Bigideas Algebra 2 Chapter 1 Answers

Chapter 1 Chapter 1 Maintaining Mathematical Proficiency (p. 1) 1. 5 23 + 7 = 5 8 + 7 ⋅ 9. y ⋅ (−4, 5) = 40 + 7 (−5, 4) (−2, 5) (−1, 4) = 47 2. 4 − 2(3 + 2 =4− 2)2 2(5)2 −6 −4 −2 x = 4 − 2(25) = 4 − 50 10. Sample answer: Consider the expression 50 ÷ 25 = −46 3. 48 ÷ 42 + 3 —5 = 48 ÷ 16 + ⋅ 3 —5 3 = 3 + —5 = 15 3 — + —5 5 18 =— 5 = 33—5 4. 50 ÷ 52 ⋅ 2 = 50 ÷ 25 ⋅ 2 = 2 ⋅2 5. ⋅ ⋅ Chapter 1 Mathematical Practices (p. 2) 1. =4 1 —2 ( 22 ⋅ 2. When the order of operations are followed (multiply and divide from left to right), the expression gives 50÷ 25 2 = 2 2 = 4. However, when they are not followed, then the result might be 50 ÷ 25 2 = 50 ÷ 50 = 1. Following the order in which transformations are given is also important. For example, translating the point (3, 2) up 3 units and then reflecting in the x-axis, the new coordinate is (3, −5). Reflecting in the x-axis and then translating up 3 units, the new coordinate is (3, 1). 4 10 −10 1 + 22 ) = —2 (4 + 22) 10 −6 6 −4 −10 1 = —2 (26) The square viewing window makes the line look steeper. = 13 2. 10 10 1 6. —1 (6 + 18) − 22 = —6 (24) − 22 6 −10 10 1 = —6 (24) − 4 −11 =0 7. 7 −2 −10 =4−4 The square viewing window makes the graph look narrower. 3. 10 2 y −6 (0, 6) (3, 6) −10 6 10 4 −6 −10 2 −4 −2 The square viewing window makes the parabola look wider. (0, 0) (3, 0) x 4. 10 −10 8. 5 10 y −1 (0, 4) −10 (−2, 2) −4 −2 2 (1, 3) 2 8 −1 The standard viewing window makes the curve look flatter. 4 x Algebra 2 Worked-Out Solutions 1 Chapter 1 5. 10 −10 2. Sample answer: Most of the parent functions go through 2 −6 the origin. Most are either symmetric about the y-axis or the origin. 6 10 3. a. The equation for the graph is y = ∣ x ∣. — b. The equation for the graph is y = √x . −6 −10 c. The equation for the graph is y = 1. The square viewing window makes the curve look wider. d. The equation for the graph is y = 2x. 6. 10 e. The equation for the graph is y = x 3. 4 f. The equation for the graph is y = x. −10 −6 10 6 1 x g. The equation for the graph is y = —. h. The equation for the graph is y = x 2. −4 −10 The square viewing window makes the curve look wider. Height Width 8 − (−2) 8 − (−8) 10 16 5 8 7. — = — = — = — 1.1 Monitoring Progress (pp. 4 –7) 1. The function g belongs to the family of quadratics. The graph is translated right and is wider than the graph of the parent quadratic function. The domain of each function is all real numbers and the range of each function is y ≥ 0. no; The height-to-width ratio is 5 to 8. 2 Height 8 − (−2) 10 8. — = — = — = — 8 − (−7) 15 3 Width yes; The height-to-width ratio is 2 to 3. 2. 2 Height 8 − (−2) 10 9. — = — = — = — Width 9 − (−6) 15 y 6 4 g(x) = x + 3 3 −6 Height 3 − (−3) 6 3 10. — = — = — = — Width 2 − (−2) 4 −4 −2 2 no; The height-to-width ratio is 3 to 2. −6 So, the graph of g(x) = x + 3 is a translation 3 units up of the parent linear function. yes; The height-to-width ratio is 2 to 3. 3 − (−3) 4 − (−4) 8 4 6 x −4 Height 3 − (−3) 6 2 11. — = — = — = — 5 − (−4) 9 3 Width Width 4 −2 2 6 3 Height 12. — = — = — = — y=x 2 yes; The height-to-width ratio is 2 to 3. 3. 8 y no; The height-to-width ratio is 3 to 4. 1.1 Explorations (p. 3) 1. a. absolute value; The graph is a “V” shape. 4 y = x2 2 b. square root; The domain is x ≥ 0 and the range is y ≥ 0. c. constant; The y-value is the same for all x-values. d. exponential; The function grows faster as x increases. e. cubic; The domain and range are both all real numbers, and the function grows quickly indicating an exponent. f. linear; The function is always increasing at the same rate. g. reciprocal; The function has a vertical asymptote at the y-axis, and the y-values become smaller as x increases. h. quadratic; The graph is a parabola. 2 Algebra 2 Worked-Out Solutions h(x) = (x − 2)2 −4 −2 2 4 x −2 So, the graph of h(x) = (x − 2)2 is a translation 2 units right of the parent quadratic function. Chapter 1 4. 8. y 11 y=x 4 2 −3 y= x −4 −2 2 4 The graph of h(x) = − —41 x + 5 is a reflection in the x-axis n(x) = − x −2 followed by a vertical shrink and a translation 5 units up of the parent linear function. −4 So, the graph of n(x) = ∣x ∣ is a reflection in the x-axis of the parent absolute value function. 5. 9. −5 −2 2 4 9 2 −3 d(x) = 3(x − 5) − 1 y=x 2 The graph of h(x) = 3(x − 5)2 − 1 is a translation 5 units right followed by a vertical stretch and then a translation 1 unit down of the parent quadratic function. x g(x) = 3x 10. So, the graph of g(x) = 3x is a vertical stretch of the parent linear function. 6. 11 y = x2 y 4 −4 11 h(x) = −14 x + 5 −3 x y y 16 12 8 4 10 0 8 h(x) = 3 2 2x 0 10 20 30 40 50 x The data appear to lie on a line. So, you can model the data with a linear function. The graph shows that the tank will be empty after 50 minutes. 6 4 y = x2 1.1 Exercises (pp. 8 –10) 2 Vocabulary and Core Concept Check −4 −2 So, the graph of h(x) = quadratic function. 7. 1. The function f (x) = x2 is the parent function of f (x) = 2x2 − 3. 4 x 2 3 —2 x 2 is a vertical stretch of the parent 2. The question that is different is “What are the vertices of the figure after a translation of 6 units up, followed by a reflection in the x-axis?” The coordinates are: (1, −2) → (1, −4) y 8 (3, −2) → (3, −4) y= x 6 (3, −4) → (3, −2) 4 c(x) = 0.2 x 2 (1, −4) → (1, −2) The other three questions result in the coordinates: (1, −2) → (3, 4) −6 −4 −2 2 4 6 x So, the graph of h(x) = 0.2∣ x ∣ is a vertical shrink of the parent absolute value function. (3, −2) → (5, 4) (3, −4) → (5, 2) (1, −4) → (3, 2) Algebra 2 Worked-Out Solutions 3 Chapter 1 Monitoring Progress and Modeling with Mathematics 10. y 3. The function f belongs to the family of absolute value functions. The graph of f (x) = 2∣x + 2 ∣ − 8 is a horizontal translation 2 units left followed by a vertical stretch and a vertical translation 8 units down of the parent absolute value function. The domain of each function is all real numbers, but the range of f is y ≥ −8, and the range of the parent function is y ≥ 0. −4 So, the graph of f (x) = x − 6 is a vertical translation 6 units down of the parent linear function. 11. y 8 6 4 graph of f (x) = 5x − 2 is a vertical stretch followed by a vertical translation 2 units down of the parent linear function. The domain and range of each function is all real numbers. y = x2 Temperature (°F) 2 −2 2 f(x) = x − 1 So, the graph of f (x) = x2 − 1 is a vertical translation 1 unit down of the parent quadratic function. 12. y 8 y 60 y = x2 6 56 52 48 4 2 y = 2x + 43 44 40 0 0 2 4 6 −6 x −4 −2 h(x) = (x + 4) Time (hours) The type of function that can model the data is a linear function. The temperature is increasing by the same amount at each interval. 8. The type of function that is used as a model is a quadratic 2 2 x −2 So, the graph of h(x) = (x + 4)2 is a horizontal translation 4 units left of the parent quadratic function. 13. y y = x 4 function. 2 9. y g(x) = x − 5 g(x) = x + 4 4 −4 y=x 2 −2 2 4 −2 2 4 6 8 10 x So, the graph of g(x) = ∣x − 5 ∣ is a horizontal translation 5units right of the parent absolute value function. x −2 −4 So, the graph of g(x) = x + 4 is a vertical translation 4 units up of the parent linear function. 4 x 2 −2 6. The function f belongs to the family of constant functions. 7. x f(x) = x − 6 5. The function f belongs to the family of linear functions. The The graph of f (x) = 3 is a vertical translation 2 units up of the parent constant function. The domain of each function is all real numbers, but the range of f is y = 3, and the range of the parent function is y = 1. 4 −4 4. The function f belongs to the family of quadratic functions. The graph of f (x) = −2x2 + 3 is a reflection in the x-axis followed by a vertical stretch and a vertical translation 3 units up of the parent absolute value function. The domain of each function is all real numbers, but the range of f is y ≤ 3, and the range of the parent quadratic function is y ≥ 0. y=x 4 Algebra 2 Worked-Out Solutions Chapter 1 14. y 10 19. f(x) = 4 + x y y=x 4 8 2 f(x) = 13 x 6 −4 2 4 y = x 2 −6 −4 4 x −2 −2 2 4 −4 6 x 1 So, the graph of f (x) = —3 x is a vertical shrink of the parent linear function. So, the graph of f (x) = 4 + ∣ x∣ is a vertical translation 4units up of the parent absolute value function. 20. 15. 8 y = x2 y y 4 y=x 2 4 −4 −4 x 4 h(x) = −x −2 2 x 4 2 g(x) = 4x −8 −5 So, the graph of h(x) = −x2 is a reflection in the x-axis of the parent quadratic function. 16. So, the graph of g(x) = 4x is a vertical stretch of the parent linear function. y 21. 4 y=x 2 −4 8 y 6 4 −2 2 4 x y = x2 −2 2 f(x) = 2x −4 −2 g(x) = −x So, the graph of g(x) = −x is a reflection in the x-axis of the parent linear function. 17. 4 y f(x) = 3 −2 2 4 22. 8 6 4 x 2 So, the graph of f (x) = 3 is a vertical translation 2 units up of the parent constant function. 2 −4 −2 y 4 h(x) = 13 x −6 −4 −2 2 4 2 6 x 1 So, the graph of h(x) = 3— x2 is a vertical shrink of the parent quadratic function. y=1 2 −3 y y = x2 −2 18. 2 x So, the graph of f (x) = 2x2 is a vertical stretch of the parent quadratic function. y=1 2 −4 2 x f(x) = −2 So, the graph of f (x) = −2 is a vertical translation 3 units down of the parent constant function. Algebra 2 Worked-Out Solutions 5 Chapter 1 23. y 28. 8 y=x 4 h(x) = −x + 5 2 2 −4 4 −4 So, the graph of h(x) = −x + 5 is a reflection in the x-axis followed by a vertical translation 5 units up of the parent linear function. −4 3 So, the graph of h(x) = —4 x is a vertical shrink of the parent linear function. 29. g(x) = y = |x| 9 h(x) = −3|x| − 1 4 3x −6 So, the graph of h(x) = −3∣ x∣ − 1 is a reflection in the x-axis followed by a vertical stretch and then a vertical translation 1 unit down of the parent absolute value function. y=x 2 −2 6 −9 y −4 11 x −2 24. y=x −7 h(x) = 43 x 2 4 x −2 30. 4 −4 y = |x| 6 −6 4 So, the graph of g(x) = —3 x is a vertical stretch of the parent linear function. f(x) = 34 |x| + 1 −4 25. y 3 So, the graph of f (x) = —4 ∣x ∣ + 1 is a vertical shrink followed by a vertical translation 1 unit up of the parent absolute value function. 4 h(x) = 3x −4 y = x −2 2 4 31. 10 x y = x2 So, the graph of h(x) = 3 ∣ x∣ is a vertical stretch of the parent absolute value function. 26. y g(x) = 12 x2 − 6 y = x 1 2 −2 f(x) = 12x 2 −7 So, the graph of g(x) = —2 x2 − 6 is a vertical shrink followed by a vertical translation 6 units down of the parent quadratic function. 4 −4 12 −12 4 x So, the graph of f (x) = ∣ x∣ is a vertical shrink of the parent absolute value function. 32. 8 1 — 2 y = x2 9 −9 f(x) = 4x2 − 3 27. 5 f(x) = 3x + 2 −4 y=x −7 7 −5 So, the graph of f (x) = 3x + 2 is a vertical stretch followed by a vertical translation 2 units up of the parent linear function. 6 Algebra 2 Worked-Out Solutions So, the graph of f (x) = 4x2 − 3 is a vertical stretch followed by a vertical translation 3 units down of the parent quadratic function. Chapter 1 33. 44. Function f is in the family of quadratic functions. The domain 7 is all real numbers and the range is y ≤ 6. y = x2 −10 10 f(x) = −(x + 3)2 + 41 45. The type of function that can model the data is an absolute value function because the data are linear and there are positive speeds for the positive and negative displacements. The speed of the car 20 yards past the intersection is estimated to be 8 miles per hour. −7 1 So, the graph of f (x) = −(x + 3)2 + —4 is a translation 3 units left followed by a reflection in the x-axis and then a vertical 1 translation —4 unit up of the parent quadratic function. 34. 46. Sample answer: y 3 4 y = |x| 2 y = x2 4 −4 g(x) = −|x − 1| − 21 −4 −2 2 −3 4 x y = −x 2 − 1 So, the graph of g(x) = − ∣x − 1 ∣ − —2 is a translation 1 unit 1 −4 right followed by a reflection in the x-axis and then a vertical 1 translation 2— unit down of the parent absolute value function. The transformation of the parent quadratic function is a reflection in the x-axis followed by a vertical translation 1 unit down. 35. The error is there is no vertical shrink of the parent quadratic function. The graph is a reflection in the x-axis followed by a vertical stretch of the parent quadratic function. y 47. 4 36. The error is that a translation to the right of 3 units is represented by a subtraction of 3 in the expression for the function. The graph is a translation 3 units right of the parent absolute function, so the function is f (x) = ∣x − 3 ∣. f(x) = x − 4 2 −2 2 4 6 x −2 37. (x, y) → (x, y − 2) A(2, 1) → A′(2, −1) B(−1, −2) → B′(−1, −4) C(2, −3) 38. → C′(2, −5) (x, y) → (x, −y) A(−1, 3) → A′(−1, −3) B(1, 3) → B′(1, −3) −4 g(x) = x − 4 The graphs are not equivalent. The graph of f is a translation to the right, whereas the graph of g is a translation down of the parent absolute value function. 1 48. a. The graph of g is a vertical shrink because the factor is —2 . b. Multiply the y-coordinate of each point of f by −2. C(−1, 1) → C′(−1, −1) 49. Your friend is correct. Shifting the parent linear function down D(−3, 1) → D′(−3, −1) 2 units will create the same graph as shifting it 2 units right. 39. Function g is in the family of absolute value functions. The domain is all real numbers and the range is y ≥ −1. 40. Function h is in the family of absolute value functions. The domain is all real numbers and the range is y ≥ 2. 41. Function g is in the family of linear functions. The domain is all real numbers and the range is all real numbers. 42. Function f is in the family of linear functions. The domain is all real numbers and the range is all real numbers. 43. Function f is in the family of quadratic functions. The domain is all real numbers and the range is y ≥ −2. 50. A linear function can be used to model the distance the swimmer travels. If the person has a 10-meter head start, the type of transformation is a vertical translation. The graph will be shifted up 10 units to represent the head start. 51. a. The type of function modeled by the equation is a quadratic function. b. The value for t when the ball is released from the hand is zero, because 0 seconds have passed. c. The ball is 5.2 feet above the ground when it is released from the hand, because this corresponds to t = 0. 52. An absolute value function can be used to model the data. In this situation, the x-intercept represents the number of hours that have passed at the moment the computer has 0 battery life remaining. Algebra 2 Worked-Out Solutions 7 Chapter 1 53. a. The function f (x) = 2 ∣ x ∣ − 3 is a vertical translation of the parent function. The graph will be shifted 3 units down. b. The function f (x) = (x − 8)2 is a horizontal translation of the parent function. The graph will be shifted 8 units right. c. The function f (x) = ∣ x + 2 ∣ + 4 is both a horizontal translation and vertical translation of the parent function. The graph will be shifted 2 units left and 4 units up. d. The function f (x) = 4x 2 is neither a horizontal translation nor vertical translation of the parent function. The graph will have a vertical stretch. 54. a. f (x) = 3x + 1; The graph will intersect the x-axis at 1 x = −—3 . b. f (x) = ∣ 2x − 6 ∣ − 2; The graph will intersect the x-axis at x = 2 and x = 4. c. f (x) = −1 ⋅ x2 + 1; The graph will intersect the x-axis at x = 1 and x = −1. d. f (x) = 0; This is the x-axis. Maintaining Mathematical Proficiency 55. f (x) = ∣ x + 2 ∣ ? −3 = ∣1 + 2 ∣ ? −3 = ∣3 ∣ −3 ≠ 3 So, (1, −3) is not a solution. 56. f (x) = ∣ x ∣ − 3 ? −5 = ∣−2 ∣ − 3 ? −5 = 2 − 3 −5 ≠ −1 So, (−2, −5) is not a solution. 57. f (x) = x − 3 ? 2=5−3 2=2✓ So, (5, 2) is a solution. 58. f (x) = x − 4 ? 8 = 12 − 4 8=8✓ So, (12, 8) is a solution. 59. To find the x-intercept let y = 0, then solve for x. y=x 0=x x=0 To find the y-intercept let x = 0, then solve for y. y=x y=0 So, the x-intercept is (0, 0) and the y-intercept is (0, 0). 8 Algebra 2 Worked-Out Solutions 60. To find the x-intercept let y = 0, then solve for x. y=x+2 0=x+2 x = −2 To find the y-intercept let x = 0, then solve for y. y=x+2 y=0+2 y=2 So, the x-intercept is (−2, 0) and the y-intercept is (0, 2). 61. To find the x-intercept let y = 0, then solve for x. 3x + y = 1 3x + 0 = 1 3x = 1 x = —31 To find the y-intercept let x = 0, then solve for y. 3x + y = 1 3(0) + y = 1 y=1 ( ) So, the x-intercept is 3—1, 0 and the y-intercept is (0, 1). 62. To find the x-intercept let y = 0, then solve for x. x − 2y = 8 x − 2(0) = 8 x=8 To find the y-intercept let x = 0, then solve for y. x − 2y = 8 0 − 2y = 8 −2y = 8 y = −4 So, the x-intercept is (8, 0) and the y-intercept is (0, −4). 1.2 Explorations (p. 11) 1. The graph of y = ∣ x∣ + k is a vertical translation of the parent function f (x) = ∣x ∣. If k is positive, the graph is shifted up k units. If k is negative, the graph is shifted down k units. 2. The graph of y = ∣ x + h∣ is a horizontal translation of the parent function f (x) = ∣x ∣. If h is positive, the graph is shifted right h units. If h is negative, the graph is shifted left h units. 3. The graph of y = −∣ x∣ is a reflection in the x-axis of the parent function f (x) = ∣ x∣. Chapter 1 4. The graph of y = f (x) + k is a vertical translation of the parent function f ; the graph of y = f (x + h) is a horizontal translation of the parent function f ; and the graph of y = −f (x) is a reflection in the x-axis of the parent function f. — 5. a. The graph of y = √x − 4 is a vertical translation 4units down of the parent function f (x) = √— x. — b. The graph of y = √x + 4 is a horizontal translation 4units left of the parent function f (x) = √— x. — c. The graph of y = −√x is a reflection in the x-axis of the parent function f (x) = √—x . d. The graph of y = x 2 + 1 is a vertical translation 1 unit up of the parent function f (x) = x2. e. The graph of y = (x − 1)2 is a horizontal translation 1unit right of the parent function f (x) = x2. f. The graph of y = −x 2 is a reflection in the x-axis of the parent function f (x) = x2. 6. A vertical shrink by a factor of —1 multiplies each output value 3 by —31. g(x) = —31 f (x) = —31 ( ∣ x∣ − 3 ) = —31 ∣ x ∣ − 1 The transformed function is g(x) = —31 ∣x ∣ − 1. 7. A translation 6 units down is a vertical translation that adds −6 to each output value, then a reflection in the x-axis changes the sign. g(x) = −( f (x)− 6) = −( ∣ x∣ − 6 ) = −∣ x ∣ + 6 The transformed function is g(x) = −∣ x ∣ + 6. 8. The profit function is 1.2 Monitoring Progress (pp. 13–15) 1. A translation 5 units up is a vertical translation that adds 5 to each output value. ⋅ p(x) = 0.9 3x − 50 = 2.7x − 50. To find the profit for 100 downloads, evaluate p when x = 100. g(x) = f (x) + 5 p(100) = 2.7(100) − 50 = 220 = 3x + 5 The translated function is g(x) = 3x + 5. 2. A translation 4 units to the right is a horizontal translation that subtracts 4 from each input value. g(x) = f (x − 4) = ∣x − 4 ∣ − 3 The translated function is g(x) = ∣x − 4 ∣ − 3. 3. A reflection in the x-axis changes the sign of each output value. g(x) = −f (x) = −(−∣ x + 2 ∣ − 1) Your profit is $220 for 100 downloads. 1.2 Exercises (pp. 16–18) Vocabulary and Core Concept Check 1. The function g(x) = ∣5x ∣ − 4 is a horizontal shrink of the function f (x) = ∣ x ∣ − 4. 2. The transformation “Stretch the graph of f (x) = x + 3 vertically by a factor of 2” does not belong with the other three because the other three produce the same function, f (x) = 2x + 5. Monitoring Progress and Modeling with Mathematics = ∣x + 2 ∣ + 1 The reflected function is g(x) = ∣x + 2 ∣ + 1. 4. A reflection in the y-axis changes the sign of each input value. 3. A translation 4 units left is a horizontal translation that subtracts −4 from each input value. g(x) = f (x + 4) = (x + 4) − 5 g(x) = f (−x) =x−1 = —21 (−x) + 1 1 = −—2 x + 1 The transformed function is g(x) = x − 1. 1 The reflected function is g(x) = −—2 x + 1. 5. A horizontal stretch by a factor of 2 multiplies each input value by 12— . ( ) = 4 ( — x )+ 2 g(x) = f —21 x 1 2 4. A translation 2 units right is a horizontal translation that subtracts 2 from each input value. g(x) = f (x − 2) = (x − 2) + 2 =x The transformed function is g(x) = x. = 2x + 2 The transformed function is g(x) = 2x + 2. Algebra 2 Worked-Out Solutions 9 Chapter 1 5. A translation 2 units down is a vertical translation that adds −2 to each output value. g(x) = f (x) − 2 12. A reflection in the x-axis changes the sign of each output value. g(x) = −f (x) (1 = − —2 x − 3 = (∣ 4x + 3 ∣ + 2 ) − 2 = ∣4x + 3 ∣ The transformed function is g(x) = ∣ 4x + 3 ∣. 6. A translation 6 units up is a vertical translation that adds 6 to each output value. g(x) = f (x) + 6 = (2x − 9) + 6 1 The transformed function is g(x) = −—2 x + 3. 13. A reflection in the y-axis changes the sign of each input value. g(x) = f (−x) = ∣ 6(−x)∣ − 2 = 2x − 3 = ∣ −6x ∣ − 2 = ∣ 6x∣ − 2 The transformed function is g(x) = 2x − 3. 7. A translation 3 units right is a horizontal translation that ) 1 = −—2 x + 3 The transformed function is g(x) = ∣6x ∣ − 2. subtracts 3 from each input value. g(x) = f (x − 3) = 4 − ∣ (x − 3) + 1∣ = 4 − ∣ x − 2∣ 14. A reflection in the y-axis changes the sign of each input value. g(x) = f (−x) = ∣ 2(−x) − 1∣ + 3 The transformed function is g(x) = 4 − ∣ x − 2 ∣. 8. A translation 1 unit up is a vertical translation that adds 1 to = ∣ −2x − 1 ∣ + 3 The transformed function is g(x) = ∣−2x − 1 ∣ + 3. each output value. g(x) = f (x) + 1 = (∣ 4x ∣ + 5) + 1 = ∣4x ∣ + 6 15. A reflection in the y-axis changes the sign of each input value. g(x) = f (−x) = −3 + ∣ (−x) − 11∣ The transformed function is g(x) = ∣ 4x ∣ + 6. 9. A horizontal translation 3 units right or a vertical translation = −3 + ∣ −x − 11∣ The transformed function is g(x) = −3 + ∣ −x − 11 ∣ . 3 units up will produce the function g from the function f. 16. A reflection in the y-axis changes the sign of each input 10. The transformation needed to model the situation using the value. function f is a vertical translation. The new model for the net income is g(x) = 4000x − 12,000. To find how many weeks it will take to pay off the extra expenses, set g equal to 0 and solve for x. g(x) = f (−x) 4000x − 12,000 = 0 The transformed function is g(x) = x + 1. 4000x = 12,000 x=3 It will take 3 weeks to pay off the extra expenses. = −(−x) + 1 =x+1 17. A vertical stretch by a factor of 5 multiplies each output value by 5. g(x) = 5f (x) = 5(x + 2) 11. A reflection in the x-axis changes the sign of each output value. g(x) = −f (x) = −(−5x + 2) = 5x − 2 The transformed function is g(x) = 5x − 2 . = 5x + 10 The transformed function is g(x) = 5x + 10. 18. A vertical shrink by a factor of 1— multiplies each output value 2 by 1—2. 1 g(x) = —2 f (x) 1 = 2— (2x + 6) =x+3 The transformed function is g(x) = x + 3. 10 Algebra 2 Worked-Out Solutions Chapter 1 19. A horizontal shrink by a factor of 1— multiplies each input 2 value by 2. g(x) = f (2x) ) 1 g(x) = f 2— (x − 2) = ∣4x ∣ + 4 The transformed function is g(x) = ∣4x ∣ + 4. 20. A horizontal stretch by a factor of 4 multiplies each input value by —41 . ( ) = ∣( — x) + 3∣ g(x) = f —41 x ∣( ) ∣ 1 = ∣ —2 x − 2 ∣ 1 = 2—x − 2 ∣ ∣ 1 The transformed function is g(x) = —2 x − 2 . 30. A reflection in the y-axis changes the sign of each input value and a translation 3 units right is a horizontal translation that subtracts 3 from each input value. 1 4 g(x) = f ((−x) − 3 ) = ∣—14 x + 3∣ The transformed function is g(x) = ∣ 1 —4x + 3 ∣. 21. A vertical shrink by a factor of —1 multiplies each output value 2 by 1—2 . ( subtracts 2 from each input value and a horizontal stretch by a factor of 2 multiplies each input value by —21. ( = ∣2(2x) ∣ + 4 g(x) = —21 f (x) = —21 −2 29. A translation 2 units right is a horizontal translation that = ∣x − 3 ∣ The transformed function is g(x) = ∣x − 3 ∣. 31. A reflection in the x-axis changes the sign of each output value and a translation 8 units down is a vertical translation that adds −8 to the output value. g(x) = −f (x) − 8 = − ∣ x∣ − 8 ∣ x − 4 ∣ + 2) = − ∣x − 4 ∣ + 1 The transformed function is g(x) = −∣ x − 4 ∣ + 1. 22. A horizontal stretch by a factor of —1 multiplies each input The transformed function is g(x) = −∣ x ∣ − 8. 32. A translation 3 units down is a vertical translation that adds value by 3. −3 to each output value and a translation 1 unit right is a horizontal translation that subtracts 1 from each input value. g(x) = f (3x) g(x) = f (x − 1) − 3 3 = 6 − (3x) = 6 − 3x The transformed function is g(x) = 6 − 3x. 23. C; There is a horizontal translation to the left. = ∣ x − 1∣ − 3 The transformed function is g(x) = ∣x − 1 ∣ − 3. 33. The error is that 3 was added rather than subtracted to represent the translation 3 units right. The correct function is g(x) = ∣ x − 3∣ + 2. 24. A; The graph has been stretched vertically. 25. D; There is a vertical translation up. 26. B; The graph has been shrunk horizontally. 27. A vertical stretch by a factor of 2 multiplies each output value by 2 and a translation 1 unit is a vertical translation that adds 1 to each output value. g(x) = 2f (x) + 1 = 2(x) + 1 = 2x + 1 The transformed function is g(x) = 2x + 1. 28. A translation 3 units down is a vertical translation that adds −3 to each output value and a vertical shrink by a factor of 1—3 multiplies each output value by —31. g(x) = —31 ( f(x) − 3) = —31 (x − 3) = —31 x − 1 34. A vertical stretch is given by y = af(x), not y = f (ax). The correct expression for the function is g(x) = 5(x − 6) = 5x − 30. 35. Your friend is incorrect. The order in which transformations of functions are performed is important. For example, the functions f (x) = −x + 6 and g(x) = −(x + 6) both represent a reflection in the x-axis and a horizontal translation by 6, but the orders in which the transformations are performed are different. 36. The transformation is a horizontal shrink by a factor of —1. By 2 using this transformation, the sales for 2010 would be cut in half according to the model. 37. The transformation is a horizontal translation 6 units left. The area of the shaded triangle is 1 1 A = —2bh = —2(6)(3) = 9 square units. 38. The transformation is a vertical translation 5 units up. The area 1 1 of the shaded triangle is A = —2bh = —2 (6)(3) = 9 square units. The transformed function is g(x) = —31 x − 1. Algebra 2 Worked-Out Solutions 11 Chapter 1 39. The transformation is a reflection in the x-axis. The area of 1 1 the shaded triangle is A =2— bh = 2— (8)(4) = 16 square units. 45. To have a reflection in the x-axis followed by a translation 1 unit left and 1 unit up, a = −2, b = 1, and c = 0. The reflection will change the sign of the 2 in front of the absolute value, the translation to the left will introduce a −1 on the inside of the absolute value, and the translation up will then have the 0 added to the absolute value. 40. The transformation is a reflection in the y-axis. The area of 1 1 the shaded triangle is A =—2 bh = —2 (10)(5) = 25 square units. 41. a. For a vertical translation, find the difference of the y-intercepts, c − b. So, the expression is f (x) + (c − b). b. For a horizontal translation, find the difference of the x-intercepts. ( ) )) ( 46. f (x) = x + 4; x = 3 f (3) = (3) + 4 = 3 + 4 = 7 c−b b c −— − − — = − — m m m So, the expression is c−b c−b f x − −— + b = f x + — + b. m m ( ( Maintaining Mathematical Proficiency 47. f (x) = 4x − 1; x = −1 f (−1) = 4(−1) − 1 ) = −4 − 1 = −5 42. a. Reflecting the graph of f in the y-axis changes the sign of the slope and the x-intercept, and the y-intercept does not change. 48. f (x) = −x + 3; x = 5 f (5) = −(5) + 3 b. Shrinking the graph of f vertically by a factor of —1 3 = −5 + 3 reduces the slope and y-intercept by a factor of —31, and the x-intercept does not change. c. Stretching the graph of f horizontally by a factor of 2 reduces the slope by —21, doubles the value of the = −2 49. f (x) = −2x − 2; x = −1 x-intercept, and the y-intercept does not change. f (−1) = −2(−1) − 2 43. The order of the transformations is a vertical stretch by a =2−2 factor of 4 followed by a vertical translation 2 units down, and then a reflection in the x-axis. A 4 was multiplied in only one place, so that was performed first. The vertical translation subtracts a 2, but it is added in the expression, so this had to take place second so that the reflection would cause the change in the signs. =0 50. 12 8 44. The shift to the right means that you started later than the 4 day you planned to. For example, you may have started on June 15, which is 14 days later. The function representing this change in plans is d = 72∣ x − 44∣. 51. d 2100 16 y 4 8 12 4 8 12 16 x y 25 20 15 10 5 1800 1500 1200 x 1.1–1.2 What Did You Learn? (p. 19) 900 1. After graphing the data, the shape indicates that the absolute value function models the data. 600 2. The number of weeks is between 3 and 4, so the value should 300 be rounded up to 4 weeks, because 3 is not enough. 10 20 30 40 50 60 70 x 1.1–1.2 Quiz (p. 20) 1. The function belongs to the family of linear functions. The 1 1 graph of g(x) = —3 x − 1 is a vertical shrink by a factor of —3 followed by a vertical translation 1 unit down of the parent linear function. 12 Algebra 2 Worked-Out Solutions Chapter 1 2. The function belongs to the family of quadratic functions 8. The graph of g(x) = 2(x + 1)2 is a horizontal translation of 1 unit left followed by a vertical stretch by a factor of 2 of the parent quadratic function. y 10 y = x2 8 6 3. The function belongs to the family of absolute value functions. The graph of g(x) = ∣ x + 1 ∣ − 2 is a horizontal translation 1 unit left followed by a vertical translation 2 units down of the parent absolute value function. 4. 4 2 f(x) = 41x + 1 2 −6 y −4 −2 4 2 4 6 x −2 f(x) = 32 2 −4 2 y=1 −2 The graph of f (x) = —41 x2 + 1 is a vertical shrink by a factor of 1—4 followed by a vertical translation 1 unit up of the parent quadratic function. x 3 1 The graph of f (x) = —2 is a vertical translation —2 unit up of the parent constant function. 5. 9. y 6 4 y 4 −6 −4 y=x 2 y=x 2 −2 2 −4 −2 2 4 6 x −2 x 4 f(x) = −12 x −4 f(x) = 3x −4 −6 The graph of f (x) = 3x is a vertical stretch by a factor of 3 of the parent linear function. 1 The graph of f (x) = − —2x − 4 is a vertical shrink by a factor of —21 followed by a reflection in the x-axis and then a vertical translation 4 units down of the parent linear function. 6. y 10. A translation 3 units up is a vertical translation that adds 3 to each output value. y = x2 2 f(x) = 2(x − 1) −4 −2 2 4 2 6 x The graph of f (x) = 2(x − is a vertical stretch by a factor of 2 followed by a horizontal translation of 1 unit right of the parent quadratic function. 1)2 8 −4 11. A vertical shrink by a factor of —1 multiplies each output value 2 by 12— . = —21 (−3∣x − 4 ∣ ) = −—2 ∣ x − 4∣ 3 y = x −8 The transformed function is g(x) = 2x + 4. g(x) = —21 f (x) y 4 −12 = (2x + 1) + 3 = 2x + 4 −2 7. g(x) = f (x) + 3 4 8 12 x −4 The transformed function is g(x) = − —2 ∣ x − 4 ∣. 3 12. A reflection in the x-axis changes the sign of each output value. f(x) = −x + 2 − 7 −12 g(x) = −f (x) = −3∣ x + 5∣ The transformed function is g(x) = −3∣ x + 5 ∣. The graph of f (x) = −∣ x + 2∣ − 7 is a reflection in the x-axis followed by a horizontal translation 2 units left and a vertical translation 7 units down of the parent absolute value function. Algebra 2 Worked-Out Solutions 13 Chapter 1 13. A translation 4 units left is a horizontal translation that adds the discount of $30 relates to a vertical translation 30 units down. The function for the amount that senior citizens pay is given by g(x) = 1—2(20x + 80) − 30 = 10x + 10. So, if a senior citizen camped for 3 days the cost would be g(3) = 10(3) + 10 = $40. g(x) = f (x + 4) 2 = —3 (x + 4) − —3 1 = = 1 —x 3 1 —3 x + 19. Half the price relates to a vertical shrink by a factor of 1— and 2 4 to each input value. 2 4 —3 − —3 2 —3 + 1 2 The transformed function is g(x) = —3x + 3—. 1.3 Explorations (p. 21) 1. a. A linear function that models the value V of the copier, where t is the number of years after it was purchased, is V(t) = −1250t + 12,000. 14. A translation 2 units down is a vertical translation that subtracts 2 from each output value and a horizontal shrink by a factor of —32 multiplies each input value by —23. b. y 12,000 ( ) 3 g(x) = f —2 x − 2 − 2. subtracts 9 from each output value and a reflection in the y-axis changes the sign of each input value. = −x − 9 6000 4000 0 The transformed function is g(x) = −x − 9. 0 2 4 6 8 10 x Time (years) 16. A reflection in the x-axis changes the sign of each output value, a vertical stretch by a factor of 4 multiplies each output value by 4, a translation 7 units down adds −7 to each output value, and a translation 1 unit right subtracts 1 from each input value. The depreciation is called straight line depreciation because the value decreases, or depreciates, at a constant rate. c. The slope represents the value the copier decreases each year, which is $1250. 2. a. The corresponding graph is B. The slope, −20, is the g(x) = 4( −f (x − 1) )− 7 = −4 ∣x − 1 ∣ − 7 The transformed function is g(x) = −4∣x − 1 ∣ − 7. 17. A translation 1 unit down is a vertical translation that subtracts 1 from each output value, a translation 2 units left is a horizontal translation that adds 2 to each input value, and a vertical shrink by a factor of 1—2 multiplies each output value by —21. 1 g(x) = —2[ f (x + 2) − 1 ] amount the loan is reduced per week and the y-intercept is the original amount of the loan, 200. b. The corresponding graph is C. The slope, 2, is the amount earned per unit produced per hour and the y-intercept is the base amount earned per hour, 12.5. c. The corresponding graph is A. The slope, 0.565, is the amount paid for each mile driven and the y-intercept is the amount paid per day for food, 30. d. The corresponding graph is D. The slope, −100, is the value the computer decreased per year and the y-intercept is the amount paid when the computer was purchased, 750. 1 1 = —2∣ x + 2 ∣ − —2 The transformed function is g(x) = ∣ x + 2 ∣ − 1 —2 1 —2. 3. The y-intercept of a linear function represents the original or base amount, and the slope represents the change over time. y 10,000 7,500 5,000 2,500 0 2 4 6 8 x The data appear to lie on a straight line. So, a linear function can be used to model the data. The estimated mileage after 1 year is 13,800 miles. 14 8000 2000 g(x) = f (−x)−9 0 Value (dollars) The transformed function is g(x) = 3 —x 2 15. A translation 9 units down is a vertical translation that 18. V(t) = 12,000 − 1250t 10,000 3 = —2x − 2 Algebra 2 Worked-Out Solutions Chapter 1 five-day period a. 1 2 3 4 5 6 A Day 0 1 2 3 4 B Inches of snow 5 4 3 2 1 Inches of snow S(t) = 5 − t 40 10 20 30 x The data show a linear relationship. Use the points (22, 130) and (30, 154) to write an equation of the line. y2 − y1 154 − 130 24 m=— = —=3 x2 − x1 = — 30 − 22 8 y − 130 = 3(x − 22) 2 y − 130 = 3x − 66 2 y = 3x + 64 4 t Days 1.3 Monitoring Progress (pp. 22– 25) y2 − y1 3 18 − 15 10 0 − 10 2 1 and the y-intercept is 18. So the equation of the line is 3 y = −— x + 18. The balance decreases $300 per payment 10 and the initial amount is $18,000. The amount remaining 3 after 36 months is y = − —(36) + 18 = 7.2, or $7200. 10 — 1. The slope of the line is m = — x − x =— = − 2. In this situation, the rental fee is $500 and the cost for the 140 students is $1400. An equation that models the cost for using Maple Ridge is y = 10x + 500, where y is the cost and x is the number of students. So, Maple Ridge will always cost less than Sunview Resort because the rental cost is less for Maple Ridge with the same cost per student. Maple Ridge will cost less if there are more than 100 students when compared to the cost of Lakeside Inn. So, an equation of a line is y = 3x + 64. Use this equation to estimate the height of the woman. y = 3(40) + 64 = 184 The approximate height of a woman with a 40-centimeter humerus is 184 centimeters. b. Using the linear regression feature on a graphing calculator, the line of best fit for the data is given by the equation y = 3.05x + 63.53. A female that has a humerus 40 centimeters long will have an approximate height of y = 3.05(40) + 63.53 ≈ 185.53 centimeters. The result in part (a) is relatively close to the height given by the linear regression line. 1.3 Exercises (pp. 26 –28) Vocabulary and Core Concept Check 1 1. The linear equation y = —2x + 3 is written in slope-intercept form. 2. When a line of best fit has a correlation coefficient of −0.98, this means that the slope is negative. y 1400 y = 10x + 600 Monitoring Progress and Modeling with Mathematics 2 3. From the graph, the slope is m = — = 0.2 and the y-intercept 10 1200 Total cost (dollars) 80 y − y1 = m(x − x1) 4 0 0 120 Humerus length (centimeters) where t is the time (in days) and S(t) is the snowfall amount (in inches). S y 160 0 0 b. A function that models the depreciation is S(t) = 5 − t, c. 3. a. Height (centimeters) 4. Sample answer: the amount of snow on the ground over a y = 10x + 500 1000 800 y = mx + b = 0.2x + 0. The equation is y = 0.2x. The slope indicates that the tip increases $0.20 for every dollar spent on the meal. 600 400 is b = 0. Using slope-intercept form, an equation of the line is y = 12x + 300 200 0 0 20 40 60 80 100 120 x Number of students Algebra 2 Worked-Out Solutions 15 Chapter 1 1 −3 90 30 y-intercept is b = 12. Using slope-intercept form, an 4. From the graph, the slope is m = — = −— and the equation of the line is y = mx + b 9. 1. Understand the Problem You are given an equation that represents the total cost for an advertisement at the Greenville Journal and a table of values showing total costs for advertisements at the Daily Times. You need to compare costs. 2. Make a Plan Write an equation that models the total 1 y = − —x + 12. 30 1 The equation is y = − —x + 12. The slope indicates that the 30 1 amount of fuel in the gasoline tank decreases by — gallon 30 per mile driven. 100 5. From the graph, the slope is m = — = 50 and the y-intercept 2 is b = 100. Using slope-intercept form, an equation of the cost of advertisements at the Daily Times. Then compare the slopes to determine which newspaper charges less per line. Finally, equate the cost expressions and solve to determine the number of lines for which the total costs are equal. 30 − 27 3. Solve the Problem The slope is m = — = 3. 5−4 Using point-slope form, the equation to represent the total cost for advertisements at Daily Times is line is y − y1 = m(x − x1) y = mx + b y − 27 = 3(x − 4) y = 50x + 100. The equation is y = 50x + 100. The slope indicates that the savings account balance increases by $50 per week. 6. From the graph, the slope is m = —6 = 1.5 and the y-intercept is 4 b = 0. Using slope-intercept form, an equation of the line is y = mx + b y = 1.5x + 0. The equation is y = 1.5x. The slope indicates that the height of the tree increases by 1.5 feet per year. 165 − 55 110 7. From the graph, the slope is m = — = — = 55 3−1 2 and the y-intercept is b = 0. Using slope-intercept form, an equation of the line is y = mx + b y = 55x + 0. The equation is y = 55x. The slope indicates that the typing rate is 55 words per minute. 300 − 180 120 8. From the graph, the slope is m = — = — = −60. 3−5 −2 Using slope-intercept form, an equation of the line is y − y1 = m(x − x1) y − 300 = −60(x − 3) y = −60x + 480. The equation is y = −60x + 480. The slope indicates that the water level in the swimming pool decreases by 60 cubic feet per hour. y = 3x + 15. Equate the cost expressions and solve. 2x + 20 = 3x + 15 5=x Comparing the slopes of the equations, the Greenville Journal costs $2 per line, which is less than the $3 per line that the Daily Times charges. The total costs are the same if there are 5 lines in an advertisement. 10. 1. Understand the Problem You have to write an equation that represents the linear relationship between Fahrenheit and Celsius and calculate several temperatures. 2. Make a Plan Use the point-slope form to write an equation that gives degrees Fahrenheit in terms of degrees Celsius. Then, substitute the given temperature for x in the equation to calculate y. Finally, rewrite the equation by solving for x. 3. Solve the Problem The slope is 32 − 212 −180 9 m = — = — = — . Using point-slope form, 0 − 100 −100 5 the equation is y − y1 = m(x − x1) 9 y − 32 = — (x − 0) 5 9 y = —x + 32. 5 (a) An equation that gives degrees Fahrenheit in terms of 9 degrees Celsius is y = — x + 32. 5 (b) Substitute 22 for x. 9 y = —(22) + 32 5 = 71.6 The outside temperature is 71.6°F. 16 Algebra 2 Worked-Out Solutions Chapter 1 9 y = — x + 32 5 9 y − 32 = — x 5 5 x = —( y − 32) 9 An equation that gives degrees Celsius in terms of 5 degrees Fahrenheit is x = —9 ( y − 32). (d) Substitute 83 for y. 5 x = —( 83 − 32) 9 ≈ 28.33 The temperature of the hotel pool is about 28.33°F. 11. The slope was correctly used in the situation, however the intercept was not used correctly. In the situation the starting balance is $100, so after 7 years the balance is $170. 12. The slope is incorrect in the situation. The slope is 11, so the income is $11 per hour. 13. Sample answer: Step 1 Draw a scatter plot of the data. The data show a linear relationship. Step 2 Sketch the line that most closely appears to fit the data. One possible line is shown. Calories burned y Step 2 Sketch the line that most closely appears to fit the data. One possible line is shown. y 10 y = 0.5x − 1.5 8 6 4 2 0 0 4 8 12 16 20 x Months Step 3 Choose two points on the line. For the line shown, you might choose (9, 3) and (13,5). Step 4 Write the equation of the line. First, find the slope. y2 − y1 5 − 3 = 2 = 0.5 m =— — x2 − x1 = — 13 − 9 4 Use point-slope form to write an equation. Use (x1, y1) = (9, 3). y − y1 = m(x − x1) y − 3 = 0.5(x − 9) y = 0.5x − 1.5 Use the equation to estimate the length of hair. 40 y = 0.5(15)−1.5 = 6 The approximate hair length after 15 months is 6 inches. 20 0 Step 1 Draw a scatter plot of the data. The data show a linear relationship. y − 3 = 0.5x − 4.5 y = 4.2x + 1.8 60 14. Sample answer: Hair length (inches) (c) Solve the equation for x. 0 4 8 12 16 x Minutes walking Step 3 Choose two points on the line. For the line shown, you might choose (1, 6) and (6, 27). Step 4 Write the equation of the line. First, find the slope. y2 − y1 27 − 6 21 = — = 4.2 m=— x2 − x1 = — 6−1 5 Use the point-slope form to write an equation. Use (x1, y1) = (1, 6). y − y1 = m(x − x1) y − 6 = 4.2(x − 1) y − 6 = 4.2x − 4.2 y = 4.2x + 1.8 Use the equation to estimate the number of calories burned. y = 4.2(15) + 1.8 = 64.8 The approximate number of calories burned when you walk for 15 minutes is 64.8 calories. Algebra 2 Worked-Out Solutions 17 Chapter 1 17. Enter the data into two lists. Use the linear regression 15. Sample answer: Step 1 Draw a scatter plot of the data. The data show a linear relationship. Step 2 Sketch the line that most closely appears to fit the data. One possible line is shown. Battery life (%) 80 y = −4.55x + 91 40 18. Sample answer: Step 1 Draw a scatter plot of the data. The data show a linear relationship. 20 0 y = 380.03(15) + 11,290 = 16,990.45 The approximate average annual tuition cost in the year 2020 is $16,990.45. The annual tuition increases by about $380 each year and the cost of tuition in 2005 is about $11,290. y 60 feature. The line of best fit is y = 380.03x + 11,290. Use the equation to estimate the annual tuition cost in 2020 (x = 15). 0 4 8 12 16 Step 2 Sketch the limit that most closely appears to fit the data. One possible line is shown. 20 x Hours y Step 3 Choose two points on the line. For the line shown, you might choose (9, 50) and (20, 0). Use point-slope form to write an equation. Use (x1, y1) = (20, 0). y − y1 = m(x − x1) Tickets sold Step 4 Write the equation of the line. First, find the slope. y2 − y1 0 − 50 −50 = — ≈ −4.55 m=— x2 − x1 = — 20 − 9 11 400 y = −10x + 620 300 200 100 0 0 4 8 12 16 20 24 28 x Ticket price (dollars) y − 0 = −4.55(x − 20) y = −4.55x + 91 Use the equation to estimate the battery life. Step 3 Choose two points on the line. For the line shown, you might choose (17, 450) and (22, 400). y = −4.55(15) + 91 = 22.75 The approximate battery life after 15 hours is 23%. Step 4 Write the equation of the line. First, find the slope. y2 − y1 400 − 450 −50 = — = −10 m =— x2 − x1 = — 22 − 17 5 Use point-slope form to write an equation. Use (x1, y1) = (17, 450). 16. Draw a scatter plot of the data. The data does not show a linear relationship. y − y1 = m(x − x1) y y − 450 = −10(x − 17) 120 Heart rate (bpm) y − 450 = −10x + 170 100 y = −10x + 620 Use the equation to estimate the number of tickets sold. 80 y = −10(85) + 620 60 = −230 The approximate number of tickets sold when the price is $85 is −230. This does not seem reasonable because the number of tickets sold is less than zero. 40 20 0 0 2 4 6 8 10 Shoe size 12 x 19. Enter the data into two lists. Use the linear regression feature. The line of best fit is y = 0.42x + 1.44. The correlation coefficient is r ≈ 0.61. This represents a weak positive correlation. 20. Enter the data into two lists. Use the linear regression feature. The line of best fit is y = 0.88x + 1.69. The correlation coefficient is r ≈ 0.88. This represents a strong positive correlation. 18 Algebra 2 Worked-Out Solutions Chapter 1 21. Enter the data into two lists. Use the linear regression feature. The line of best fit is y = −0.45x + 4.26. The correlation coefficient is r ≈ −0.67. This represents a weak negative correlation. 22. Enter the data into two lists. Use the linear regression feature. The line of best fit is y = −1.04x + 5.68. The correlation coefficient is r ≈ −0.93. This represents a strong negative correlation. 23. Enter the data into two lists. Use the linear regression feature. The line of best fit is y = 0.61x + 0.10. The correlation coefficient is r ≈ 0.95. This represents a strong positive correlation. 24. Enter the data into two lists. Use the linear regression feature. The line of best fit is y = −0.48x + 4.08. The correlation coefficient is r ≈ −0.91. This represents a strong negative correlation. 28. Consider the possible locations of the three points. (1) A and B are two different points on the line and C also lies on the line. (2) A and B are two different points of the line and C does not lie on the line. Sample answer: Let A be (0, 4) and B be (4, 0). Then a point that is the same distance to A and to B is the midpoint of the line segment between A and B, which is ( x2 + x1 y2 + y1 ) ( 0+4 4+0 2 2 ) — , — = — , — = (2, 2). 2 2 So, C is (2, 2). Because points A, B, and C all lie on the same line, the equation of the line through A and C and the equation of the line through B and C are both y = −x + 4. Next, let C be a point that is not on the line y = −x + 4. The origin is 4 units from A and 4 units from B. So, let C be (0, 0). The equation of the line through A and C is x = 0. The equation of the line through B and C is y = 0. 29. As x increases, y increases, so z decreases. Therefore, the 25. a. Sample answer: height and weight; temperature and ice cream sales; Correlation is positive because as the first goes up, so does the second. b. Sample answer: miles driven and gas remaining; hours used and battery life remaining; Correlation is negative because as the first goes up, the second goes down. c. Sample answer: age and length of hair; typing speed and shoe size; There is no relationship between the first and second. correlation between x and z is negative. 30. The equation is D. The equation of a line that is perpendicular to the graph of y = −4x + 1 and passes 1 through (8, −5) has a slope of m = —4. So, the equation is given by y − y1 = m(x − x1) 1 y − (−5) = —4(x − 8) y+5 26. a. To determine the slope of the line, use the points (0, 30) and (24, 0). So, the slope is y2 − y1 0 − 30 −30 5 = — = − —. m =— x2 − x1 = — 4 24 − 0 24 The slope represents the amount the loan is reduced each month. y 1 = —4 x 1 = —4 x −2 − 7. b. The domain of the function is 0 ≤ x ≤ 24 and the range is 0 ≤ y ≤ 30. The domain represents the term of the balance of the loan from start to finish and the range represents the amount left to pay on the loan. c. The equation of the line that models the amount left to 5 pay on the loan is y = −—x + 30. The amount left to pay 4 5 after 12 months is y = −—(12) + 30 = 15, or $1500. 4 27. Your friend is incorrect. Because r = 0.3 is closer to 0 than 1, the line of best fit will not make good predictions. Algebra 2 Worked-Out Solutions 19 Chapter 1 31. The path that will give the shortest distance to the river is given by the line that is perpendicular to the graph of y = 3x + 2 and that passes through the point (2, 1). The 1 slope of a perpendicular line to y = 3x + 2 is m = −—3. So, the equation of the line perpendicular to the graph of y = 3x + 2 and passing through the point (2, 1) is given by y − y1 = m(x − x1) Maintaining Mathematical Proficiency 33. Solve the system using elimination. { There is no need to change coefficients because the variable y differs only in sign. 3x + y = 7 1 y − 1 = − —3(x − 2) 1 2 1 5 −2x − y = 9 y − 1 = − —3 x + —3 x = 16 y = − —3 x + —3 . The point where the two lines intersect has an x-value given by 1 3x + 2 = −—3x + —35 9x + 6 = −x + 5 So, x = 16. Now, back-substitute x = 16 into one of the original equations of the system that contains the variable x. By back-substituting x = 16 into Equation 1, you can solve for y. 3x + y = 7 10x + 6 = 5 3(16) + y = 7 10x = −1 1 x = −— 10 . 48 + y = 7 ( ) 1 . So, the y-value of the intersection is y = 3 − — 10 + 2 = — 10 Thus, the closest point on the line y = 3x + 2 to the point 1 17 (2, 1) is the point − — . Use the Distance Formula to find 10 , — 10 the shortest distance to the river. ( ) √[ ( ) ] [ 1 2 − −— 10 ] 17 2+ 1− — 10 2 y = −41 17 ——— d= 3x + y = 7 Equation 1 −2x − y = 9 Equation 2 7 =— — ≈ 2.2 √10 So, you must travel about 2.2 miles. 32. a. A positive correlation does make sense because the number of computers per capita and the average life expectancy have both increased over time, which would relate to a positive slope. b. It is not reasonable to conclude that giving residents of a country computers will lengthen their lives. There is a correlation, but there is not a causation between the two quantities. So, the solution is x = 16 and y = −41. 34. Solve the system using elimination. { 4x + 3y = 2 Equation 1 2x − 3y = 1 Equation 2 There is no need to change coefficients because the variable y differs only in sign. 4x + 3y = 2 2x − 3y = 1 6x = 3 Solving the equation 6x = 3 produces x = —21 . Now backsubstitute x = 1—2 into one of the original equations of the system that contains the variable y. By back-substituting x = 1—2 into Equation 1, you can solve for y. 4x + 3y = 2 () 4 —21 + 3y = 2 2 + 3y = 2 3y = 0 y=0 So, the solution is x = —21 and y = 0. 20 Algebra 2 Worked-Out Solutions Chapter 1 35. Solve the system using substitution. { 37. Solve the system using elimination. 2x + 2y = 3 Equation 1 x = 4y − 1 Equation 2 Substitute 4y − 1 from Equation 2 for x into Equation 1 and solve the resulting single-variable equation for y. 2x + 2y = 3 —2 x + 4y = 4 1 Equation 1 2x − y = 1 Equation 2 You can obtain coefficients of x that differ only in sign by multiplying Equation 2 by 4. 1 1 —2x + 4y = 4 → —2 x + 4y = 4 2(4y − 1) + 2y = 3 2x − y = 1 → 8x − 4y = 4 8y − 2 + 2y = 3 17 — x=8 2 10y − 2 = 3 16 17 Solving the equation — . Now, x = 8 produces x = — 2 17 10y = 5 y { 16 1 = —2 1 Finally, you can solve for x by back-substituting y = —2 into the equation x = 4y − 1. x = 4y − 1 into one of the original or revised back-substitute x = — 17 equations of the system that contains the variable x. By 16 into Equation 2, you can solve for y. back-substituting x = — 17 2 () 1 x = 4 —2 − 1 x=2−1 ( )−y=1 16 — 17 32 — 17 −y=1 15 −y = −— 17 x=1 15 1 So, the solution is x = 1 and y = —2. y =— 17 16 15 . So, the solution is x = — and y = — 17 17 36. Solve the system using substitution. { y=1+x Equation1 2x + y = −2 Equation 2 Substitute 1 + x from Equation 1 for y into Equation 2 and solve the resulting single-variable equation for x. 2x + y = −2 2x + (1 + x) = −2 3x + 1 = −2 38. Solve the system using substitution. { y=x−4 Equation 1 4x + y = 26 Equation 2 Substitute x − 4 from Equation 1 for y into Equation 2 and solve the resulting single-variable equation for x. 4x + y = 26 4x + (x − 4) = 26 5x − 4 = 26 3x = −3 5x = 30 x = −1 Finally, you can solve for y by back-substituting x = −1 into the equation y = 1 + x. y=1+x y = 1 + (−1) y=0 So, the solution is x = −1 and y = 0. x=6 Finally, you can solve for y by back-substituting x = 6 into the equation y = x − 4. y=x−4 y=6−4 y=2 So, the solution is x = 6 and y = 2. 1.4 Explorations (p. 29) 1. a. The system matches graph B; The two equations have the same slope but different y-intercepts, so their graphs are parallel lines. The system is inconsistent. b. The system matches graph C; The two equations have different slopes and different y-intercepts, so there is a point that they will intersect. The system is consistent. c. The system matches graph A; The two equations have the same slope and same y-intercept, so they will produce the same graph. The system is consistent. Algebra 2 Worked-Out Solutions 21 Chapter 1 2. a. Solve x − y = 1 for x, which gives x = y + 1. Then substitute y + 1 for x into 2x + y = 5 and solve the resulting equation for y. 2x + y = 5 2(y + 1) + y = 5 2y + 2 + y = 5 3y + 2 = 5 3y = 3 y=1 So, y = 1. Back-substitute y = 1 into x = y + 1 giving x = 1 + 1 = 2. Thus, the solution of the system is (2, 1). The solution found graphically is (2, 1), the point where the two lines intersect. b. Solve x + 3y = 1 for x, which gives x = −3y + 1. Then substitute −3y + 1 for x into −x + 2y = 4 and solve the resulting equation for y. 5. Add Equation 1 to Equation 3. x+y+z=1 −x − y + z = −1 _______________ 2z = 0 So, 2z = 0, or z = 0. This produces the system x+y=1 x−y=3 −x − y = −1. Now use the new forms of Equation 1 and Equation 2 with the elimination method. x+y=1 x−y=3 2x = 4 So, 2x = 4, or x = 2. Finally, substitute the values of x and z into Equation 1 to solve for y. −x + 2y = 4 x+y+z=1 −(−3y + 1) + 2y = 4 2+y+0=1 3y − 1 + 2y = 4 5y − 1 = 4 5y = 5 y=1 So, y = 1. Back-substitute y = 1 into x = −3y + 1 giving x = −3(1) + 1 = −2. Thus, the solution of the system is (−2, 1). The solution found graphically is (−2, 1), the point where the two lines intersect. c. Solve x + y = 0 for x, which gives x = −y. Then substitute −y for x into 3x + 2y = 1 and solve the resulting equation for y. 3x + 2y = 1 3(−y) + 2y = 1 −3y + 2y = 1 −y = 1 y = −1 So, y = −1. Back-substitute y = −1 into x = −y giving x = 1. Thus, the solution of the system is (1, −1). The solution found graphically is (1, −1), the point where the two lines intersect. 3. If the graph of a system is parallel lines, then there is no solution. If the graph of a system shows two lines that intersect or coincide, then there is at least one solution. 4. Use elimination to eliminate a variable in equations 1 and 2. Eliminate the same variable from equations 1 and 3. Using the two new resulting equations, solve that system for both variables. Substitute those values into one of the original equations to find the third variable. y = −1 So, the solution of the linear system is (2, −1, 0). 1.4 Monitoring Progress (pp. 32–33) 1. Step 1 Rewrite the system as a linear system in two variables. x − 2y + z = −11 −x + 2y + 4z = −9 __________________ 5z = −20 z = −4 −3x + 6y − 3z = 33 3x + 2y − z =7 ______________________ 8y − 4z = 40 8y − 4(−4) = 40 8y + 16 = 40 y=3 Step 2 Substitute y = 3 and z = −4 into an original equation and solve for x. x − 2y + z = −11 x − 2(3) + (−4) = −11 x − 6 − 4 = −11 x = −1 The solution is (−1, 3, −4). 2. Step 1 Rewrite the system as a linear system in two variables. −4x − 4y + 4z = 4 4x + 4y − 4z = −2 __________________ 0=2 Because you obtain a false equation, the original system has no solution. 22 Algebra 2 Worked-Out Solutions Chapter 1 3. Step 1 Rewrite the system as a linear system in two variables. 2. You know that a linear system in three variables has infinitely many solutions when you obtain a result of 0 = 0 when solving. x+y+z=8 x−y+z=8 ____________ Monitoring Progress and Modeling with Mathematics 2x + 2z = 16 x +z=8 3. Step 1 Rewrite the system as a linear system in two variables. x + y − 2z = 5 −x + 2y + z = 2 x+y+ z=8 −2x − y − 2z = −16 __________________ −x 3y − − z = −8 z= 7 −2x + 4y + 2z = 4 Step 2 Solve the new linear system for both of its variables. 2x + 3y − z = 9 7y + z = 13 x+z=8 −x − z = −8 ___________ 0=0 Step 2 Solve the new linear system for both of its variables. 3y − z = 7 Because you obtain the identity 0 = 0, the system has infinitely many solutions. Step 3 Solve new Equation 1 for z to obtain z = 8 − x. Then substitute 8 − x for z in original Equation 1 to obtain y = 0. A solution of the system can be represented by any ordered triple of the form (x, 0, 8 − x). 4. The result of Example 3 is that the solution can be expressed as (x, x + 3, 0). In this case, y = x + 3. So to write the solution in terms of y rather than x, solve for x in y = x + 3, which gives x = y − 3. Thus, the solution can be expressed as (y − 3, y, 0). 7y + z = 13 10y = 20 y=2 7y + z = 13 7(2) + z = 13 z = −1 Step 3 Substitute y = 2 and z = −1 into an original equation and solve for x. x + y − 2z = 5 x + 2 − 2(−1) = 5 x=1 5. The system of equations is y=x Equation 1 x + y + z = 10,000 Equation 2 75x + 55y + 30z = 356,000. Equation 3 Rewrite the system as a linear system in two variables by substituting x for y in Equations 2 and 3. The solution is (1, 2, −1). 4. Step 1 Rewrite the system as a linear system in two variables. x + 4y − 6z = −1 −x + 2y − 4z = 5 _________________ 6y − 10z = 4 x + y + z = 10,000 75x + 55y + 30z = 356,000 2x − y + 2z = −7 x + x + z = 10,000 75x + 55(x) + 30z = 356,000 2x + z = 10,000 130x + 30z = 356,000 −2x + 4y − 8z = 10 __________________ Solve the new linear system for both of its variables. −60x − 30z = 300,000 130x + 30z = 356,000 ___________________ 70x = 56,000 x = 800 y = 800 z = 8400 Of the 10,000 tickets sold, 8400 are lawn seats. So, there are 17,000 − 8400 = 8600 lawn seats still available. 1.4 Exercises (pp. 34–36) Vocabulary and Core Concept Check 1. The solution of a system of three linear equations is expressed as an ordered triple. 3y − 6z = 3 Step 2 Solve the new linear system for both of its variables. 6y − 10z = 4 −6y + 12z = −6 _______________ 2z = − 2 z = −1 y = −1 Step 3 Substitute y = −1 and z = −1 into an original equation and solve for x. x + 4y − 6z = − 1 x + 4(−1) − 6(−1) = −1 x − 4 + 6 = −1 x = −3 The solution is (−3, −1, −1). Algebra 2 Worked-Out Solutions 23 Chapter 1 5. Step 1 Rewrite the system as a linear system in 7. Step 1 Rewrite the system as a linear system in two variables. two variables. 2x + y − z = 9 5x + 7y + z = 4 ______________ 2x + 2y + 5z = −1 −2x + y − z = −2 __________________ 7x + 8y = 13 3y + 4z = −3 4x + 2y − 2z = 18 2x + 2y + 5z = −1 −x + 6y + 2z = −17 __________________ −2x − 4y + 3z = −14 ___________________ 3x + 8y =1 Step 2 Solve the new linear system for both of its variables. 7x + 8y = 13 −2y + 8z = −15 Step 2 Solve the new linear system for both of its variables. −6y − 8z = 6 − 3x − 8y = −1 _____________ 4x = 12 − 2y + 8z = −15 ______________ −8y = −9 9 x=3 y = —8 51 z = −— 32 y = −1 Step 3 Substitute x = 3 and y = −1 into an original equation and solve for z. 9 2x + y − z = 9 2x − y + z = 2 2(3) + (−1) − z = 9 9 2x − —8 + − — 32 = 2 6−1−z=9 ( ) 51 151 x =— 64 z = −4 The solution is (3, −1, −4). 6. Step 1 Rewrite the system as a linear system in 51 8. Step 1 Rewrite the system as a linear system in two variables. 3x + 2y − z = 8 −3x + 4y + 5z = −14 ___________________ 3x + 2y − 3z = −2 7x − 2y + 5z = −14 __________________ 10x + 2z = −16 5x + z = −8 6y + 4z = −6 3x − 9y + 12z = −42 _____________________ −5y + 17z = −56 Step 2 Solve the new linear system for both of its variables. 30y + 20z = −30 − 30y + 102z = −336 __________________ 122z = −366 z = −3 2x + 4y + Step 3 Substitute z = −3 and y = 1 into an original equation and solve for x. x − 3y + 4z = −14 x − 3(1) + 4(−3) = −14 x − 3 − 12 = −14 z=6 14x − 4y + 10z = −28 ___________________ 16x + 11z = −22 Step 2 Solve the new linear system for both of its variables. 16x + 11z = −22 −55x − 11z = 88 ________________ −39x = 66 22 x = −— 13 6 z =— 13 y=1 22 Algebra 2 Worked-Out Solutions 6 Step 3 Substitute x = −— 13 and z = — 13 into an original equation and solve for y. 2x + 4y + z = 6 ( ) 22 6 =6 2 −— 13 + 4y + — 13 x=1 The solution is (1, 1, −3). 24 ) ( 151 9 , — , −— The solution is — 32 . 61 8 two variables. −3x + 4y + 5z = −14 51 Step 3 Substitute y = —8 and z = − — 32 into an original equation and solve for x. 29 y=— 13 ( 22 ) 29 6 ,— The solution is −— 13, — 13 13 . Chapter 1 9. The entire second equation should be multiplied by 4, not just the x-term. 4x − y + 2z = −18 −4x + 8y + 4z = 44 ___________________ 7y + 6z = 26 10. The entire first equation should be multiplied by 3, not just one side. 12x − 3y + 6z = −54 3x + 3y − 4z = 44 __________________ 15x + 2z = −10 Step 3 Solve Equation 2 for x to obtain x = z − 1. A solution of the system can be represented by any ordered triple of the form (z − 1, 1, z). 14. Step 1 Rewrite the system as a linear system in two variables. 2x + 4y − 2z = 6 −2x − y + z = −1 __________________ 3y − z = 5 −6x − 3y + 3z = −3 6x − 3y − z = −7 ___________________ −6y + 2z = −10 11. Step 1 Rewrite the system as a linear system in two variables. −6x + 2y − 4z = −8 6x − 2y + 4z = −8 __________________ 0 = −16 Because you obtain a false equation, the original system has no solution. 12. Step 1 Rewrite the system as a linear system in two variables. 5x + y − z = 6 x+ y+z=2 ______________ 6x + 2y =8 Step 2 Solve the new linear system for both of its variables. −12x − 4y = −16 12x + 4y = 10 _______________ 0 = −6 Because you obtain a false equation, the original system has no solution. Step 2 Solve the new linear system for both of its variables. −6y − 2z = 10 −6y + 2z = −10 ______________ 0=0 Because you obtain the identity 0 = 0, the system has infinitely many solutions. Step 3 Solve new Equation 2 for z to obtain z = 3y − 5. Then substitute 3y − 5 for z in original Equation 1 to obtain y = x + 2. Then z = 3x + 1. A solution of the system can be represented by any ordered triple of the form (x, x + 2, 3x + 1). 15. Step 1 Rewrite the system as a linear system in two variables. x + 2y + 3z = 4 3x − 2y + z = −12 _________________ 4x + 4z = −8 x+ z = −2 x + 2y + 3z = 4 13. Step 1 Rewrite the system as a linear system in two variables. −x − 3y + z = −2 x+ y−z=0 ________________ −2y = −2 −3x − 3y + 3z = 0 3x + 2y − 3z = −1 ___________________ −y = −1 Step 2 You obtain a system in one variable. Solve the new linear system for y. −2x − 2y − 4z = −14 ___________________ −x − z = −10 Step 2 Solve the new linear system for both of its variables. x + z = −2 −x − z = −10 ____________ 0 = −12 Because you obtain a false equation, the system has no solution. −2y = −2 −y = −1 _________ −3y = −3 y=1 The system has infinitely many solutions. Algebra 2 Worked-Out Solutions 25 Chapter 1 16. Step 1 Rewrite the system as a linear system in two variables. Step 5 Substitute p = 5 and s = 3 into an original equation and solve for d. −2x − 3y + z = −6 2x + 2y − 2z = 10 _________________ 5 + d + 3(3) = 15 p + d + 3s = 15 −y − z = 4 d=1 −7x − 7y + 7z = −35 So, one pizza costs $5, a liter of soda costs $1, and a salad costs $3. 7x + 8y − 6z = 31 ___________________ y + z = −4 18. Step 1 Write a verbal model for the situation. Step 2 Solve the new linear system for both of its variables. y + z = −4 ___________ 0=0 Because you obtain the identity 0 = 0, the system has infinitely many solutions. Step 3 Solve new Equation 3 for z to obtain z = −y − 4. Then substitute −y − 4 for z in original Equation 2 to obtain x = −2y + 1. A solution of the system can be represented by any ordered triple of the form (−2y + 1, y, −y − 4). 17. Step 1 Write a verbal model for the situation. 2 ⋅ Cost of pizza, p + Cost of soda, d + Cost of salad, s = 14 Cost of pizza, p + Cost of soda, d + 3 Cost of salad, s = 15 ⋅ 3 ⋅ Cost of pizza, p + ⋅ Cost of salad, s Step 2 Write a system of equations. S+ = 1300 S+ 2C = 1400 S+ + C = 1600 Step 3 Rewrite the system as a linear system in two variables. S+ = 1300 −S − 2C = −1400 _____________________ − 2C = −100 S + 2C = 1400 −S − − C = −1600 _____________________ − + C = −200 Cost of soda, d +2 = 22 Step 2 Write a system of equations. Step 4 Solve the new linear system for both of its variables. − 2C = −100 2p + d + s = 14 − + C = −200 _________________ p + d + 3s = 15 −C = −300 3p + d + 2s = 22 C = 300 Step 3 Rewrite the system as a linear system in two variables. 2p + d + s = 14 −p − d − 3s = −15 _________________ p− 2s = −1 p + d + 3s = 15 −3 p − d − 2s = −22 __________________ −2p + s = −7 Step 4 Solve the new linear system for both of its variables. p − 2s = −1 −4p + 2s = −14 __________________ −3p = −15 p=5 s=3 26 Cost of sofa, S + Cost of love seat, = 1300 Cost of sofa, S + 2 Cost of chairs, C = 1400 Cost of sofa, S + Cost of love seat, + Cost of chair, C = 1600 ⋅ −y − z = 4 Algebra 2 Worked-Out Solutions = 500 Step 5 Substitute solve for S. = 500 into an original equation and S + = 1300 S + 500 = 1300 S = 800 So, a sofa costs $800, a love seat costs $500, and a chair costs $300. Chapter 1 19. Step 1 Solve the first equation for y. y = 1 + 2x − 6z Step 2 Rewrite the system as a linear system in two variables by substituting for y from Step 1 into Equations 2 and 3. 21. Step 1 Solve the first equation for y. y=4−x−z Step 2 Rewrite the system as a linear system in two variables by substituting for y from Step 1 into Equations 2 and 3. 3x + 2y + 5z = 16 5x + 5y + 5z = 12 3x + 2(1 + 2x − 6z) + 5z = 16 5x + 5(4 − x − z) + 5z = 12 3x + 2 + 4x − 12z + 5z = 16 5x + 20 − 5x − 5z + 5z = 12 7x − 7z = 14 x−z=2 7x + 3y − 4z = 11 7x + 3(1 + 2x − 6z) − 4z = 11 7x + 3 + 6x − 18z − 4z = 11 13x − 22z = 8 Step 3 Solve the new linear system for both of its variables, x and z, and then use the values to solve for y. −13x + 13z = −26 20 = 12 Because you obtain a false equation, the original system has no solution. 22. Step 1 Solve the first equation for x. x = −2y − 1 Step 2 Rewrite the system as a linear system in two variables by substituting for x from Step 1 into Equations 2 and 3. −x + 3y + 2z = −4 13x − 22z = 8 ________________ −(−2y − 1) + 3y + 2z = −4 −9z = −18 2y + 1 + 3y + 2z = −4 z=2 5y + 2z = −5 x=4 −x + y − 4z = 10 y = −3 The solution is (4, −3, 2). 20. Step 1 Solve the first equation for x. x = 6y + 2z − 8 Step 2 Rewrite the system as a linear system in two variables by substituting for x from Step 1 into Equations 2 and 3. − x + 5y + 3z = 2 −(6y + 2z − 8)+ 5y + 3z = 2 −6y − 2z + 8 + 5y + 3z = 2 −y + z = −6 3x − 2y − 4z = 18 3(6y + 2z − 8) − 2y − 4z = 18 −(−2y − 1) + y − 4z = 10 2y + 1 + y − 4z = 10 3y − 4z = 9 Step 3 Solve the new linear system for both of its variables, y and z, and then use the values to solve for x. 10y + 4z = −10 3y − 4z = 9 ______________ 13y = −1 1 y = −— 13 30 z = −— 13 x= 11 −— 13 ( 11 1 ) 30 — — The solution is −— 13, −13 , −13 . 18y + 6z − 24 − 2y − 4z = 18 16y + 2z = 42 8y + z = 21 Step 3 Solve the new linear system for both of its variables, y and z, and then use the values to solve for x. y−z=6 8y + z = 21 __________ 9y = 27 y=3 z = −3 x=4 The solution is (4, 3, −3). Algebra 2 Worked-Out Solutions 27 Chapter 1 23. Step 1 Substitute 5 for z from Equation 3 into Equation 2 to solve for y. y=2 y + 2z = 13 −y + 7z = 5 y + 2(5) = 13 −2 + 7z = 5 y + 10 = 13 y=3 Step 2 Substitute 5 for z and 3 for y into Equation 1 to solve for x. 2x − 3y + z = 10 2x − 3(3) + 5 = 10 2x − 9 + 5 = 10 2x = 14 x=7 The solution is (7, 3, 5). 24. Step 1 Substitute 4 for x from Equation 1 into Equation 2 to solve for y. 7z = 7 z=1 x=3 The solution is (3, 2, 1). 26. Step 1 Solve the first equation for y. y = 2x − z − 15 Step 2 Rewrite the system as a linear system in two variables by substituting for y from Step 1 into Equations 2 and 3. 4x + 5y + 2z = 10 4x + 5(2x − z − 15) + 2z = 10 4x + 10x − 5z − 75 + 2z = 10 x + y = −6 14x − 3z = 85 4 + y = −6 −x − 4y + 3z = −20 y = −10 Step 2 Substitute 4 for x and −10 for y into Equation 3 to solve for x. −4x − 3y + 2z = 26 4(4) − 3(−10) + 2z = 26 16 + 30 + 2z = 26 2z = −20 z = −10 The solution is (4, −10, −10). 25. Step 1 Solve the first equation for x. x=4−y+z Step 2 Rewrite the system as a linear system in two variables by substituting for x from Step 1 into Equations 2 and 3. 3x + 2y + 4z = 17 3(4 − y + z) + 2y + 4z = 17 12 − 3y + 3z + 2y + 4z = 17 −y + 7z = 5 −x + 5y + z = 8 −(4 − y + z) + 5y + z = 8 −4 + y − z + 5y + z = 8 6y = 12 y=2 28 Step 3 Solve the new linear system for the variables x and z. Algebra 2 Worked-Out Solutions −x − 4(2x − z − 15) + 3z = −20 −x − 8x + 4z + 60 + 3z = −20 −9x + 7z = −80 Step 3 Solve the new linear system for both of its variables, x and z, and then use the values to solve for y. 98x − 21z = 595 −27x + 21z = −240 _________________ 71x = 355 x=5 z = −5 y=0 The solution is (5, 0, −5). Chapter 1 27. Step 1 Solve the first equation for y. 29. Step 1 Write a verbal model for the situation. y = 5 − 4x − 5z Step 2 Rewrite the system as a linear system in two variables by substituting for y from Step 1 into Equations 2 and 3. 8x + 2y + 10z = 10 + Right-handed, r + Ambidextrous, a =1 Right-handed, r = Left-handed, Right-handed, r = 9 ( Left-handed, Left-handed, 1 — 10 ⋅ ⋅ + Ambidextrous, a ) Step 2 Write a system of equations. 8x + 2(5 − 4x − 5z) + 10z = 10 8x + 10 − 8x − 10z + 10z = 10 0=0 r 10 r = 9( —= Because you obtain the identity 0 = 0, the system has infinitely many solutions. −x − y − 2z = −2 variables by substituting for Equations 1 and 3. −x − (5 − 4x − 5z) − 2z = −2 from Equation 2 into −x − 5 + 4x + 5z − 2z = −2 5x + 3z = 3 r 10 — +r+a=1 5 z = 1 − —3 x 5 Step 3 Substitute 1 − —3 x for z in revised Equation 1 to 13 x. A solution of the system can be represented obtain y = — 3 ( 11 10 11r + 10a = 10 — r+ a = 1 ) 5 13 by any ordered triple of the form x, — x, 1 − —3x . 3 r = 9( + a) 28. Step 1 Solve the first equation for z. z = x + 2y − 3 Step 2 Rewrite the system as a linear system in two variables by substituting for z from Step 1 into Equations 2 and 3. 2x + 4y − 2z = 6 2x + 4y − 2(x + 2y − 3) = 6 6y + 6 = 6 6y = 0 y=0 −x − 2y + z = −6 −x − 2y + (x + 2y − 3) = −6 −3 = −6 Because you obtain a false equation, the system has no solution. r = 9 + 9a r r = 9 — + 9a 10 ( ) 1 10 9a − — r = 0 90a − r = 0 Step 4 Solve the new linear system for both of its variables, r and a, and then use the values to solve for . −99r − 90a = −90 −r + 90a = 0 ________________ −100r = −90 9 r =— 10 1 a =— 100 9 =— 100 So, the percent of people who are ambidextrous is 1%. Algebra 2 Worked-Out Solutions 29 Chapter 1 30. Step 1 Write a verbal model for the situation. First place, f + Second place, s + Third place, t = 20 5 First place, f + 3 Second place, s + 1 Third place, t = 68 Second place, s = First place, f + Third place, t Step 2 Write a system of equations. ⋅ ⋅ ⋅ 32. Sample answer: When solving a system that has four linear equations and four variables, use the same method of elimination to rewrite the system in three linear equations using three variables. Then use the method of elimination that is used to solve a system of three linear equations in three variables. 33. Because the perimeter is 65 feet, the following system of equations provides the relationships between the three sides of the triangle. 1 = —3 m n= Step 3 Rewrite the system as a linear system in two variables by substituting for s from Equation 3 into Equations 1 and 2. + m − 15 + m + n = 65 Step 1 Rewrite the system of equations. 1 − —3 m = 0 − − m + n = −15 + m + n = 65 Step 2 Add Equation 2 to Equation 3. − − m + n = −15 + m + n = 65 _________________ 2n = 50 Step 4 Solve the new linear system for both of its variables, f and t, and then use the values to solve for s. −f − t = −10 2f + t = 17 ____________ f=7 t=3 s = 10 So, 7 athletes finished in first place, 10 athletes finished in second place, and 3 athletes finished in third place. 31. Sample answer: It is more convenient to use the elimination method rather than the substitution method when there are more than two equations, in particular when one variable has the same coefficient or its opposite in each equation. An example is 3x + 2y − 4z = −5 2x + 2y + 3z = 8 5x − 2y − 7z = −9. 30 Algebra 2 Worked-Out Solutions n = 25 Step 3 Substitute n = 25 into the third equation to form a system of two equations and solve. 1 − —3m = 0 − − m = −40 ______________ 4 −—3m = −40 m = 30 = 10 So, the lengths of the three sides of the triangle are = 10 feet, m = 30 feet, and n = 25 feet. Chapter 1 34. Because the sum of the angle measures in a triangle is 180°, the following system of equations provides the relationship between the three interior angle measures of a triangle. A + B + C = 180 2A − C = B A+B=C Step 1 Rewrite the linear system. A + B + C = 180 2A − B − C = 0 A+B−C=0 Step 2 Add Equation 1 to Equation 2, then solve for A. A + B + C = 180 2A − B − C = 0 ________________ 3A = 180 A = 60 37. 1. Understand the Problem Find the number of square feet that each bedroom, kitchen, and bathroom will have. 2. Make a Plan Write a verbal model using the information so that you can write a system of linear equations to solve. 3. Solve the Problem Step 1 Write a verbal model. Bedroom, b + Kitchen, k + Bathroom, t = 840 Bedroom, b = Kitchen, k 2 ( Kitchen, k + Bathroom, t ) = 980 Step 2 Write a linear system. ⋅ b=k Step 3 Rewrite the system as a linear system in two variables by substituting for b from Equation 2 into Equation 1. Step 3 Add Equation 1 to Equation 3, then substitute A = 60 into the result. A + B + C = 180 A+B−C=0 _______________ 2A + 2B = 180 2(60) + 2B = 180 B = 30 Step 4 Substitute A = 60 and B = 30 into an original equation and solve for C. A + B + C = 180 60 + 30 + C = 180 C = 90 So, m∠ A = 60°, m∠ B = 30°, and m∠ C = 90°. 35. a. Sample answer: If a = 1, b = 1, and c = 1, then the system has no solution. In this case, use elimination of Equation 1 and Equation 2, which results in a false equation. b. Sample answer: If a = 10, b = 9, and c = 8, then the system has one solution. In this case, no equations are equivalent and no combinations of equations will lead to a false equation. c. Sample answer: If a = 5, b = 5, and c = 5, then the system has infinitely many solutions. In this case, Equation 1 and Equation 2 are equivalent. 36. Your friend is incorrect. It is possible for two of the equations to intersect at a line while the third forms a parallel plane to one of the others. Therefore, the point (x, y, z) satisfies two of the equations but not the third. Step 4 Solve the new linear system for both of its variables, k and t, and then use the values to solve for b. 2k + t = 840 −2k − 2t = −980 _______________ t = 140 k = 350 b = 350 So, 350 square feet of carpet is needed for each bedroom. 38. By inspection, because the constants in all three equations are 0, one solution is (0, 0, 0). Use the method of elimination to see whether there are other solutions. Add −2 times Equation 2 to Equation 1. 4x + y + z = 0 −4x − y − 4z = 0 _______________ −3z = 0 z=0 Substitute z = 0 into the first and second equations. 4x + y = 0 −4x − y = 0 ___________ 0=0 Because you obtain the identity 0 = 0, the system has infinitely many solutions. Solve new Equation 1 for x 1 to obtain x = −—4 y. So, a solution of the system can be 1 represented by any ordered triple of the form −—4 y, y, 0 . ( Algebra 2 Worked-Out Solutions ) 31 Chapter 1 39. a. Step 1 Write a verbal model. 5 ⋅ ( 2.5 ⋅ Roses, r + 4 ⋅ Lilies, + 2 ⋅ Irises, i ) = 160 Roses, r + Lilies, + Irises, i = 12 Roses, r = 2 ( Lilies, + Irises, i ) ⋅ Step 2 Write a system of equations. + 2i) = 160 41. In order for the system of equations to have only one solution, (−1, 2, −3) will satisfy all three equations. So, a, b, and c can be determined by substituting −1 for x, 2 for y, and −3 for z in all three equations. x + 2y − 3z = a −1 + 2(2) − 3(−3) = a 12 = a r+ + i) b. Step 3 Rewrite the system as a linear system in two variables by substituting for r from Equation 3 into Equations 1 and 2. + 2i) = 160 −x − y + z = b −(−1) − 2 + (−3) = b −4 = b 2x + 3y − 2z = c 2(−1) + 3(2) − 2(−3) = c 2.5r + 4 + 2i = 32 2.5 [ 2( + i) ] + 4 + 2i = 32 5 + 5i + 4 + 2i = 32 9 + 7i = 32 + i = 12 2 + 2i + + i = 12 3 +i=4 Step 4 Solve the new linear system for both of its variables, and i, and then use the values to solve for r. 42. The arrangement 3x + 2y + (−5)z = −30 produces the solution (−3, 2, 5) and the arrangement 2x + (−5)y + 3z = −30 produces the solution (−699, −288, −24). First arrangement: x − 3y + 6z = 21 Equation 1 3x + 2y − 5z = −30 Equation 2 2x − 5y + 2z = −6 Equation 3 9 + 7i = 32 −3x + 9y − 18z = −63 −9 ______________ −2i = −4 3x + 2y − 5z = −30 ____________________ i=2 =2 r=8 So, there should be 8 roses, 2 lilies, and 2 irises in each bouquet. c. If there is no limitation on the total cost, then there are five possible solutions, each of which is based on the total cost. The solutions are 8 roses, 0 lilies, and 4 irises at a cost of $140 8 roses, 1 lily, and 3 irises at a cost of $150 8 roses, 2 lilies, and 2 irises at a cost of $160 8 roses, 3 lilies, and 1 iris at a cost of $170 8 roses, 4 lilies, and 0 irises at a cost of $180 40. a. Because all three circles intersect at one single point, the system of equations has one solution. b. Because all three circles do not intersect at any point, the system of equations has no solution. 32 10 = c So, a = 12, b = −4, and c = 10. Algebra 2 Worked-Out Solutions 11y − 23z = −93 New Equation 2 −2x + 6y − 12z = −42 2x − 5y + 2z = −6 ____________________ y − 10z = −48 11y − 23z = −93 −11y + 110z = 528 _________________ 87z = 435 z=5 y − 10(5) = −48 y=2 x − 3(2) + 6(5) = 21 x = −3 The solution is (−3, 2, 5). New Equation 3 Chapter 1 Second arrangement: 47. (4 − y)2 = (4 − y)(4 − y) x − 3y + 6z = 21 Equation 1 = 16 − 4y − 4y + y2 2x − 5y + 3z = −30 Equation 2 = y2 − 8y + 16 2x − 5y + 2z = −6 Equation 3 −2x + 6y − 12z = −42 subtracts −2 from each input value. 2x − 5y + 3z = −30 ____________________ y − 9z = −72 g(x) = f(x + 2) New Equation 2 = ∣x + 2 ∣ − 5 The transformed function is g(x) = ∣ x + 2 ∣ − 5. −2x + 5y − 3z = 30 2x − 5y + 2z = −6 ____________________ 49. A reflection in the x-axis changes the sign of each output −z = 24 z = −24 48. A translation 2 units left is a horizontal translation that value. New Equation 3 y − 9(−24) = −72 g(x) = −f(x) = −( ∣x∣ − 5 ) = −∣ x ∣ + 5 y = −288 x − 3(−288) + 6(−24) = 21 x = −699 The solution is (−699, −288, −24). 43. The linear system represented by the three pictures is a+t=g The transformed function is g(x) = −∣ x ∣ + 5. 50. A translation 4 units up is a vertical translation that adds 4 to each output value. g(x) = f(x) + 4 = ( ∣ x ∣ − 5) + 4 = ∣x ∣ − 1 t+b=a 2g = 3b. Solve for b in Equation 2. b=a−t Substitute b into Equation 3. 3 g = —2 b 3 = —2 (a − t) Substitute g into Equation 1 and express a in terms of t. a+t=g 3 a + t = —2(a − t) 2a + 2t = 3a − 3t 5t = a So, 5 tangerines are required to balance one apple. Maintaining Mathematical Proficiency 44. (x − 2)2 = (x − 2)(x − 2) = x2 − 2x − 2x + 4 = x2 − 4x + 4 45. (3m + 1)2 = (3m + 1)(3m + 1) = (3m)2 + 3m + 3m + 1 The transformed function is g(x) = ∣ x ∣ − 1. 51. A vertical stretch by a factor of 3 multiplies each output value by 3. g(x) = 3f(x) = 3(∣ x∣ − 5 ) = 3∣ x ∣ − 15 The transformed function is g(x) = 3 ∣x ∣ − 15. 1.3–1.4 What Did You Learn? (p. 37) 1. Sample answer: Because the y-intercept is 0, use one of the points in the equation y = mx. 2. Sample answer: An equation can be written stating that the total number of placers was 20. A second equation can be written stating that the total number of points was 68. A third equation can be written stating that the number of second place finishers was equal to the number of first and third place finishers combined. 3. Sample answer: If a = b = c = 5, the first two equations will be the same and a solution will exist. So, choose any number so long as a, b, and c are not 5 for the system to have no solution. = 9m2 + 6m + 1 46. (2z − 5)2 = (2z − 5)(2z − 5) = (2z)2 − 10z − 10z + 25 = 4z2 − 20z + 25 Algebra 2 Worked-Out Solutions 33 Chapter 1 Chapter 1 Review (pp. 38–40) 1. 5. y 6 y y= x 6 4 4 2 f(x) = x + 3 y=x 2 −6 −4 −2 x −4 −2 2 2 4 6 x −2 4 f(x) = − x − 3 −4 −2 −6 The graph of f is a translation 3 units up of the parent linear function. 2. y −8 The graph of f is a reflection in the x-axis and a translation 3 units down of the parent absolute value function. y = x 4 6. y 2 −4 8 −2 2 x 4 y = x2 −6 −4 −4 2 4 x −4 The graph of g is a translation 1 unit down of the parent absolute value function. 3. 4 g(x) = x − 1 −2 y g(x) = −3(x + 3) 2 The graph of g is a vertical stretch by a factor of 3 followed by a reflection in the x-axis and translation 3 units left of the parent quadratic function. 4 y = x2 2 h(x) = 12 x −4 −2 2 2 7. Step 1 First write a function h that represents the reflection of f. 4 x 1 — 2 The graph of h is a vertical shrink by a factor of of the parent quadratic function. 4. y 6 h(x) = −f(x) = −∣ x ∣ Step 2 Then write a function g that represents the translation 4 units left of h. g(x) = h(x + 4) h(x) = 4 = −∣ x + 4 ∣ y=1 2 −4 −2 2 4 The transformed function is g(x) = −∣ x + 4∣. x 8. Step 1 First write a function h that represents the vertical shrink of f. The graph of h is a translation 3 units up of the parent constant function. 1 h(x) = —2 f(x) = —2∣ x ∣ 1 Step 2 Then write a function g that represents the translation 2 units up of h. g(x) = h(x) + 2 1 = —2∣ x ∣ + 2 The transformed function is g(x) = —2∣ x∣ + 2. 1 34 Algebra 2 Worked-Out Solutions Chapter 1 9. Step 1 First write a function h that represents the translation 3 units down of f. Step 3 Substitute y = −2 and z = 1 into an original equation and solve for x. h(x) = f(x) − 3 x+y+z=3 =x−3 Step 2 Then write a function g that represents a reflection in the y-axis of h. x−2+1=3 g(x) = h(−x) = −x − 3 The transformed function is g(x) = −x − 3. 10. Enter the data into two lists. Use the linear regression feature. The line of best fit is y = 0.03x + 1.23. x=4 The solution is (4, −2, 1). 13. Step 1 Rewrite the system as a linear system in two variables. 2x − 5y − z = 17 5x + 5y +15z = 95 __________________ 7x + 14z = 112 11. Having traveled 3.5 miles in 10 minutes corresponds to the point (10, 3.5) and traveling 10.5 miles in 30 minutes corresponds to the point (30, 10.5). Write an equation of the line that passes through the points (10, 3.5) and (30, 10.5). First, find the slope. y2 − y1 7 10.5 − 3.5 = — = 0.35 m=— x2 − x1 = — 20 30 − 10 Use point-slope form to write an equation. Use (x1, y1) = (10, 3.5). y − y1 = m(x − x1) x + 2z = 16 −6x − 6y −18z = −114 −4x + 6y + z = −20 _____________________ −10x − 17z = −134 Step 2 Solve the new linear system for both of its variables. 10x + 20z = 160 −10x − 17z = −134 _________________ 3z = 26 y − 3.5 = 0.35(x − 10) y − 3.5 = 0.35x − 3.5 y = 0.35x Use the equation to estimate how far you ride your bike in 45 minutes. y = 0.35(45) = 15.75 After 45 minutes of riding your bike, you have traveled 15.75 miles. 12. Step 1 Rewrite the system as a linear system in two variables by substituting for x from Equation 3 into Equations 1 and 2. x+y+z=3 4z + y + z = 3 y + 5z = 3 −x + 3y + 2z = −8 −4z + 3y + 2z = −8 3y − 2z = −8 Step 2 Solve the new linear system for both of its variables. 3y + 15z = 9 −3y + 2z = 8 ______________ 17z = 17 z= 1 y + 5(1) = 3 y = −2 26 z =— 3 x+2 ( ) = 16 26 — 3 4 x = −—3 4 26 into an original Step 3 Substitute x = −—3 and z = — 3 equation and solve for x. x + y + 3z = 19 4 ⋅ 26 = 19 −—3 + y + 3 — 3 17 y = −— 3 ( 4 17 ) 26 The solution is −—3, − — . 3,— 3 14. Step 1 Rewrite the system as a linear system in two variables. −x − y − z = −2 2x − 3y + z = 11 _________________ x − 4y =9 4x − 6y + 2z = 22 −3x + 2y − 2z = −13 ___________________ x − 4y = 9 Step 2 Solve the new linear system for both of its variables. x − 4y = 9 −x + 4y = −9 ____________ 0=0 Because you obtain the identity 0 = 0, the system has infinite solutions. Algebra 2 Worked-Out Solutions 35 Chapter 1 15. Step 1 Rewrite the system as a linear system in 17. Step 1 Rewrite the system as a linear system in two variables. two variables. x + 4y − 2z = 3 −3x − 3y − 3z = −18 −x − 3y − 7z = −1 _________________ 3x + 3y + 4z = 28 ___________________ y − 9z = 2 z = 10 −2x − 8y + 4z = −6 x + 2z = 4 2x + 9y − 13z = 2 ___________________ x + 2(10) = 4 x = −16 y − 9z = −4 Step 2 Substitute x = −16 and z = 10 into an original equation and solve for y. Step 2 Solve the new linear system for both of its variables. y − 9z = 2 x+y+z=6 −y + 9z = 4 ___________ −16 + y + 10 = 6 0=6 Because you obtain a false equation, the system has no solution. 16. Step 1 Rewrite the system as a linear system in y = 12 The solution is (−16, 12, 10). 18. Step 1 Write a verbal model for the situation. Students, s + Adults, a + Children, c = 600 two variables. 5x − 5y + 15z = 30 −6x + 6y − 15z = −27 ____________________ −x + y = 3 ⋅ Students, s 3 ⋅ + 7 Adults, a + 2 Adults, a = Students, s + 150 ⋅ Children, c = 3150 Step 2 Write a system of equations. Step 2 Solve the new linear system for both of its variables. −x + y = 3 x − 2y = 5 ___________ −y = 8 y = −8 Step 3 Rewrite the system as a linear system in two variables by substituting for a from Equation 3 into Equations 1 and 2. x − 2(−8) = 5 x = −11 Step 3 Substitute x = −11 and y = −8 into an original equation and solve for z. x − y + 3z = 6 −11 + 8 + 3z = 6 3z = 9 z=3 The solution is (−11, −8, 3). Step 4 Solve the new linear system for both of its variables, s and c, and then use the values to solve for a. −2s − c = −450 5s + c = 1050 ______________ 3s = 600 s = 200 c = 50 a = 200 + 150 a = 350 So, 200 student tickets, 350 adult tickets, and 50 children under 12 tickets were sold. 36 Algebra 2 Worked-Out Solutions Chapter 1 4. Step 1 Rewrite the system as a linear system in two Chapter 1 Test (p. 41) variables. 1. Using the points (2, 400) and (3, 600), write the equation of −3x − 6y − 15z = −6 the line passing through the points. First find the slope. y2 − y1 600 − 400 200 m = — = — = — = 200 x2 − x1 3−2 1 3x + 6y − 3z = 9 ___________________ −18z = 3 Use point-slope form to write an equation. Use (x1, y1) = (2, 400). 1 z = −—6 1 Step 2 Substitute z = − —6 into the first equation and solve for y. y − y1 = m(x − x1) y − 400 = 200(x − 2) ( ) 1 1 y = —2 − —6 y − 400 = 200x − 400 y = 200x = The slope, $200, is the amount the balance increases each week and the y-intercept is the original amount in the account, $0. 1 1 — Step 3 Substitute y = −— 12 and z = − 6 into an original equation and solve for x. x + 2y + 5z = 2 ( 2. Using the points (0, 50) and (20, 40), write the equation of 1 ( ) ) 1 — x + 2 −— 12 + 5 − 6 = 2 the line passing through the points. First find the slope. y2 − y1 50 − 40 10 m = — = —= — = −0.5 x2 − x1 −20 0 − 20 Use point-slope form to write an equation. Use (x1, y1) = (0, 50). 1 −— 12 x=3 ( 1 1 ) — The solution is 3, −— 12, −6 . 5. Step 1 Rewrite the system as a linear system in two variables. y − y1 = m(x − x1) −2x + 2y − 10z = −6 y − 50 = −0.5(x − 0) 2x + 3y − z = 2 ____________________ y − 50 = −0.5 5y − 11z = −4 y = −0.5x + 50 4x + 6y − 2z = The slope, −$0.50, is the amount the price decreases for each 1% of discount and the y-intercept is the initial price, $50. 4 −4x − y − 9z = −8 ____________________ 5y − 11z = −4 3. Step 1 Rewrite the system as a linear system in two variables. Step 2 Solve the new linear system for both of its variables. −2x + y + 4z = 5 5y − 11z = −4 2x + 6y − 2z = 4 ____________________ − 5y + 11z = 4 _______________ 7y + 2z = 9 0= −4x + 2y + 8z = 10 0 Because you obtain the identity 0 = 0, the system has infinitely many solutions. 4x + y − 6z = 11 _________________ 3y + 2z = 21 11 4 −7y − 2z = −9 Step 3 Solve new Equation 2 for y to obtain y = — z − —5 . 5 4 11 Substitute — z − —5 for y into original Equation 1 to obtain 5 14 11 . A solution of the system can be represented x = −—5 z + — 5 3y + 2z = 21 ______________ 11 11 by any ordered triple of the form − — , —z −—54 , z . 5 z+— 5 5 Step 2 Solve the new linear system for both of its variables. ( −4y = 12 6. y = −3 y 14 ) y= x 4 z = 15 Step 3 Substitute y = −3 and z = 15 into an original equation and solve for x. −2x + y + 4z = 5 −2x + (−3) + 4(15) = 5 x = 26 The solution is (26, −3, 15). 2 f(x) = x − 1 −4 −2 2 4 6 x −2 The graph of f is the graph of the parent absolute value function translated right 1 unit. Algebra 2 Worked-Out Solutions 37 Chapter 1 7. 8 Step 5 Substitute b = 25 into the first original equation and solve for d. 6 d = 3(25) y = 75 4 y = x2 So, 75 doughnuts, 30 muffins, and 25 bagels are made. 2 f(x) = (3x) −2 2 x 2 The graph of f is the graph of the parent quadratic value function with a horizontal shrink by a factor of —31. 8. y 13. The graph that will model the second fountain will have a horizontal shrink of the graph of f by a factor of —21. So, g(t) = f(2t) = 14—∣ 2t − 20 ∣. The depth of the first fountain after 4 minutes is f(4) = 1—4 ∣ 4 − 20 ∣ = 4 feet. The depth of the second fountain after 4 minutes is g(4) = —41 ∣2(4) − 20 ∣ = 3 feet. Chapter 1 Standards Assessment (pp. 42–43) 6 f(x) = 4 1. a. The graph of g is a vertical translation 5 units up of the graph of f. b. The graph of g is a vertical shrink by a factor of —1 of the y=1 2 2 graph of f. x c. The graph of g is a horizontal shrink by a factor of —1 of the The graph of f is the graph of the parent constant function translated up 3 units. d. The graph of g is a vertical translation 8 units down of the −4 −2 2 4 3 graph of f. graph of f. 9. The graph of g is B; g(x) = 2x + 3 e. The graph of g is a reflection in the x-axis of the graph 10. The graph of g is C; g(x) = 3x − 2 f. The graph of g is a reflection in the y-axis of the graph 11. The graph of g is A; g(x) = −2x − 3 12. Step 1 Write a verbal model for the situation. Doughnuts, d = 3 1 ⋅ Doughnuts, d + 1.5 ⋅ Bagels, b ⋅ Muffins, m + 1.2 ⋅ Bagels, b = 150 Doughnuts, d + Muffins, m + Bagels, b = 130 Step 2 Write a system of equations. of f. of f. 2. a. Step 1 Draw a scatter plot of the data. The data shows a linear relationship. Step 2 Sketch a line that most closely appears to fit the data. One possibility is shown. y 40,000 30,000 y = 1440x + 36,208 20,000 10,000 0 Step 3 Rewrite the system as a linear system in two variables by substituting d = 3b from Equation 1 into Equations 2 and 3. m + 4b = 130 Step 4 Solve the new linear system for both of its variables. 1.5m + 4.2b = 150 −1.5m − 6b = −195 __________________ −1.8b = −45 b = 25 m = 30 38 Algebra 2 Worked-Out Solutions 0 2 4 x Choose two points on the line. For the line shown, you might choose (0, 36,208) and (2, 39,088). Step 4 Write the equation of the line. First, find the slope. y2 − y1 39,088 − 36,208 2880 m = — = —— = — = 1440 x2 − x1 2−0 2 Use the point-slope form of an equation. Use (x1, y1) = (0, 36,208). y − y1 = m(x − x1) y − 36,208 = 1440(x − 0) y − 36,208 = 1440x y = 1440x + 36,208 b. The slope, $1440, is the amount the tuition increases each year and the y-intercept is the tuition cost in 2010, $36,208. Chapter 1 c. Use the equation in part (a) to estimate the cost of tuition in 2015. y = 1440(5) + 36,208 = 43,408 The approximate cost of tuition for a private school in 2015 is $43,408. 3. Your friend is incorrect. The slope of the line of best fit is negative, thus the correlation coefficient is negative. 19 524 Step 3 Substitute y = — and z = — into an original 61 61 equation and solve for x. −x + 2y + z = 11 ( ) 524 19 −x + 2 — + — = 11 61 61 396 x=— 61 396 524 19 The solution is —, —, — . 61 61 61 ( 4. System A Step 1 Rewrite the system as a linear system in two variables. ) So, the linear systems are ordered as B, C, A, according to the number of solutions from least to greatest. 2x + 4y − z = 7 −2x + 12y + 24z = 26 ___________________ 16y + 23z = 33 −14x − 28y + 7z = −49 14x + 28y − 7z = 49 _____________________ 0=0 Because you have obtained the identity 0 = 0, the system has infinite solutions. 5. a. Write a verbal model for the situation. 30 ⋅ Comedy, c + 60 ⋅ Drama, d ⋅ + 60 Reality-based, r = 360 Comedy, c + Drama, d + Reality-based, r = 7 Drama, d = 2 ⋅ Comedy, c The system of equations that models the situation is 30c + 60d + 60r = 360 c+d+r=7 System B d = 2c Step 1 Rewrite the system as a linear system in two variables. 3x − 3y + 3z = 5 −3x + 3y − 3z = 24 _________________ 0 = 29 Because you have obtained a false equation, the system has no solution. System C Step 1 Rewrite the system as a linear system in two variables. 4x − y + 2z = 18 −4x + 8y + 4z = 44 _________________ 7y + 6z = 62 −3x + 6y + 3z = 33 3x + 3y − 4z = 44 _________________ 9y − z = 77 Step 2 Solve the new linear system for both of its variables. 7y + 6z = 62 54y − 6z = 462 _____________ 61y = 524 524 y =— 61 19 z =— 61 b. Step 1 Rewrite the system as a linear system in two variables. 30c + 60d + 60r = 360 − 60c − 60d − 60r = −420 _______________________ −30c = −60 Solve the equation −30c = −60 for c to obtain c = 2. Step 2 Substitute c = 2 into the equation d = 2c. So, d = 2(2) = 4. Step 3 Substitute c = 2 and d = 4 into c + d + r = 7. So, 2 + 4 + r = 7, which gives r = 1. So, there are 2 episodes of comedies, 4 episodes of dramas, and 1 reality-based episode. c. Use a method similar to part (a) to write a system of equations that models the situation. 30c + 60d + 60r = 360 c+d+r=6 Step 1 Rewrite the system as a linear system in two variables. 30c + 60d + 60r = 360 − 60c − 60d − 60r = −360 ________________________ −30c = 0 c=0 Substitute c = 0 to write a new linear system. 60d + 60r = 360 d+r=6 Algebra 2 Worked-Out Solutions 39 Chapter 1 Step 2 Solve the new linear system for both of its variables. 60d + 60r = 360 −60d − 60r = −360 0=0 Because you obtain the identity 0 = 0, the system has infinitely many solutions. Step 3 Describe the solutions of the system using an ordered triple of the form (c, d, r). One way to do this is to solve the equation d + r = 6 for r to obtain r = 6 − d. So, the solutions of the system are of the form (0, d, 6 − d). The two DVDs have a different number of comedies because the first DVD has 2 episodes of comedy and the second DVD has none. The two DVDs may have a different number of dramas and reality-based episodes because the second DVD can have from 0 to 6 dramas and from 0 to 6 reality-based episodes. 6. Equation D models the situation. From the graph, the y-intercept is (0, 450). Another point on the line is (5, 400). So, the slope is y = mx + 450 400 = m(5) + 450 −50 = 5m −10 = m. Therefore, an equation of the line is y = −10x + 450, or 10x + y = 450. 7. Sample answer: The order of the transformations is F followed by A, then followed by C. 8. For every 1 unit increase in x, f (x) increases by 3. So, the data values for (x, f (x)) can be modeled by a linear equation with a slope of 3. Use the point (− 2, −14) to determine the equation. y + 14 = 3(x + 2) y + 14 = 3x + 6 y = 3x − 8 An equation relating x and f (x) is f (x) = 3x − 8 and f (22) = 3(22) − 8 = 58. For every 1 unit increase in x, g(x) increases by 4. So, the data values for (x, g(x)) can be modeled by a linear equation with a slope of 4. Use the point (0, −10) to determine the equation. y + 10 = 4(x − 0) y + 10 = 4x y = 4x − 10 An equation relating x and g(x) is g(x) = 4x − 10 and g(22) = 4(22) − 10 = 78. So, f (22) < g(22). 40 Algebra 2 Worked-Out Solutions