14. a. 12 − (− 4) = 8 ( − (− y x 2)) 3 + 4 = 8 + y x 0 15 16 3 3 –2 = 8 + y x 4 The equation of the line is: − = y y1 m( − ) x x1 b. Answers will vary. We select (5, 2) and 3 3 (11, 9). The slope of the line containing c. 6 points is: = 9 − 2 = 7 m 11 − 5 6 6 The equation of the line is: − − = y y1 m( x x1 3 –3 ) − 2 = 7 ( − y x 5) –6 − 2 = 7 − 35 y x 6 6 d. Using the LINear REGresssion program, the line of best fit is: = − 7 23 y x = 2.2 + 1.2 y x 6 6 c. 12 e. –3 6 0 15 –2 –6 = 1.1286 − 3.8619 y x d. Using the LINear REGression program, the line of best fit is: 8 16. a. e. 12 3 –3 0 15 –6 –2 –2 15. a. 6 b. Answers will vary. We select (–2, 7) and (1, 2). The slope of the line containing these points is: 3 –3 = 2 − 7 = − 5 = − 5 m 1 − (−2) 3 3
14. a. 12 0 15 –2 The equation of the line is: y − y1 = m( x − x1 ) y − (− 4) = 8 ( x − (− 2)) 3 y + 4 = 8 x + 16 3 3 y = 8 x + 4 b. Answers will vary. We select (5, 2) and (11, 9). The slope of the line containing 3 3 c. 6 points is: m = 9 − 2 = 7 11 − 5 6 The equation of the line is: y − y1 = m( x − x1 ) y − 2 = 7 ( x − 5) 6 –3 3 –6 y − 2 = 7 x − 35 6 6 y = 7 x − 23 6 6 d. Using the LINear REGresssion program, the line of best fit is: y = 2.2 x + 1.2 c. 12 e. 6 –3 0 15 –2 –6 d. Using the LINear REGression program, the line of best fit is: y = 1.1286 x − 3.8619 16. a. 8 e. 12 –3 3 0 15 –2 15. a. 6 –3 3 –6 –2 b. Answers will vary. We select (–2, 7) and (1, 2). The slope of the line containing these points is: m = 2 − 7 = − 5 = − 5 1 − (−2) 3 3
Downloadable-Solution-Manual-for-Precalculus-Concepts-Through-1.docx