Essentials of College Algebra For these Global Editions, the editorial team at Pearson has collaborated with educators across the world to address a wide range of subjects and requirements, equipping students with the best possible learning tools. This Global Edition preserves the cutting-edge approach and pedagog of the original, but also features alterations, customization, and adaptation from the North American version. Global Global Global Essentials of College Algebra ELEVENTH Margaret L. Lial John Hornsb David I. Schneider Callie J. Daniels ELEVENTH Lial Hornsb Schneider Daniels This is a special of an established title widel used b colleges and universities throughout the world. Pearson published this eclusive for the benefit of students outside the United States and Canada. If ou purchased this book within the United States or Canada ou should be aware that it has been imported without the approval of the Publisher or Author. Pearson Global Edition LIAL_12927585_mech.indd 1 12/8/14 7:8 pm
Essentials of College Algebra GLOBAL EDITION Margaret L. Lial American River College John Hornsb Universit of New Orleans David I. Schneider Universit of Marland Callie J. Daniels St. Charles Communit College Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montréal Toronto Delhi Meico Cit São Paulo Sdne Hong Kong Seoul Singapore Taipei Toko
SECTION 2.2 Circles 21 2.2 Eercises In Eercises 1 12, (a) find the center-radius form of the equation of each circle, and (b) graph it. See Eamples 1 and 2. 1. center 1, 2, radius 6 2. center 1, 2, radius 9 3. center 12, 2, radius 6 4. center 13, 2, radius 3 5. center 1, 42, radius 4 6. center 1, - 32, radius 7 7. center 1-2, 52, radius 4 8. center 14, 32, radius 5 9. center 15, -42, radius 7 1. center 1-3, -22, radius 6 11. center A22, 22 B, radius 22 12. center A - 23, - 23 B, radius 23 Connecting Graphs with Equations In Eercises 13 16, use each graph to determine the equation of the circle in (a) center-radius form and (b) general form. 13. 14. (3, 3) (1, 1) (5, 1) (3, 1) ( 1, 1) ( 4, 2) (2, 2) ( 1, 5) 15. 16. ( 2, 4) ( 4, 2) (, 2) ( 2, ) (, 3) (3, ) (6, 3) (3, 6) 17. Concept Check Which one of the two screens is the correct graph of the circle with center 1-3, 52 and radius 4? A. 1 B. 1 15 15 15 15 1 1 18. When the equation of a circle is written in the form 1 - h2 2 + 1 - k2 2 = m, how does the value of m indicate whether the graph is a circle, is a point, or is noneistent? Decide whether or not each equation has a circle as its graph. If it does, give the center and the radius. If it does not, describe the graph. See Eamples 3 5. 19. 2 + 2 + 6 + 8 + 9 = 2. 2 + 2 + 8-6 + 16 = 21. 2 + 2-4 + 12 = -4 22. 2 + 2-12 + 1 = -25 23. 4 2 + 4 2 + 4-16 - 19 = 24. 9 2 + 9 2 + 12-18 - 23 =
22 Chapter 2 Graphs and Functions 25. 2 + 2 + 2-6 + 14 = 26. 2 + 2 + 4-8 + 32 = 27. 2 + 2-6 - 6 + 18 = 28. 2 + 2 + 4 + 4 + 8 = 29. 9 2 + 9 2-6 + 6-23 = 3. 4 2 + 4 2 + 4-4 - 7 = Relating Concepts For individual or collaborative investigation (Eercises 31 36) The distance formula, the midpoint formula, and the center-radius form of the equation of a circle are closel related in the following problem. A circle has a diameter with endpoints 1-1, 32 and 15, -92. Find the center-radius form of the equation of this circle. ( 1, 3) 3 1 5 Work Eercises 31 36 in order to see the relationships among these concepts. (5, 9) 31. To find the center-radius form, we must find both the radius and the coordinates of the center. Find the coordinates of the center using the midpoint formula. (The center of the circle must be the midpoint of the diameter.) 32. There are several was to find the radius of the circle. One wa is to find the distance between the center and the point 1-1, 32. Use our result from Eercise 31 and the distance formula to find the radius. 33. Another wa to find the radius is to repeat Eercise 32, but use the point 15, -92 rather than 1-1, 32. Do this to obtain the same answer ou found in Eercise 32. 34. There is et another wa to find the radius. Because the radius is half the diameter, it can be found b finding half the length of the diameter. Using the endpoints of the diameter given in the problem, find the radius in this manner. You should once again obtain the same answer ou found in Eercise 32. 35. Using the center found in Eercise 31 and the radius found in Eercises 32 34, give the center-radius form of the equation of the circle. 36. Use the method described in Eercises 31 35 to find the center-radius form of the equation of the circle with diameter having endpoints 13, -52 and 1-7, 32. Find the center-radius form of the circle described or graphed. (See Relating Concepts Eercises 31 36.) 37. a circle having a diameter with endpoints 1-1, 22 and 111, 72 38. a circle having a diameter with endpoints 15, 42 and 1-3, -22 39. 4. (1, 4) ( 3, 1) (5, 1) (5, 5)
SECTION 2.2 Circles 23 Epicenter of an Earthquake Solve each problem. To visualize the situation, use graph paper and a pair of compasses to carefull draw the graphs of the circles. See Eample 6. 41. Suppose that receiving stations X, Y, and Z are located on a coordinate plane at the points 17, 42, 1-9, -42, and 1-3, 92, respectivel. The epicenter of an earthquake is determined to be 5 units from X, 13 units from Y, and 1 units from Z. Where on the coordinate plane is the epicenter located? 42. Suppose that receiving stations P, Q, and R are located on a coordinate plane at the points 13, 12, 15, -42, and 1-1, 42, respectivel. The epicenter of an earthquake is determined to be 25 units from P, 6 units from Q, and 221 units from R. Where on the coordinate plane is the epicenter located? 43. The locations of three receiving stations and the distances to the epicenter of an earthquake are contained in the following three equations: 1-22 2 + 1-12 2 = 25, 1 + 22 2 + 1-22 2 = 16, and 1-12 2 + 1 + 22 2 = 9. Determine the location of the epicenter. 44. The locations of three receiving stations and the distances to the epicenter of an earthquake are contained in the following three equations: 1-22 2 + 1-42 2 = 25, 1-12 2 + 1 + 32 2 = 25, and 1 + 32 2 + 1 + 62 2 = 1. Determine the location of the epicenter. Concept Check Work each of the following. 45. Find the center-radius form of the equation of a circle with center 13, 22 and tangent to the -ais. (Hint: A line tangent to a circle touches it at eactl one point.) 46. Find the equation of a circle with center at 1-4, 32, passing through the point 15, 82. Write it in center-radius form. 47. Find all points 1, 2 with = that are 4 units from 11, 32. 48. Find all points satisfing + = that are 8 units from 1-2, 32. 49. Find the coordinates of all points whose distance from 11, 2 is 21 and whose distance from 15, 42 is 21. 5. Find the equation of the circle of least radius that contains the points 11, 42 and 1-3, 22 within or on its boundar. 51. Find all values of such that the distance between 13, 2 and 1-2, 92 is 12. 52. Suppose that a circle is tangent to both aes, is in the third quadrant, and has radius 22. Find the center-radius form of its equation. 53. Find the shortest distance from the origin to the graph of the circle with equation 2-16 + 2-14 + 88 =. 54. Find the coordinates of the points of intersection of the line = 1 and the circle centered at 13, 2 with radius 2. 55. Phlash Phelps is the morning radio personalit on SiriusXM Satellite Radio s Sities on Si Decades channel. Phlash is an epert on U.S. geograph and loves traveling around the countr to strange, out-ofthe-wa locations. The photo shows Curt Gilchrist (standing) and Phlash (seated) visiting a small Arizona settlement called Nothing. (Nothing is so small that it s not named on current maps.) The sign indicates that Nothing is 5 mi from Wickenburg, AZ, 75 mi from Kingman, AZ, 15 mi from Phoeni, AZ, and 18 mi from Las Vegas, NV. Discuss how the concepts of Eample 6 can be used to locate Nothing, AZ, on a map of Arizona and southern Nevada.
24 Chapter 2 Graphs and Functions 2.3 Functions n Relations and Functions n Domain and Range n Determining Whether Relations Are Functions n Function Notation n Increasing, Decreasing, and Constant Functions Relations and Functions Recall from Section 2.1 how we described one quantit in terms of another. The letter grade ou receive in a mathematics course depends on our numerical scores. The amount ou pa (in dollars) for gas at the gas station depends on the number of gallons pumped. The dollars spent on household epenses depends on the categor. We used ordered pairs to represent these corresponding quantities. For eample, 13, $1.52 indicates that ou pa $1.5 for 3 gallons of gas. Since the amount ou pa depends on the number of gallons pumped, the amount (in dollars) is called the dependent variable, and the number of gallons pumped is called the independent variable. Generalizing, if the value of the second component depends on the value of the first component, then is the dependent variable and is the independent variable. Independent variable 1, 2 Dependent variable A set of ordered pairs such as 513, 1.52, 18, 28.2, 11, 35.26 is called a relation. A special kind of relation called a function is ver important in mathematics and its applications. Relation and Function A relation is a set of ordered pairs. A function is a relation in which, for each distinct value of the first component of the ordered pairs, there is eactl one value of the second component. Note The relation from the beginning of this section representing the number of gallons of gasoline and the corresponding cost is a function since each -value is paired with eactl one -value. Eample 1 Deciding Whether Relations Define Functions Decide whether each relation defines a function. F = 511, 22, 1-2, 42, 13, 426 G = 511, 12, 11, 22, 11, 32, 12, 326 H = 51-4, 12, 1-2, 12, 1-2, 26 Solution Relation F is a function, because for each different -value there is eactl one -value. We can show this correspondence as follows. 51, -2, 36 -values of F 52, 4, 46 -values of F
SECTION 2.3 Functions 25 As the correspondence below shows, relation G is not a function because one first component corresponds to more than one second component. 51, 26 -values of G 51, 2, 36 -values of G In relation H the last two ordered pairs have the same -value paired with two different -values ( -2 is paired with both 1 and ), so H is a relation but not a function. In a function, no two ordered pairs can have the same first component and different second components. Different -values H = 51-4, 12, 1-2, 12, 1-2, 26 Not a function Same -value n Now Tr Eercises 1 and 3. -values 1 2 3 F Figure 19 -values 2 4 F is a function. H -values -values 4 2 1 H is not a function. Relations and functions can also be epressed as a correspondence or mapping from one set to another, as shown in Figure 19 for function F and relation H from Eample 1. The arrow from 1 to 2 indicates that the ordered pair 11, 22 belongs to F each first component is paired with eactl one second component. In the mapping for relation H, which is not a function, the first component -2 is paired with two different second components, 1 and. Since relations and functions are sets of ordered pairs, we can represent them using tables and graphs. A table and graph for function F are shown in Figure 2. 1 2-2 4 3 4 ( 2, 4) Graph of F Figure 2 Finall, we can describe a relation or function using a rule that tells how to determine the dependent variable for a specific value of the independent variable. The rule ma be given in words: for instance, the dependent variable is twice the independent variable. Usuall the rule is an equation, such as the one below. (3, 4) (1, 2) Dependent variable = 2 Independent variable Note Another wa to think of a function relationship is to think of the independent variable as an input and the dependent variable as an output. This is illustrated b the input-output (function) machine for the function defined b = 2. 4 (Input ) = 2 8 (Output ) Function machine In a function, there is eactl one value of the dependent variable, the second component, for each value of the independent variable, the first component.