and slope. See Example 1 and Example 2. • Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches, compressions, and reflections. See Example 3. • The y-intercept and slope of a line may be used to write the equation of a line. • The x-intercept is the point at which the graph of a linear function crosses the x-axis. See Example 4 and Example 5. • Horizontal lines are written in the form, f (x) = b. See Example 6. • Vertical lines are written in the form, x = b. See Example 7. • Parallel lines have the same slope. • Perpendicular lines have negative reciprocal slopes, assuming neither is vertical. See Example 8. • A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the x- and y-values of the given point into the equation, f (x) = mx + b, and using the b that results. Similarly, the point-slope form of an equation can also be used. See Example 9. • A line perpendicular to another line, passing through a given point, may be found
in the same manner, with the exception of using the negative reciprocal slope. See Example 10 and Example 11. • A system of linear equations may be solved setting the two equations equal to one another and solving for x. The y-value may be found by evaluating either one of the original equations using this x-value. • A system of linear equations may also be solved by finding the point of intersection on a graph. See Example 12 and Example 13. 2.3 Modeling with Linear Functions • We can use the same problem strategies that we would use for any type of function. • When modeling and solving a problem, identify the variables and look for key values, including the slope and y-intercept. See Example 1. • Draw a diagram, where appropriate. See Example 2 and Example 3. CHAPTER 2 review 189 • Check for reasonableness of the answer. • Linear models may be built by identifying or calculating the slope and using the y-intercept. • The x-intercept may be found by setting y = 0, which is setting the expression mx + b equal to 0. • The point of intersection of a system of linear equations is the point where the x- and y-values are the same. See Example 4. • A graph of the system may be used to identify the points where one line falls below (or above) the other line. 2.4 Fitting Linear Models to Data • Scatter plots show the relationship between two sets of data. See Example 1. • Scatter plots may represent linear or non-linear models. • The line of best fit may be estimated or calculated, using a calculator or statistical software. See Example 2. • Interpolation can be used to predict values inside the domain and range of the data, whereas extrapolation can be used to predict values outside the domain and range of the data. See Example 3. • The correlation coefficient, r, indicates the degree of linear relationship between data. See Example 5. • A regression line best fits the data. See Example 6. • The least squares regression line is found by minimizing the squares of the distances of points from a line passing through the data and may be used to make predictions regarding either of the variables. See Example 4. 190 CHAPTER 2 linear Functions CHAPTeR 2 ReVIeW exeRCISeS lIneAR FUnCTIOnS 1. Determine whether the algebraic equation is linear. 2. Determine whether the algebraic
equation is linear. 2x + 3y = 7 6x 2 − y = 5 3. Determine whether the function is increasing or 4. Determine whether the function is increasing or decreasing. f (x) = 7x − 2 decreasing. g(x) = −x + 2 5. Given each set of information, find a linear equation that satisfies the given conditions, if possible. Passes through (7, 5) and (3, 17) 6. Given each set of information, find a linear equation that satisfies the given conditions, if possible. x-intercept at (6, 0) and y-intercept at (0, 10) 7. Find the slope of the line shown in the graph. 8. Find the slope of the line shown in the graph. y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 21 3 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 21 3 4 5 6 x 9. Write an equation in slope-intercept form for the line shown. 10. Does the following table represent a linear function? If so, find the linear equation that models the data. x g(x) –4 18 0 –2 2 10 –12 –52 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 21 3 4 5 6 x 11. Does the following table represent a linear function? If so, find the linear equation that models the data. x 6 8 12 26 g(x) –8 –12 –18 –46 12. On June 1st, a company has $4,000,000 profit. If the company then loses 150,000 dollars per day thereafter in the month of June, what is the company’s profit nth day after June 1st? GRAPHS OF lIneAR FUnCTIOnS For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular: 13. 2x − 6y = 12 −x + 3y = 1 1 __ x − 2 14. y = 3 3x + y = − 9 CHAPTER 2 review 191 For the following exercises, find the x- and y-
intercepts of the given equation 15. 7x + 9y = −63 16. f (x) = 2x − 1 For the following exercises, use the descriptions of the pairs of lines to find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular, or neither? 17. Line 1: Passes through (5, 11) and (10, 1) Line 2: Passes through (−1, 3) and (−5, 11) 18. Line 1: Passes through (8, −10) and (0, −26) Line 2: Passes through (2, 5) and (4, 4) 19. Write an equation for a line perpendicular to f (x) = 5x − 1 and passing through the point (5, 20). 20. Find the equation of a line with a y-intercept of (0, 2) and slope − 1 __. 2 21. Sketch a graph of the linear function f (t) = 2t − 5. 22. Find the point of intersection for the 2 linear functions: x = y + 6 2x − y = 13 23. A car rental company offers two plans for renting a car. Plan A: 25 dollars per day and 10 cents per mile Plan B: 50 dollars per day with free unlimited mileage How many miles would you need to drive for plan B to save you money? MODelInG WITH lIneAR FUnCTIOnS 24. Find the area of a triangle bounded by the y-axis, the line f (x) = 10 − 2x, and the line perpendicular to f that passes through the origin. 25. A town’s population increases at a constant rate. In 2010 the population was 55,000. By 2012 the population had increased to 76,000. If this trend continues, predict the population in 2016. 26. The number of people afflicted with the common cold in the winter months dropped steadily by 50 each year since 2004 until 2010. In 2004, 875 people were inflicted. Find the linear function that models the number of people afflicted with the common cold C as a function of the year, t. When will no one be afflicted? For the following exercises, use the graph in Figure 1 showing the profit, y, in thousands of dollars, of a company in a given year, x, where x represents years since 1980. y 12,000 10,000 8,000 6,000 4,000
2,000 0 5 10 15 20 25 30 x Figure 1 27. Find the linear function y, where y depends on x, the number of years since 1980. 28. Find and interpret the y-intercept. 192 CHAPTER 2 linear Functions For the following exercise, consider this scenario: In 2004, a school population was 1,700. By 2012 the population had grown to 2,500. 29. Assume the population is changing linearly. a. How much did the population grow between the year 2004 and 2012? b. What is the average population growth per year? c. Find an equation for the population, P, of the school t years after 2004. For the following exercises, consider this scenario: In 2000, the moose population in a park was measured to be 6,500. By 2010, the population was measured to be 12,500. Assume the population continues to change linearly. 30. Find a formula for the moose population, P. 31. What does your model predict the moose population to be in 2020? For the following exercises, consider this scenario: The median home values in subdivisions Pima Central and East Valley (adjusted for inflation) are shown in Table 1. Assume that the house values are changing linearly. Year 1970 2010 Pima Central 32,000 85,000 Table 1 East Valley 120,250 150,000 32. In which subdivision have home values increased at a higher rate? 33. If these trends were to continue, what would be the median home value in Pima Central in 2015? FITTInG lIneAR MODelS TO DATA 34. Draw a scatter plot for the data in Table 2. Then determine whether the data appears to be linearly related. 0 –105 2 –50 4 1 6 55 8 105 10 160 Table 2 35. Draw a scatter plot for the data in Table 3. If we wanted to know when the population would reach 15,000, would the answer involve interpolation or extrapolation? Year Population 1990 5,600 1995 5,950 2000 6,300 2005 6,600 2010 6,900 Table 3 36. Eight students were asked to estimate their score on a 10-point quiz. Their estimated and actual scores are given in Table 4. Plot the points, then sketch a line that fits the data. Predicted Actual 6 6 7 7 7 8 8 8 Table 4 7 9 9 10 10 10 10 9 CHAPTER 2 review 193 37. Draw a best-fit line for
the plotted data. y 120 100 80 60 40 20 0 0 2 4 6 8 10 12 x For the following exercises, consider the data in Table 5, which shows the percent of unemployed in a city of people 25 years or older who are college graduates is given below, by year. Year Percent Graduates 2000 6.5 2002 7.0 Table 5 2005 7.4 2007 8.2 2010 9.0 38. Determine whether the trend appears to be linear. If so, and assuming the trend continues, find a linear regression model to predict the percent of unemployed in a given year to three decimal places. 39. In what year will the percentage exceed 12%? 40. Based on the set of data given in Table 6, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to three decimal places. 41. Based on the set of data given in Table 7, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient to three decimal places. x y 17 15 20 25 23 31 26 37 29 40 x y 10 36 12 34 15 30 18 28 20 22 Table 6 Table 7 For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs show the population and the year over the ten-year span (population, year) for specific recorded years: (3,600, 2000); (4,000, 2001); (4,700, 2003); (6,000, 2006) 42. Use linear regression to determine a function y, 43. Predict when the population will hit 12,000. where the year depends on the population, to three decimal places of accuracy. 44. What is the correlation coefficient for this model to 45. According to the model, what is the population three decimal places of accuracy? in 2014? 194 CHAPTER 2 linear Functions CHAPTeR 2 PRACTICe TeST 1. Determine whether the following algebraic equation can be written as a linear function. 2x + 3y = 7 2. Determine whether the following function is increasing or decreasing. f (x) = −2x + 5 3. Determine whether the following function is increasing or decreasing. f (x) = 7x + 9 4. Given the following set of information, find a linear equation satisfying the conditions, if possible. Passes through (5, 1) and (3, −9) 5. Given the following set of information
, find a linear equation satisfying the conditions, if possible. x-intercept at (−4, 0) and y-intercept at (0, −6) 6. Find the slope of the line in Figure 1. 7. Write an equation for line in Figure 2. y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 21 3 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 21 3 4 5 6 x Figure 1 Figure 2 8. Does Table 1 represent a linear function? If so, find 9. Does Table 2 represent a linear function? If so, find a linear equation that models the data. a linear equation that models the data. x −6 14 g(x) 0 32 2 38 4 44 Table 1 x g(x) 1 4 3 9 7 19 11 12 Table 2 10. At 6 am, an online company has sold 120 items that day. If the company sells an average of 30 items per hour for the remainder of the day, write an expression to represent the number of items that were sold n after 6 am. For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither parallel nor perpendicular: 3 __ x − 9 11. y = 4 −4x − 3y = 8 13. Find the x- and y-intercepts of the equation 2x + 7y = −14. 12. −2x + y = 3 3 __ y = 5 3x + 2 14. Given below are descriptions of two lines. Find the slopes of Line 1 and Line 2. Is the pair of lines parallel, perpendicular, or neither? Line 1: Passes through (−2, −6) and (3, 14) Line 2: Passes through (2, 6) and (4, 14) CHAPTER 2 practice test 195 15. Write an equation for a line perpendicular to f (x) = 4x + 3 and passing through the point (8, 10). 16. Sketch a line with a y-intercept of (0, 5) and slope − 5 __. 2 17. Graph of the linear function f (x) = −x + 6. 18. For the two linear functions, find the point of intersection: x = y + 2 2x −
3y = −1 19. A car rental company offers two plans for renting 20. Find the area of a triangle bounded by the y-axis, a car. Plan A: $25 per day and $0.10 per mile Plan B: $40 per day with free unlimited mileage How many miles would you need to drive for plan B to save you money? 21. A town’s population increases at a constant rate. In 2010 the population was 65,000. By 2012 the population had increased to 90,000. Assuming this trend continues, predict the population in 2018. the line f (x) = 12 − 4x, and the line perpendicular to f that passes through the origin. 22. The number of people afflicted with the common cold in the winter months dropped steadily by 25 each year since 2002 until 2012. In 2002, 8,040 people were inflicted. Find the linear function that models the number of people afflicted with the common cold C as a function of the year, t. When will less than 6,000 people be afflicted? For the following exercises, use the graph in Figure 3, showing the profit, y, in thousands of dollars, of a company in a given year, x, where x represents years since 1980. y 35,000 30,000 25,000 20,000 15,000 10,000 5,000 0 5 10 15 20 25 30 x Figure 3 23. Find the linear function y, where y depends on x, the number of years since 1980. 24. Find and interpret the y-intercept. 25. In 2004, a school population was 1250. By 2012 the population had dropped to 875. Assume the population is changing linearly. a. How much did the population drop between the year 2004 and 2012? b. What is the average population decline per year? c. Find an equation for the population, P, of the school t years after 2004. 196 CHAPTER 2 linear Functions 26. Draw a scatter plot for the data provided in Table 3. Then determine whether the data appears to be linearly related. 0 2 −450 −200 4 10 6 265 8 500 10 755 Table 3 27. Draw a best-fit line for the plotted data. 35 30 25 20 15 10 5 0 y 0 2 4 6 8 10 12 x For the following exercises, use Table 4 which shows the percent of unemployed persons 25 years or older who are college graduates in a particular city, by year. Year Percent Graduates 2000
8.5 2002 8.0 Table 4 2005 7.2 2007 6.7 2010 6.4 28. Determine whether the trend appears linear. If so, and assuming the trend continues, find a linear regression model to predict the percent of unemployed in a given year to three decimal places. 29. In what year will the percentage drop below 4%? 30. Based on the set of data given in Table 5, calculate the regression line using a calculator or other technology tool, and determine the correlation coefficient. Round to three decimal places of accuracy. x y 16 106 18 110 20 115 24 120 26 125 Table 5 For the following exercises, consider this scenario: The population of a city increased steadily over a ten-year span. The following ordered pairs shows the population (in hundreds) and the year over the ten-year span, (population, year) for specific recorded years: (4,500, 2000); (4,700, 2001); (5,200, 2003); (5,800, 2006) 31. Use linear regression to determine a function y, where the year depends on the population. Round to three decimal places of accuracy. 32. Predict when the population will hit 20,000. 33. What is the correlation coefficient for this model? 3 Polynomial and Rational Functions Figure 1 35-mm film, once the standard for capturing photographic images, has been made largely obsolete by digital photography. (credit “film”: modification of work by Horia Varlan; credit “memory cards”: modification of work by Paul Hudson) CHAPTeR OUTlIne 3.1 Complex numbers 3.2 Quadratic Functions 3.3 Power Functions and Polynomial Functions 3.4 Graphs of Polynomial Functions 3.5 Dividing Polynomials 3.6 Zeros of Polynomial Functions 3.7 Rational Functions 3.8 Inverses and Radical Functions 3.9 Modeling Using Variation Introduction Digital photography has dramatically changed the nature of photography. No longer is an image etched in the emulsion on a roll of film. Instead, nearly every aspect of recording and manipulating images is now governed by mathematics. An image becomes a series of numbers, representing the characteristics of light striking an image sensor. When we open an image file, software on a camera or computer interprets the numbers and converts them to a visual image. Photo editing software uses complex polynomials to transform images, allowing us to manipulate the image in order to
crop details, change the color palette, and add special effects. Inverse functions make it possible to convert from one file format to another. In this chapter, we will learn about these concepts and discover how mathematics can be used in such applications. 197 Polynomial and Rational Functions 198 CHAPTER 3 polynomial and rational Functions leARnInG OBjeCTIVeS In this section, you will: • • • • Express square roots of negative numbers as multiples of i. Plot complex numbers on the complex plane. Add and subtract complex numbers. Multiply and divide complex numbers. 3.1 COMPlex nUMBeRS The study of mathematics continuously builds upon itself. Negative integers, for example, fill a void left by the set of positive integers. The set of rational numbers, in turn, fills a void left by the set of integers. The set of real numbers fills a void left by the set of rational numbers. Not surprisingly, the set of real numbers has voids as well. For example, we still have no solution to equations such as x 2 + 4 = 0 Our best guesses might be +2 or −2. But if we test +2 in this equation, it does not work. If we test −2, it does not work. If we want to have a solution for this equation, we will have to go farther than we have so far. After all, to this point we have described the square root of a negative number as undefined. Fortunately, there is another system of numbers that provides solutions to problems such as these. In this section, we will explore this number system and how to work within it. expressing Square Roots of negative numbers as Multiples of i We know how to find the square root of any positive real number. In a similar way, we can find the square root of a negative number. The difference is that the root is not real. If the value in the radicand is negative, the root is said to be an imaginary number. The imaginary number i is defined as the square root of negative 1. — −1 = i √ So, using properties of radicals, i2 = ( √ We can write the square root of any negative number as a multiple of i. Consider the square root of −25. −1 )2 = −1 — √ — −25 = √ = √ = 5i — 25 ⋅ (−1) — −1 25 √ — We use 5i and not −
5i because the principal root of 25 is the positive root. A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written a + bi where a is the real part and bi is the imaginary part. For example, 5 + 2i is a complex number. So, too, is 3 + 4 √ — 3i. 5 + 2i ↗ ↖ Real part Imaginary part Imaginary numbers are distinguished from real numbers because a squared imaginary number produces a negative real number. Recall, when a positive real number is squared, the result is a positive real number and when a negative real number is squared, again, the result is a positive real number. Complex numbers are a combination of real and imaginary numbers. imaginary and complex numbers A complex number is a number of the form a + bi where • a is the real part of the complex number. • bi is the imaginary part of the complex number. If b = 0, then a + bi is a real number. If a = 0 and b is not equal to 0, the complex number is called an imaginary number. An imaginary number is an even root of a negative number. SECTION 3.1 complex numBers 199 How To… Given an imaginary number, express it in standard form. 1. Write √ 2. Express √ — −a as √ −1 as i. — — a √ — −1. 3. Write √ — a ⋅ i in simplest form. Example 1 Expressing an Imaginary Number in Standard Form Express √ — −9 in standard form. Solution √ — −9 = √ — 9 √ — −1 = 3i In standard form, this is 0 + 3i. Try It #1 Express √ — −24 in standard form. Plotting a Complex number on the Complex Plane We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them graphically. To represent a complex number we need to address the two components of the number. We use the complex plane, which is a coordinate system in which the horizontal axis represents the real component and the vertical axis represents the imaginary component. Complex numbers are the points on the plane, expressed as ordered pairs (a, b), where a represents the coordinate for the horizontal axis and b represents the coordinate for the vertical axis. Let’s consider the number −2 + 3i. The real part of
the complex number is −2 and the imaginary part is 3i. We plot the ordered pair (−2, 3) to represent the complex number −2 + 3i as shown in Figure 1. i 3 2 1 –3 –2 –1 1 2 3 r Figure 1 complex plane In the complex plane, the horizontal axis is the real axis, and the vertical axis is the imaginary axis as shown in Figure 2. imaginary real Figure 2 200 CHAPTER 3 polynomial and rational Functions How To… Given a complex number, represent its components on the complex plane. 1. Determine the real part and the imaginary part of the complex number. 2. Move along the horizontal axis to show the real part of the number. 3. Move parallel to the vertical axis to show the imaginary part of the number. 4. Plot the point. Example 2 Plotting a Complex Number on the Complex Plane Plot the complex number 3 − 4i on the complex plane. Solution The real part of the complex number is 3, and the imaginary part is −4i. We plot the ordered pair (3, −4) as shown in Figure 3. i 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 r Figure 3 Try It #2 Plot the complex number −4 − i on the complex plane. Adding and Subtracting Complex numbers Just as with real numbers, we can perform arithmetic operations on complex numbers. To add or subtract complex numbers, we combine the real parts and combine the imaginary parts. complex numbers: addition and subtraction Adding complex numbers: Subtracting complex numbers: (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) − (c + di) = (a − c) + (b − d)i How To… Given two complex numbers, find the sum or difference. 1. Identify the real and imaginary parts of each number. 2. Add or subtract the real parts. 3. Add or subtract the imaginary parts. Example 3 Adding Complex Numbers Add 3 − 4i and 2 + 5i. SECTION 3.1 complex numBers 201 Solution We add the real parts and add the imaginary parts. (a + bi) + (c + di) = (a + c) + (b + d)i (3 − 4i) + (2 + 5i) = (3 +
2) + (−4 + 5)i = 5 + i Try It #3 Subtract 2 + 5i from 3 − 4i. Multiplying Complex numbers Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately. Multiplying a Complex Numbers by a Real Number Let’s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So, for example: 3(6 + 2i) = (3 ⋅ 6) + (3 ⋅ 2i) Distribute. = 18 + 6i Simplify. How To… Given a complex number and a real number, multiply to find the product. 1. Use the distributive property. 2. Simplify. Example 4 Multiplying a Complex Number by a Real Number Find the product 4(2 + 5i). Solution Distribute the 4. 4(2 + 5i) = (4 ⋅ 2) + (4 ⋅ 5i) = 8 + 20i Try It #4 Find the product −4(2 + 6i). Multiplying Complex Numbers Together Now, let’s multiply two complex numbers. We can use either the distributive property or the FOIL method. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. Using either the distributive property or the FOIL method, we get Because i2 = −1, we have (a + bi)(c + di) = ac + adi + bci + bdi2 (a + bi)(c + di) = ac + adi + bci − bd To simplify, we combine the real parts, and we combine the imaginary parts. (a + bi)(c + di) = (ac − bd) + (ad + bc)i 202 CHAPTER 3 polynomial and rational Functions How To… Given two complex numbers, multiply to find the product. 1. Use the distributive property or the FOIL method. 2. Simplify. Example 5 Multiplying a Complex Number by a Complex Number Multiply (4 + 3i)(2 − 5i). Solution Use (a + bi)(c + di) = (ac − bd) + (ad + bc)i (4 + 3i)(2 − 5i) = (4 ⋅ 2 −
3 ⋅ (−5)) + (4 ⋅ (−5) + 3 ⋅ 2)i = (8 + 15) + (−20 + 6)i = 23 − 14i Try It #5 Multiply (3 − 4i)(2 + 3i). Dividing Complex numbers Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of a + bi is a − bi. Note that complex conjugates have a reciprocal relationship: The complex conjugate of a + bi is a − bi, and the complex conjugate of a − bi is a + bi. Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another. Suppose we want to divide c + di by a + bi, where neither a nor b equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply. Multiply the numerator and denominator by the complex conjugate of the denominator. c + di _____ a + bi where a ≠ 0 and b ≠ 0 (c + di) ______ ⋅ (a + bi) (a − bi) _______ = (a − bi) (c + di)(a − bi) ____________ (a + bi)(a − bi) Apply the distributive property. Simplify, remembering that i2 = −1. = ca − cbi + adi − bdi2 ________________ a2 − abi + abi − b2i2 = ca − cbi + adi − bd( − 1) ____________________ a2 − abi + abi − b2(− 1) = (ca + bd) + (ad − cb)i __________________ a2 + b2 SECTION 3.1 complex numBers 203 the complex conjugate The complex conjugate of
a complex number a + bi is a − bi. It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged. • When a complex number is multiplied by its complex conjugate, the result is a real number. • When a complex number is added to its complex conjugate, the result is a real number. Example 6 Finding Complex Conjugates Find the complex conjugate of each number. a. 2 + i √ — 5 1 __ b. − i 2 Solution a. The number is already in the form a + bi. The complex conjugate is a − bi, or 2 − i √ — 5. 1 1 __ __ b. We can rewrite this number in the form a + bi as 0 − i. The complex conjugate is a − bi, or 0 + i. This can 2 2 1 __ i. be written simply as 2 Analysis Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by i. How To… Given two complex numbers, divide one by the other. 1. Write the division problem as a fraction. 2. Determine the complex conjugate of the denominator. 3. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator. 4. Simplify. Example 7 Dividing Complex Numbers Divide (2 + 5i) by (4 − i). Solution We begin by writing the problem as a fraction. (2 + 5i) _______ (4 − i) Then we multiply the numerator and denominator by the complex conjugate of the denominator. (2 + 5i) _______ ⋅ (4 − i) (4 + i) ______ (4 + i) To multiply two complex numbers, we expand the product as we would with polynomials (the process commonly called FOIL). (2 + 5i) _______ ⋅ (4 − i) (4 + i) ______ (4 + i) = 8 + 2i + 20i + 5i2 ______________ 16 + 4i − 4i − i2 = 8 + 2i + 20i + 5(−1) __________________ 16 + 4
i − 4i − (−1) Because i2 = −1 = 3 + 22i ______ 17 = 3 __ 17 + 22 ___ i 17 Separate real and imaginary parts. Note that this expresses the quotient in standard form. 204 CHAPTER 3 polynomial and rational Functions Example 8 Substituting a Complex Number into a Polynomial Function Let f(x) = x2 − 5x + 2. Evaluate f (3 + i). Solution Substitute x = 3 + i into the function f(x) = x2 − 5x + 2 and simplify. f (3 + i) = (3 + i)2 − 5(3 + i) + 2 Substitute 3 + i for x. = (3 + 6i + i2) − (15 + 5i) + 2 Multiply. = 9 + 6i + (−1) −15 −5i + 2 Substitute −1 for i2. = −5 + i Combine like terms. Analysis We write f (3 + i) = −5 + i. Notice that the input is 3 + i and the output is −5 + i. Try It #6 Let f(x) = 2x2 − 3x. Evaluate f (8 − i). Example 9 Substituting an Imaginary Number in a Rational Function Let f(x) =. Evaluate f (10i). 2 + x _____ x + 3 Solution Substitute x = 10i and simplify. 2 + 10i _______ 10i + 3 2 + 10i _______ 3 + 10i 2 + 10i _______ ⋅ 3 + 10i 3 − 10i _______ 3 − 10i 6 − 20i + 30i − 100i2 __________________ 9 − 30i + 30i − 100i2 6 − 20i + 30i − 100( −1) ______________________ 9 − 30i + 30i − 100( −1) 106 + 10i ________ 109 106 ____ 109 + 10 ___ i 109 Try It #7 Substitute 10i for x. Rewrite the denominator in standard form. Prepare to multiply the numerator and denominator by the complex conjugate of the denominator. Multiply using the distributive property or the FOIL method. Substitute −1 for i 2. Simplify. Separate the real and imaginary parts. Let f(x) =. Evaluate f (−i). x + 1
_____ x − 4 Simplifying Powers of i The powers of i are cyclic. Let’s look at what happens when we raise i to increasing powers. i1 = i i2 = −1 i3 = i2 ⋅ i = −1 ⋅ i = −i i4 = i3 ⋅ i = −i ⋅ i = −i2 = −(−1) = 1 i5 = i4 ⋅ i = 1 ⋅ i = i SECTION 3.1 complex numBers 205 We can see that when we get to the fifth power of i, it is equal to the first power. As we continue to multiply i by itself for increasing powers, we will see a cycle of 4. Let’s examine the next 4 powers of i. i6 = i5 ⋅ i = i ⋅ i = i2 = −1 i7 = i6 ⋅ i = i2 ⋅ i = i3 = −i i8 = i7 ⋅ i = i3 ⋅ i = i4 = 1 i9 = i8 ⋅ i = i4 ⋅ i = i5 = i Example 10 Simplifying Powers of i Evaluate i35. Solution Since i4 = 1, we can simplify the problem by factoring out as many factors of i4 as possible. To do so, first determine how many times 4 goes into 35: 35 = 4 ⋅ 8 + 3. i35 = i4 ⋅ 8 + 3 = i4 ⋅ 8 ⋅ i3 = (i4)8 ⋅ i3 = 18 ⋅ i3 = i3 = −i Q & A… Can we write i35 in other helpful ways? As we saw in Example 10, we reduced i35 to i3 by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of i35 may be more useful. Table 1 shows some other possible factorizations. Factorization of i35 i34 ⋅ i i33 ⋅ i2 i31 ⋅ i4 i19 ⋅ i16 Reduced form (i2)17⋅ i i33 ⋅ (−1) i31 ⋅ 1 i19 ⋅(i4)4 Simplified form (−1)17 ⋅ i −i33 i31
i19 Table 1 Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method. Access these online resources for additional instruction and practice with complex numbers. • Adding and Subtracting Complex numbers (http://openstaxcollege.org/l/addsubcomplex) • Multiply Complex numbers (http://openstaxcollege.org/l/multiplycomplex) • Multiplying Complex Conjugates (http://openstaxcollege.org/l/multcompconj) • Raising i to Powers (http://openstaxcollege.org/l/raisingi) 206 CHAPTER 3 polynomial and rational Functions 3.1 SeCTIOn exeRCISeS VeRBAl 1. Explain how to add complex numbers. 2. What is the basic principle in multiplication of complex numbers? 3. Give an example to show the product of two imaginary numbers is not always imaginary. 4. What is a characteristic of the plot of a real number in the complex plane? AlGeBRAIC For the following exercises, evaluate the algebraic expressions. 5. If f (x) = x2 + x − 4, evaluate f (2i). 6. If f (x) = x3 − 2, evaluate f (i). 7. If f (x) = x2 + 3x + 5, evaluate f (2 + i). 8. If f (x) = 2x2 + x − 3, evaluate f (2 − 3i). 9. If f (x) =, evaluate f (5i). x + 1 _____ 2 − x 10. If f (x) =, evaluate f (4i). 1 + 2x ______ x + 3 GRAPHICAl For the following exercises, determine the number of real and nonreal solutions for each quadratic function shown. 11. y 12. y x x For the following exercises, plot the complex numbers on the complex plane. 13. 1 − 2i 15. i nUMeRIC 14. −2 + 3i 16. −3 − 4i For the following exercises, perform the indicated operation and express the result as a simplified complex number. 17. (3 + 2i) + (5 − 3i) 18. (−2 − 4i) + (1 + 6i) 19. (−5 + 3i) − (6 − i) 20. (2 − 3i
) − (3 + 2i) 21. (−4 + 4i) − (−6 + 9i) 22. (2 + 3i)(4i) 23. (5 − 2i)(3i) 26. (2 + 3i)(4 − i) 29. (3 + 4i)(3 − 4i) 32. −5 + 3i ______ 2i 24. (6 − 2i)(5) 25. (−2 + 4i)(8) 27. (−1 + 2i)(−2 + 3i) 28. (4 − 2i)(4 + 2i) 30. 3 + 4i _____ 2 33. 6 + 4i _____ i 31. 6 − 2i _____ 3 34. 2 − 3i _____ 4 + 3i SECTION 3.1 section exercises 207 35. 3 + 4i _____ 2 − i 36. 2 + 3i _____ 2 − 3i 37. √ — −9 + 3 √ — −16 38. − √ — −4 − 4 √ — −25 41. i8 TeCHnOlOGY 39. — −12 2 + √ _________ 2 40. — −20 4 + √ _________ 2 42. i15 43. i22 For the following exercises, use a calculator to help answer the questions. 44. Evaluate (1 + i)k for k = 4, 8, and 12. Predict the 45. Evaluate (1 − i)k for k = 2, 6, and 10. Predict the value if k = 16. value if k = 14. 46. Evaluate (1 + i)k − (1 − i)k for k = 4, 8, and 12. Predict the value for k = 16. 47. Show that a solution of x6 + 1 = 0 is — 3 √ ____ 2 1 __ + i. 2 48. Show that a solution of x8 − 1 = 0 is — 2 √ ____ 2 + — 2 √ ____ i. 2 exTenSIOnS For the following exercises, evaluate the expressions, writing the result as a simplified complex number. 4 1 __ __ + 49. i3 i 52. i−3 + 5i7 55. (3 + i)2 _______ (1 + 2i)2 58. 3 + 2i _____ − 1 + 2i 2 − 3i _____ 3
+ i 50. 1 __ i11 − 1 __ i21 51. i7(1 + i2) 53. (2 + i)(4 − 2i) ___________ (1 + i) 54. (1 + 3i)(2 − 4i) ____________ (1 + 2i) 56. 3 + 2i _____ 2 + i + (4 + 3i) 57. 4 + i ____ i + 3 − 4i _____ 1 − i 208 CHAPTER 3 polynomial and rational Functions leARnInG OBjeCTIVeS In this section, you will: • • • • Recognize characteristics of parabolas. Understand how the graph of a parabola is related to its quadratic function. Determine a quadratic function’s minimum or maximum value. Solve problems involving a quadratic function’s minimum or maximum value. 3.2 QUADRATIC FUnCTIOnS Figure 1 An array of satellite dishes. (credit: Matthew Colvin de Valle, Flickr) Curved antennas, such as the ones shown in Figure 1 are commonly used to focus microwaves and radio waves to transmit television and telephone signals, as well as satellite and spacecraft communication. The cross-section of the antenna is in the shape of a parabola, which can be described by a quadratic function. In this section, we will investigate quadratic functions, which frequently model problems involving area and projectile motion. Working with quadratic functions can be less complex than working with higher degree functions, so they provide a good opportunity for a detailed study of function behavior. Recognizing Characteristics of Parabolas The graph of a quadratic function is a U-shaped curve called a parabola. One important feature of the graph is that it has an extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry. These features are illustrated in Figure 2. y 6 4 2 Axis of symmetry x-intercepts –6 –4 –2 2 4 6 x –2 y–inter
cept 4 Vertex –6 Figure 2 SECTION 3.2 Quadratic Functions 209 The y-intercept is the point at which the parabola crosses the y-axis. The x-intercepts are the points at which the parabola crosses the x-axis. If they exist, the x-intercepts represent the zeros, or roots, of the quadratic function, the values of x at which y = 0. Example 1 Identifying the Characteristics of a Parabola Determine the vertex, axis of symmetry, zeros, and y-intercept of the parabola shown in Figure 3. y 10 8 6 4 2 –4 –2 2 4 6 8 x Figure 3 Solution The vertex is the turning point of the graph. We can see that the vertex is at (3, 1). Because this parabola opens upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is x = 3. This parabola does not cross the x-axis, so it has no zeros. It crosses the y-axis at (0, 7) so this is the y-intercept. Understanding How the Graphs of Parabolas are Related to Their Quadratic Functions The general form of a quadratic function presents the function in the form f(x) = ax 2 + bx + c where a, b, and c are real numbers and a ≠ 0. If a > 0, the parabola opens upward. If a < 0, the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry. b2 − 4ac −b ± √ The axis of symmetry is defined by x = − b _______________ __, to solve 2a 2a ax 2 + bx + c = 0 for the x-intercepts, or zeros, we find the value of x halfway between them is always x = − b __ 2a equation for the axis of symmetry.. If we use the quadratic formula, x =, the — Figure 4 represents the graph of the quadratic function written in general form as y = x2 + 4x + 3. In this form, a = 1, b = 4, and c = 3. Because a > 0, the parabola opens upward. The axis of symmetry is x = −
4 ____ = −2. This 2(1) also makes sense because we can see from the graph that the vertical line x = −2 divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, (−2, −1). The x-intercepts, those points where the parabola crosses the x-axis, occur at (−3, 0) and (−1, 0). y y = x2 + 4x + 3 8 6 4 2 –4 –6 Vertex Axis of symmetry –2 –4 x-intercepts x 2 4 6 Figure 4 210 CHAPTER 3 polynomial and rational Functions The standard form of a quadratic function presents the function in the form f(x) = a(x − h)2 + k where (h, k) is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function. As with the general form, if a > 0, the parabola opens upward and the vertex is a minimum. If a < 0, the parabola opens downward, and the vertex is a maximum. Figure 5 represents the graph of the quadratic function written in standard form as y = −3(x + 2)2 + 4. Since x − h = x + 2 in this example, h = −2. In this form, a = −3, h = −2, and k = 4. Because a < 0, the parabola opens downward. The vertex is at (−2, 4). Vertex y 4 2 y = −3(x + 2)2 + 4 –6 –4 –2 2 4 6 x –2 –4 –6 –8 Figure 5 The standard form is useful for determining how the graph is transformed from the graph of y = x 2. Figure 6 is the graph of this basic function. y 10 8 6 4 2 y = x2 –6 –4 –2 2 4 6 x –2 Figure 6 If k > 0, the graph shifts upward, whereas if k < 0, the graph shifts downward. In Figure 5, k > 0, so the graph is shifted 4 units upward. If h > 0, the graph shifts toward the right and if h < 0, the graph shifts to the left. In
Figure 5, h < 0, so the graph is shifted 2 units to the left. The magnitude of a indicates the stretch of the graph. If ∣a∣ > 1, the point associated with a particular x-value shifts farther from the x-axis, so the graph appears to become narrower, and there is a vertical stretch. But if ∣a∣ < 1, the point associated with a particular x-value shifts closer to the x-axis, so the graph appears to become wider, but in fact there is a vertical compression. In Figure 5, ∣a∣ > 1, so the graph becomes narrower. The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form. a(x − h)2 + k = ax 2 + bx + c ax 2 − 2ahx + (ah2 + k) = ax 2 + bx + c For the linear terms to be equal, the coefficients must be equal. −2ah = b, so h = − b __. 2a SECTION 3.2 Quadratic Functions 211 This is the axis of symmetry we defined earlier. Setting the constant terms equal: ah 2 + k = c k = c − ah2 = c − a  − b2 __ 4a = c − 2 b  __ 2a In practice, though, it is usually easier to remember that k is the output value of the function when the input is h, so f(h) = k. forms of quadratic functions A quadratic function is a function of degree two. The graph of a quadratic function is a parabola. The general form of a quadratic function is f(x) = ax 2 + bx + c where a, b, and c are real numbers and a ≠ 0. The standard form of a quadratic function is f(x) = a(x − h)2 + k. The vertex (h, k) is located at h = − b __ 2a, k = f(h) = f  −b . ___ 2a How To… Given a graph of a quadratic function, write the equation of the function in general form. 1. Identify the horizontal shift of the parabola; this value is h. Identify the vertical shift of the
parabola; this value is k. 2. Substitute the values of the horizontal and vertical shift for h and k. in the function f(x) = a(x − h)2 + k. 3. Substitute the values of any point, other than the vertex, on the graph of the parabola for x and f (x). 4. Solve for the stretch factor, ∣a∣. 5. If the parabola opens up, a > 0. If the parabola opens down, a < 0 since this means the graph was reflected about the x-axis. 6. Expand and simplify to write in general form. Example 2 Writing the Equation of a Quadratic Function from the Graph Write an equation for the quadratic function g in Figure 7 as a transformation of f(x) = x 2, and then expand the formula, and simplify terms to write the equation in general form. y 6 4 2 –6 –4 –2 2 4 x –2 –4 Figure 7 Solution We can see the graph of g is the graph of f(x) = x 2 shifted to the left 2 and down 3, giving a formula in the form g(x) = a(x + 2)2 − 3. 212 CHAPTER 3 polynomial and rational Functions Substituting the coordinates of a point on the curve, such as (0, −1), we can solve for the stretch factor. −1 = a(0 + 2)2 − 3 2 = 4a a = 1 __ 2 1 __ (x + 2)2 − 3. In standard form, the algebraic model for this graph is g (x) = 2 To write this in general polynomial form, we can expand the formula and simplify terms. 1 __ (x + 2)2 − 3 g(x) = 2 1 __ = (x + 2)(x + 2) − 3 2 1 __ = (x 2 + 4x + 4) − 3 2 1 __ = x 2 + 2x + 2 − 3 2 1 __ = x 2 + 2x − 1 2 Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions. 1 __ (x + 2) 2 − 3. Next, Analysis We can check our work using the table feature on a graphing utility. First enter Y
1 = 2 select TBLSET, then use TblStart = − 6 and ΔTbl = 2, and select TABLE. See Table 1. x y −6 5 −4 −1 −2 −3 0 −1 2 5 Table 1 The ordered pairs in the table correspond to points on the graph. Try It #1 A coordinate grid has been superimposed over the quadratic path of a basketball in Figure 8 Find an equation for the path of the ball. Does the shooter make the basket? Figure 8 (credit: modification of work by Dan Meyer) How To… Given a quadratic function in general form, find the vertex of the parabola. 1. Identify a, b, and c. 2. Find h, the x-coordinate of the vertex, by substituting a and b into h = − b . 3. Find k, the y-coordinate of the vertex, by evaluating k = f (h) = f  − __ 2a b __. 2a SECTION 3.2 Quadratic Functions 213 Example 3 Finding the Vertex of a Quadratic Function Find the vertex of the quadratic function f (x) = 2x2 − 6x + 7. Rewrite the quadratic in standard form (vertex form). Solution The horizontal coordinate of the vertex will be at h = − b __ 2a −6 ____ 2(2) = − 6 __ = 4 3 __ = 2 k = f(h) The vertical coordinate of the vertex will be at 2 3  = f  __ 2 3  = 2  __ 2 5 __ = 2 3  + 7 − 6  __ 2 Rewriting into standard form, the stretch factor will be the same as the a in the original quadratic. Using the vertex to determine the shifts, f (x) = ax2 + bx + c f (x) = 2x2 − 6x + 7 3  f (x) = 2  x − __ 2 2 5 __ + 2 Analysis One reason we may want to identify the vertex of the parabola is that this point will inform us where the maximum or minimum value of the output occurs, (k), and where it occurs, (x). Try It #2 Given the equation g (x) = 13 + x 2 − 6x, write the equation in general form
and then in standard form. Finding the Domain and Range of a Quadratic Function Any number can be the input value of a quadratic function. Therefore, the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all y-values greater than or equal to the y-coordinate at the turning point or less than or equal to the y-coordinate at the turning point, depending on whether the parabola opens up or down. domain and range of a quadratic function The domain of any quadratic function is all real numbers. The range of a quadratic function written in general form f(x) = ax2 + bx + c with a positive a value is f (x) ≥ f  − b __ 2a  . , or  −∞, f  − b The range of a quadratic function written in general form with a negative a value is f (x) ≤ f  − b __ __ 2a 2a The range of a quadratic function written in standard form f (x) = a(x − h)2 + k with a positive a value is f (x) ≥ k; the range of a quadratic function written in standard form with a negative a value is f (x) ≤ k. , or  f  − b __ 2a , ∞ . 214 CHAPTER 3 polynomial and rational Functions How To… Given a quadratic function, find the domain and range. 1. Identify the domain of any quadratic function as all real numbers. 2. Determine whether a is positive or negative. If a is positive, the parabola has a minimum. If a is negative, the parabola has a maximum. 3. Determine the maximum or minimum value of the parabola, k. 4. If the parabola has a minimum, the range is given by f(x) ≥ k, or [k, ∞). If the parabola has a maximum, the range is given by f (x) ≤ k, or (−∞, k]. Example 4 Finding the Domain and Range of
a Quadratic Function Find the domain and range of f (x) = −5x2 + 9x − 1. Solution As with any quadratic function, the domain is all real numbers. Because a is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the x-value of the vertex. h = − b __ 2a = − 9 _____ 2(−5) = 9 __ 10 The maximum value is given by f (h). 9 f  __ 10 2 9   = −5  __ 10 61 ___ 20 = 9 + 9  __ 10  − 1 The range is f (x) ≤ 61 ___ 20, or  −∞, 61 . ___ 20 Try It #3 Find the domain and range of f (x) = 2  x − 4  __ 7 2 + 8 __. 11 Determining the Maximum and Minimum Values of Quadratic Functions The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola. We can see the maximum and minimum values in Figure 9. y f (x) = (x − 2)2 + 1 6 4 2 (2, 1) (–3, 4) y 6 4 2 g(x) = −(x + 3)2 + 4 –6 –4 –2 2 4 6 x –6 –4 –2 2 4 6 x –2 –4 –6 (a) Minimum value of 1 occurs at x = 2 Figure 9 Minimum value of 4 occurs at x = −3 –2 –4 –6 (b) There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue. SECTION 3.2 Quadratic Functions 215 Example 5 Finding the Maximum Value of a Quadratic Function A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side. a. Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length L. b. What dimensions should she make her garden to maximize the enclosed
area? Solution Let’s use a diagram such as Figure 10 to record the given information. It is also helpful to introduce a temporary variable, W, to represent the width of the garden and the length of the fence section parallel to the backyard fence. Garden L W Backyard Figure 10 a. We know we have only 80 feet of fence available, and L + W + L = 80, or more simply, 2L + W = 80. This allows us to represent the width, W, in terms of L. W = 80 − 2L Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so A = LW = L(80 − 2L) A(L) = 80L − 2L2 This formula represents the area of the fence in terms of the variable length L. The function, written in general form, is A(L) = −2L2 + 80L. b. The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since a is the coefficient of the squared term, a = −2, b = 80, and c = 0. To find the vertex: h = − 80 _ 2(−2) = 20 k = A(20) and = 80(20) − 2(20)2 = 800 The maximum value of the function is an area of 800 square feet, which occurs when L = 20 feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet. Analysis This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in Figure 11. (20, 800) A y 1000 900 800 700 600 500 400 300 200 100 ) A ( a e r A 0 10 30 20 Length (L) Figure 11 40 x 50 216 CHAPTER 3 polynomial and rational Functions How To… Given an application involving revenue, use a quadratic equation to find the
maximum. 1. Write a quadratic equation for revenue. 2. Find the vertex of the quadratic equation. 3. Determine the y-value of the vertex. Example 6 Finding Maximum Revenue The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue? Solution Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, p for price per subscription and Q for quantity, giving us the equation Revenue = pQ. Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently p = 30 and Q = 84,000. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, p = 32 and Q = 79,000. From this we can find a linear equation relating the two quantities. The slope will be m = 79,000 − 84,000 _____________ 32 − 30 = −5,000 _______ 2 = −2,500 This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y-intercept. Q = −2,500p + b Substitute in the point Q = 84,000 and p = 30 84,000 = −2,500(30) + b Solve for b b = 159,000 This gives us the linear equation Q = −2,500p + 159,000 relating cost and subscribers. We now return to our revenue equation. Revenue = pQ Revenue = p(−2,500p + 159,000) Revenue = −2,500p2 + 159,000p We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex. h = − 159,000 _________ 2(−2,500) = 31.
8 The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function. maximum revenue = −2,500(31.8)2 + 159,000(31.8) = 2,528,100 Analysis This could also be solved by graphing the quadratic as in Figure 12. We can see the maximum revenue on a graph of the quadratic function. SECTION 3.2 Quadratic Functions 217 (31.80, 2,528.1) y 3,000 2,500 2,000 1,500 1,000 500 ) 10 20 30 40 50 60 70 80 x Price (p) Figure 12 Finding the x- and y-Intercepts of a Quadratic Function Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the y-intercept of a quadratic by evaluating the function at an input of zero, and we find the x-intercepts at locations where the output is zero. Notice in Figure 13 that the number of x-intercepts can vary depending upon the location of the graph. y 5 4 3 2 1 –3 –2 –1 –1 –2 –3 y 5 4 3 2 1 –1 –1 –2 –3 21 3 4 5 x –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 21 3 4 5 x 21 3 4 5 x –3 –2 No x-intercept One x-intercept Two x-intercepts Figure 13 number of x-intercepts of a parabola How To… Given a quadratic function f (x), find the y- and x-intercepts. 1. Evaluate f (0) to find the y-intercept. 2. Solve the quadratic equation f (x) = 0 to find the x-intercepts. Example 7 Finding the y- and x-Intercepts of a Parabola Find the y- and x-intercepts of the quadratic f(x) = 3x2 + 5x − 2. Solution We find the y-intercept by evaluating f (0). So the y-intercept is at (0, −2). For the x-intercepts, we find all solutions of
f(x) = 0. 0 = 3x2 + 5x − 2 f(0) = 3(0)2 + 5(0) − 2 = −2 In this case, the quadratic can be factored easily, providing the simplest method for solution. 1 __, 0  and (−2, 0). So the x-intercepts are at  3 0 = (3x − 1)(x + 2) 0 = 3x − 1 1 __ x = 3 0 = x + 2 or x = −2 218 CHAPTER 3 polynomial and rational Functions Analysis By graphing the function, we can confirm that the graph crosses the y-axis at (0, −2). We can also confirm that the graph crosses the x-axis 1 __, 0  and (−2, 0). See Figure 14. at  3 y (–2, 0) 1 f (x) = 3x2 + 5x − 2, 0) ) 1 3 –3 –2 –1 1 2 3 x (0, –2) –1 –2 –3 –4 –5 Figure 14 Rewriting Quadratics in Standard Form In Example 7, the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form. How To… Given a quadratic function, find the x-intercepts by rewriting in standard form. 1. Substitute a and b into h = − b __. 2a 2. Substitute x = h into the general form of the quadratic function to find k. 3. Rewrite the quadratic in standard form using h and k. 4. Solve for when the output of the function will be zero to find the x-intercepts. Example 8 Finding the x-Intercepts of a Parabola Find the x-intercepts of the quadratic function f (x) = 2x2 + 4x − 4. Solution We begin by solving for when the output will be zero. 0 = 2x2 + 4x − 4 Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form. We know that a = 2. Then we solve for h and k. f (x) = a(x
− h)2 + k h = − b __ 2a = − 4 ___ 2(2) = −1 k = f (−1) = 2(−1)2 + 4(−1) − 4 = −6 So now we can rewrite in standard form. We can now solve for when the output will be zero. f(x) = 2(x + 1)2 − 6 0 = 2(x + 1)2 − 6 6 = 2(x + 1)2 The graph has x-intercepts at (−1 − √ 3, 0) and (−x + 1)2 3 x = −1 ± √ 3, 0). — — 3 SECTION 3.2 Quadratic Functions 219 Analysis We can check our work by graphing the given function on a graphing utility and observing the x-intercepts. See Figure 15. y 6 4 2 (0.732, 0) (−2.732, 0) –6 –4 –2 2 4 6 x –2 –4 –6 Figure 15 Try It #4 In a separate Try It, we found the standard and general form for the function g(x) = 13 + x2 − 6x. Now find the y- and x-intercepts (if any). Example 9 Solving a Quadratic Equation with the Quadratic Formula Solve x2 + x + 2 = 0. Solution Let’s begin by writing the quadratic formula: x = — b2 − 4ac −b ± √ _______________ 2a. When applying the quadratic formula, we identify the coefficients a, b and c. For the equation x2 + x + 2 = 0, we have a = 1, b = 1, and c = 2. Substituting these values into the formula we have: x = — — b2 − 4ac −b ± √ _______________ 2a — 12 − 4 ⋅ 1 ⋅ (2) −1 ± √ ___________________ 2 ⋅ 1 1 − 8 −1 ± √ ____________ 2 −7 −1 ± √ __________ 2 7 −1 ± i √ _________ 2 — — = = = = The solutions to the equation are — 7 −1 + i √ _________ 2 and 7 −1 − i √ _________ 2 or −1 ___ 2 + — 7 i √
____ 2 — and −1 ___ 2 − — 7 i √ ____. 2 Example 10 Applying the Vertex and x-Intercepts of a Parabola A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation H(t) = −16t 2 + 80t + 40. a. When does the ball reach the maximum height? b. What is the maximum height of the ball? c. When does the ball hit the ground? Solution a. The ball reaches the maximum height at the vertex of the parabola. h = − 80 ______ 2(−16) = 80 ___ 32 5 __ = 2 = 2.5 The ball reaches a maximum height after 2.5 seconds. 220 CHAPTER 3 polynomial and rational Functions b. To find the maximum height, find the y-coordinate of the vertex of the parabola. k = H  − b  __ 2a = H(2.5) = −16(2.5)2 + 80(2.5) + 40 = 140 The ball reaches a maximum height of 140 feet. c. To find when the ball hits the ground, we need to determine when the height is zero, H(t) = 0. We use the quadratic formula. t = — −80 ± √ _______________________ 802 − 4(−16)(40) 2(−16) = — 8960 −80 ± √ _____________ −32 Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions. t = — 8960 −80 − √ _____________ −32 ≈ 5.458 or t = — 8960 −80 + √ _____________ −32 ≈ −0.458 The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See Figure 16. H 150 125 100 75 50 25 (2.5, 140) H(t) = −16t2 + 80t + 40 1 2 3 4 Figure 16. t 5 6 Try It #5 A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock’s height above ocean can be modeled by the equation
H(t) = −16t 2 + 96t + 112. a. When does the rock reach the maximum height? b. What is the maximum height of the rock? c. When does the rock hit the ocean? Access these online resources for additional instruction and practice with quadratic equations. • Graphing Quadratic Functions in General Form (http://openstaxcollege.org/l/graphquadgen) • Graphing Quadratic Functions in Standard Form (http://openstaxcollege.org/l/graphquadstan) • Quadratic Function Review (http://openstaxcollege.org/l/quadfuncrev) • Characteristics of a Quadratic Function (http://openstaxcollege.org/l/characterquad) SECTION 3.2 section exercises 221 3.2 SeCTIOn exeRCISeS VeRBAl 1. Explain the advantage of writing a quadratic 2. How can the vertex of a parabola be used in solving function in standard form. real world problems? 3. Explain why the condition of a ≠ 0 is imposed in the definition of the quadratic function. 5. What two algebraic methods can be used to find the horizontal intercepts of a quadratic function? AlGeBRAIC 4. What is another name for the standard form of a quadratic function? For the following exercises, rewrite the quadratic functions in standard form and give the vertex. 6. f (x) = x 2 − 12x + 32 7. g(x) = x 2 + 2x − 3 8. f (x) = x 2 − x 9. f (x) = x 2 + 5x − 2 10. h(x) = 2x 2 + 8x − 10 11. k(x) = 3x 2 − 6x − 9 12. f (x) = 2x 2 − 6x 13. f (x) = 3x 2 − 5x − 1 For the following exercises, determine whether there is a minimum or maximum value to each quadratic function. Find the value and the axis of symmetry. 14. y(x) = 2x 2 + 10x + 12 15. f(x) = 2x 2 − 10x + 4 17. f(x) = 4x 2 + x − 1 18. h(t) = −4t 2 + 6t − 1 16.
f(x) = −x 2 + 4x + 3 1 __ x 2 + 3x + 1 19. f(x) = 2 20. f(x) = − 1 __ x 2 − 2x + 3 3 For the following exercises, determine the domain and range of the quadratic function. 21. f(x) = (x − 3)2 + 2 22. f(x) = −2(x + 3)2 − 6 23. f(x) = x 2 + 6x + 4 24. f(x) = 2x 2 − 4x + 2 25. k(x) = 3x 2 − 6x − 9 For the following exercises, solve the equations over the complex numbers. 26. x 2 = −25 29. x 2 + 27 = 0 32. x 2 + 8x + 25 = 0 35. x 2 − 10x + 26 = 0 38. x(x − 2) = 10 41. 5x 2 + 6x + 2 = 0 44. x 2 − 2x + 4 = 0 27. x 2 = −8 30. x 2 + 2x + 5 = 0 33. x2 − 4x + 13 = 0 36. x 2 − 6x + 10 = 0 39. 2x 2 + 2x + 5 = 0 42. 2x 2 − 6x + 5 = 0 28. x 2 + 36 = 0 31. x 2 − 4x + 5 = 0 34. x 2 + 6x + 25 = 0 37. x(x − 4) = 20 40. 5x 2 − 8x + 5 = 0 43. x 2 + x + 2 = 0 For the following exercises, use the vertex (h, k) and a point on the graph (x, y) to find the general form of the equation of the quadratic function. 45. (h, k) = (2, 0), (x, y) = (4, 4) 46. (h, k) = (−2, −1), (x, y) = (−4, 3) 47. (h, k) = (0, 1), (x, y) = (2, 5) 48. (h, k) = (2, 3), (x, y) = (5, 12) 49. (h, k) = (−5, 3), (x, y) = (2, 9) 50
. (h, k) = (3, 2), (x, y) = (10, 1) 51. (h, k) = (0, 1), (x, y) = (1, 0) 52. (h, k) = (1, 0), (x, y) = (0, 1) 222 CHAPTER 3 polynomial and rational Functions GRAPHICAl For the following exercises, sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts. 53. f(x) = x 2 − 2x 54. f(x) = x 2 − 6x − 1 55. f(x) = x 2 − 5x − 6 56. f(x) = x 2 − 7x + 3 57. f(x) = −2x 2 + 5x − 8 58. f(x) = 4x 2 − 12x − 3 For the following exercises, write the equation for the graphed function. y 5 4 3 2 1 –4 –3 –2 –1 –1 –2 –3 –4 –5 59. 626 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 –6 –7 nUMeRIC 60 616 –5 –4 –3 –2 –1 –1 –2 1 2 3 4 x 63. y 643 –2 –1 –1 –2 –2 –1 –1 –7 –6 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 For the following exercises, use the table of values that represent points on the graph of a quadratic function. By determining the vertex and axis of symmetry, find the general form of the equation of the quadratic function. 65. x −2 −1 y 5 2 68. x −2 −1 y −8 − 66. x −2 −1 y 1 0 69. x −2 − TeCHnOlOGY For the following exercises, use a calculator to find the answer. 67. x −2 −1 y −2 1 0 2 1 2 1 −2 70. Graph on the same set of axes the functions 1 __ f (x) = x2, f(x) = 2x 2, and f(x) = x 2. What appears 3 to be the effect of changing the coefficient? 71. Graph on the same set of
axes f(x) = x 2, f(x) = x 2 + 2 and f(x) = x 2, f(x) = x 2 + 5 and f(x) = x 2 − 3. What appears to be the effect of adding a constant? 72. Graph on the same set of axes f(x) = x 2, f(x) = (x − 2)2, f(x − 3)2, and f(x) = (x + 4)2. What appears to be the effect of adding or subtracting those numbers? 73. The path of an object projected at a 45 degree angle with initial velocity of 80 feet per second is given −32 ____ by the function h(x) = (80)2 x 2 + x where x is the horizontal distance traveled and h(x) is the height in feet. Use the [TRACE] feature of your calculator to determine the height of the object when it has traveled 100 feet away horizontally. SECTION 3.2 section exercises 223 74. A suspension bridge can be modeled by the quadratic function h(x) = 0.0001x 2 with −2000 ≤ x ≤ 2000 where ∣ x ∣ is the number of feet from the center and h(x) is height in feet. Use the [TRACE] feature of your calculator to estimate how far from the center does the bridge have a height of 100 feet. exTenSIOnS For the following exercises, use the vertex of the graph of the quadratic function and the direction the graph opens to find the domain and range of the function. 75. Vertex (1, −2), opens up. 77. Vertex (−5, 11), opens down. 76. Vertex (−1, 2) opens down. 78. Vertex (−100, 100), opens up. For the following exercises, write the equation of the quadratic function that contains the given point and has the same shape as the given function. 79. Contains (1, 1) and has shape of f(x) = 2x 2. 80. Contains (−1, 4) and has the shape of f(x) = 2x 2. Vertex is on the y-axis. Vertex is on the y-axis. 81. Contains (2, 3) and has the shape of f(x) = 3x 2. 82. Contains (1, −3) and has the shape of
f(x) = −x 2. Vertex is on the y-axis. Vertex is on the y-axis. 83. Contains (4, 3) and has the shape of f(x) = 5x 2. 84. Contains (1, −6) has the shape of f(x) = 3x 2. Vertex Vertex is on the y-axis. has x-coordinate of −1. ReAl-WORlD APPlICATIOnS 85. Find the dimensions of the rectangular corral producing the greatest enclosed area given 200 feet of fencing. 86. Find the dimensions of the rectangular corral split into 2 pens of the same size producing the greatest possible enclosed area given 300 feet of fencing. 87. Find the dimensions of the rectangular corral producing the greatest enclosed area split into 3 pens of the same size given 500 feet of fencing. 88. Among all of the pairs of numbers whose sum is 6, find the pair with the largest product. What is the product? 89. Among all of the pairs of numbers whose difference is 12, find the pair with the smallest product. What is the product? 90. Suppose that the price per unit in dollars of a cell phone production is modeled by p = $45 − 0.0125x, where x is in thousands of phones produced, and the revenue represented by thousands of dollars is R = x ⋅ p. Find the production level that will maximize revenue. 91. A rocket is launched in the air. Its height, in meters above sea level, as a function of time, in seconds, is given by h(t) = −4.9t 2 + 229t + 234. Find the maximum height the rocket attains. 92. A ball is thrown in the air from the top of a building. Its height, in meters above ground, as a function of time, in seconds, is given by h(t) = −4.9t2 + 24t + 8. How long does it take to reach maximum height? 93. A soccer stadium holds 62,000 spectators. With a 94. A farmer finds that if she plants 75 trees per acre, ticket price of $11, the average attendance has been 26,000. When the price dropped to $9, the average attendance rose to 31,000. Assuming that attendance is linearly related to ticket price, what ticket price would maximize revenue? each tree will yield 20 bushels of fruit. She
estimates that for each additional tree planted per acre, the yield of each tree will decrease by 3 bushels. How many trees should she plant per acre to maximize her harvest? 224 CHAPTER 3 polynomial and rational Functions leARnInG OBjeCTIVeS In this section, you will: • • • • Identify power functions. Identify end behavior of power functions. Identify polynomial functions. Identify the degree and leading coefficient of polynomial functions. 3.3 POWeR FUnCTIOnS AnD POlYnOMIAl FUnCTIOnS Suppose a certain species of bird thrives on a small island. Its population over the last few years is shown in Table 1. Figure 1 (credit: Jason Bay, Flickr) Year Bird Population 2009 800 2010 897 Table 1 2011 992 2012 1,083 2013 1,169 The population can be estimated using the function P(t) = −0.3t3 + 97t + 800, where P(t) represents the bird population on the island t years after 2009. We can use this model to estimate the maximum bird population and when it will occur. We can also use this model to predict when the bird population will disappear from the island. In this section, we will examine functions that we can use to estimate and predict these types of changes. Identifying Power Functions In order to better understand the bird problem, we need to understand a specific type of function. A power function is a function with a single term that is the product of a real number, a coefficient, and a variable raised to a fixed real number. (A number that multiplies a variable raised to an exponent is known as a coefficient.) As an example, consider functions for area or volume. The function for the area of a circle with radius r is A(r) = πr 2 and the function for the volume of a sphere with radius r is V(r) = 4 __ πr 3 3 Both of these are examples of power functions because they consist of a coefficient, π or 4 _ 3 π, multiplied by a variable r raised to a power. power function A power function is a function that can be represented in the form f (x) = kx p where k and p are real numbers, and k is known as the coefficient. SECTION 3.3 power Functions and polynomial Functions 225 Q & A… Is f (x) = 2x a
power function? No. A power function contains a variable base raised to a fixed power. This function has a constant base raised to a variable power. This is called an exponential function, not a power function. Example 1 Identifying Power Functions Which of the following functions are power functions? f (x) = 1 f (x) = x f (x) = x 2 f (x) = x 3 f (x) = 1 __ x 1 __ x2 x f (x) = √ x f (x) = f (x) = 3 √ — — Constant function Identify function Quadratic function Cubic function Reciprocal function Reciprocal squared function Square root function Cube root function Solution All of the listed functions are power functions. The constant and identity functions are power functions because they can be written as f (x) = x0 and f (x) = x1 respectively. The quadratic and cubic functions are power functions with whole number powers f (x) = x 2 and f (x) = x 3. The reciprocal and reciprocal squared functions are power functions with negative whole number powers because they can be written as f (x) = x −1 and f (x) = x −2. The square and cube root functions are power functions with fractional powers because they can be written as f (x) = x1/2 or f (x) = x 1/3. Try It #1 Which functions are power functions? f (x) = 2x 2 ⋅ 4x 3 g(x) = −x 5 + 5x 3 − 4x h(x) = 2x5 − 1 _ 3x2 + 4 Identifying end Behavior of Power Functions Figure 2 shows the graphs of f (x) = x 2, g(x) = x 4 and h(x) = x 6, which are all power functions with even, whole-number powers. Notice that these graphs have similar shapes, very much like that of the quadratic function in the toolkit. However, as the power increases, the graphs flatten somewhat near the origin and become steeper away from the origin. f (x) = x6 y g(x) = x4 h(x) = x2 4 3 2 1 –2 –1 1 2 x Figure 2 even-power functions 226 CHAPTER 3 polynomial and rational Functions To describe the behavior as numbers become larger and larger, we use the idea of
infinity. We use the symbol ∞ for positive infinity and −∞ for negative infinity. When we say that “x approaches infinity,” which can be symbolically written as x → ∞, we are describing a behavior; we are saying that x is increasing without bound. With the even-power function, as the input increases or decreases without bound, the output values become very large, positive numbers. Equivalently, we could describe this behavior by saying that as x approaches positive or negative infinity, the f (x) values increase without bound. In symbolic form, we could write Figure 3 shows the graphs of f (x) = x 3, g(x) = x 5, and h(x) = x 7, which are all power functions with odd, whole-number powers. Notice that these graphs look similar to the cubic function in the toolkit. Again, as the power increases, the graphs flatten near the origin and become steeper away from the origin. as x → ±∞, f (x) → ∞ g(x) = x5 f (x) = x3 h(x) = x7 1 2 x y 4 2 –2 –4 –2 –1 Figure 3 Odd-power functions These examples illustrate that functions of the form f (x) = xn reveal symmetry of one kind or another. First, in Figure 2 we see that even functions of the form f (x) = xn, n even, are symmetric about the y-axis. In Figure 3 we see that odd functions of the form f (x) = xn, n odd, are symmetric about the origin. For these odd power functions, as x approaches negative infinity, f (x) decreases without bound. As x approaches positive infinity, f (x) increases without bound. In symbolic form we write as x → −∞, f (x) → −∞ as x → ∞, f (x) → ∞ The behavior of the graph of a function as the input values get very small (x → −∞) and get very large (x → ∞) is referred to as the end behavior of the function. We can use words or symbols to describe end behavior. Figure 4 shows the end behavior of power functions in the form f (x) = kxn where n is a non-negative integer depending on the power and the constant. Even power y Odd power y Positive constant k >
0 x x x → −∞, f (x) → ∞ and x → ∞, f (x) → ∞ y x → −∞, f (x) → −∞ and x → ∞, f (x) → ∞ y Negative constant k < 0 x x x → −∞, f (x) → −∞ and x → ∞, f (x) → −∞ x → −∞, f (x) → ∞ and x → ∞, f (x) → −∞ Figure 4 SECTION 3.3 power Functions and polynomial Functions 227 How To… Given a power function f (x) = kx n where n is a non-negative integer, identify the end behavior. 1. Determine whether the power is even or odd. 2. Determine whether the constant is positive or negative. 3. Use Figure 4 to identify the end behavior. Example 2 Identifying the End Behavior of a Power Function Describe the end behavior of the graph of f (x) = x8. Solution The coefficient is 1 (positive) and the exponent of the power function is 8 (an even number). As x approaches infinity, the output (value of f (x)) increases without bound. We write as x → ∞, f (x) → ∞. As x approaches negative infinity, the output increases without bound. In symbolic form, as x → −∞, f (x) → ∞. We can graphically represent the function as shown in Figure 5. y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –5 –4 –3 –2 21 3 4 5 x Example 3 Identifying the End Behavior of a Power Function Describe the end behavior of the graph of f (x) = −x 9. Figure 5 Solution The exponent of the power function is 9 (an odd number). Because the coefficient is −1 (negative), the graph is the reflection about the x-axis of the graph of f (x) = x 9. Figure 6 shows that as x approaches infinity, the output decreases without bound. As x approaches negative infinity, the output increases without bound. In symbolic form, we would write as x → −∞, f (x) → ∞ as x → ∞, f (x) → −∞ 1 –1 –2 –3 –4 –5 –6
–7 –8 –5 –4 –3 –2 21 3 4 5 x Figure 6 228 CHAPTER 3 polynomial and rational Functions Analysis We can check our work by using the table feature on a graphing utility. x −10 −5 0 5 10 f (x) 1,000,000,000 1,953,125 0 −1,953,125 −1,000,000,000 Table 2 We can see from Table 2 that, when we substitute very small values for x, the output is very large, and when we substitute very large values for x, the output is very small (meaning that it is a very large negative value). Try It #2 Describe in words and symbols the end behavior of f (x) = −5x 4. Identifying Polynomial Functions An oil pipeline bursts in the Gulf of Mexico, causing an oil slick in a roughly circular shape. The slick is currently 24 miles in radius, but that radius is increasing by 8 miles each week. We want to write a formula for the area covered by the oil slick by combining two functions. The radius r of the spill depends on the number of weeks w that have passed. This relationship is linear. We can combine this with the formula for the area A of a circle. A(r) = πr 2 Composing these functions gives a formula for the area in terms of weeks. r(w) = 24 + 8w Multiplying gives the formula. A(w) = A(r(w)) = A(24 + 8w) = π(24 + 8w)2 A(w) = 576π + 384πw + 64πw 2 This formula is an example of a polynomial function. A polynomial function consists of either zero or the sum of a finite number of non-zero terms, each of which is a product of a number, called the coefficient of the term, and a variable raised to a non-negative integer power. polynomial functions Let n be a non-negative integer. A polynomial function is a function that can be written in the form f (x) = an x n +... + a2 x 2 + a1 x + a0 This is called the general form of a polynomial function. Each ai is a coefficient and can be any real number, but an cannot = 0. Each product ai x i is a term of a polynomial function. Example 4

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.