Lesson 16.1 Step It Out Do the Math (Example 1) Identify the parent function. The parent function is y = cos x because the y-intercept is maximum a value. Identify the amplitude. From the graph, you can see that the maximum value is 2.5 and the minimum value is - 2.5 . Halfway between these values is the midline, which is at y = 0. The amplitude a of the graph is 2.5 . Identify the period. Then solve for b. The graph repeats itself in intervals of 2πb = π π . So, the period of the graph is π . → b = 0.5 Write an equation. ( y = 2.5 cos ) 2 x Do the Math (Example 2) Identify the frequency. 1 = 2000 Hz frequency = _ 1 = _ period 0.0005 period _ = 0.0005 a = 2, b = _ 2π 2π (( 5 )) P(t) = 2 sin _ 2π t = 4 sin 4000 πt 0.0005 y 4 3 2 Graph the function. 1 0 -1 x 0.00025 0.0005 0.00075 0.001 -2 -3 -4 -5 Algebra 2 AK150 Journal and Practice Workbook © Houghton Mifflin Harcourt Publishing Company Write an equation. Lesson 16.1 More Practice 1. 1 ; b = a = _ 2 1 _ 1 ; amplitude = 3 1 ; period = _ 2 2π _ 3 4. 1 ; amplitude = 5, period = 2 a = -5, b = _ π y y 6 4 0.5 2 x 0 x 0 π 2π 5. 6. a = 2; b = 2; amplitude = 2; period = 4 π 7. y 5 y 2 x 0 2π -1 © Houghton Mifflin Harcourt Publishing Company amplitude = 2; period = _ 2π ; y = 2 cos 5x 5 3 amplitude = _ , period = π; y = _ 3 sin 2x 4 4 3 1 4π 6π 1 8π 0 -2 -1 -3 -2 0.009 -5 2 y = 5 sin(500πx) 1 -1 0.006 -4 y 0 x 0.003 -3 a = -3; b = 4; amplitude = 3; period = 8 π 3 8π 4 2 3. 6π -6 -1 3 4π -4 -0.5 2. 2π -2 frequency = _ 1 = 250 Hz 0.004 x 4π 8π 12π 8. -2 Jeremy’s error was in multiplying instead of dividing by the horizontal stretch factor. x y = 4 cos _ 2 -3 Algebra 2 AK151 Journal and Practice Workbook 4πx 3 9A. y = -8 cos _ 10. y = 1.4 sin 400πx B. amplitude: 8; period: 1.5 C. 10 2 y y 1 5 0 0 x 0.01 0.02 x 2.5 5 -1 -5 -2 11. period = 2πb = 14 -10 7 → _ π 1 = _ π b=_ 14 = _ 7 2π b The correct answer choice is D. © Houghton Mifflin Harcourt Publishing Company Algebra 2 AK152 Journal and Practice Workbook Lesson 16.2 Step It Out Do the Math (Example 1) Identify the period. Then solve for b, and substitute into the function rule. The graph repeats itself in intervals of graph is πb = π _ 3 . So, the period of the π _ 3 . π _ 3 → b = ( f(x) = a tan ) _ 1 3 x 3 Use the point on the graph to find a. 2 ( ( = a tan π _ 3 - 12 ( ) ( ) π = a tan - _ 4 )) = a -1 -2 = a Write the equation. ( y = -2 tan 3 ) x © Houghton Mifflin Harcourt Publishing Company Do the Math (Example 2) Subtract two consecutive asymptotes to find the period of the tangent function. ( ) π = _ 3π period of f(x): π - - _ 2 2 ( ) π = _ period of g(x): _ π - - _ 3π 4 4 2 The period of f(x) is less than the period of g(x). Do the Math (Example 3) The height h of the climber and the distance 50 feet form two legs of a right triangle. h . Use this information to write tan θ = __ 50 Solve for h to write the function. The tangent function h = 50 tan θ gives the height of the climber as a function of the observer’s viewing angle. Evaluating the function for θ = 80° gives h ≈ 283.6, which is a height of 283.6 feet. The cliff is 300 feet high, so this viewing angle would occur when the climber is at about 284 feet. Algebra 2 AK153 Journal and Practice Workbook Lesson 16.2 More Practice 1. 1 ; b = _ a = _ 1 ; period = _ π 2 3 3 4 5. y 2 πb = _ 2π → b = _ 3 3 3 x f(x) = a tan _ 2 ( ) 2 -π 3 -π 6 π 6 0 -2 ( ( )) π 3 − _ 4 = a tan _ 2 6 ( ) −4 = a y 6. x π 0 2π () x f(x) = a tan _ 3 -4 ( 3π 1 )(− _ −2.5 = a tan ((_ 4 )) 3 y ( ) x -π 8 π 8 0 -2 2.5 = a x y = 2.5 tan _ 3 π 4 7. -4 The distance from the observer to the base of the cliff is 50 meters. The cliff is 150 meters high. 4. 1 ; b = _ 1 a = − _ 1 ; period = _ 2π 2 2 4 The climber is climbing at a rate of 1.5 m/min. The height of the climber h and the view angle h θ is related by the equation tan θ = __ 50 . y The height of the climber h is a function h = 1.5t of time t. 2 x -1 -0.5 0 0.5 tan θ = _ 1.5t 50 _ tan θ = 3t 100 1 -2 100 tan θ = 3t -4 Algebra 2 tan θ =t _ 100 3 AK154 Journal and Practice Workbook © Houghton Mifflin Harcourt Publishing Company π −2.5 = a tan − _ 4 −2.5 = a(−1) 2 -π 4 ) 3π , −2.5. A point on the graph is − _ 4 1 ; period = _ a = - 5; b = _ π 4 4 4 The graph repeats itself in intervals of 3π. The period of the graph is 3π. πb = 3π → b = 3 -2 3. ( ) y = −4 tan _ 3x 2 2 -2π -π ) π 4 = a tan − _ 4 4 = a(−1) a = 3; b = 4; period = 4π 4 ( π , 4 . A point on the graph is − _ 6 πx 3 -4 2. The graph repeats itself in intervals of __ 2π . 3 2π __ The period of the graph is 3 . yes; The viewing angle would reach 70° when h , or 137.4 ≈ h, when the climber is about tan 70° = __ 50 137 meters above the ground. 8. The domain of the function is ⎧ ⎫ π ⎨ x | x ≠ b( __ + nπ)⎬ . It is not affected by a. 2 ⎩ ⎭ The range of the function is all real numbers, It is not affected by the values of a and b. period of the tangent function. ( ) 3 = _ 3 - ( − _ 6 = 3 period of g(x): _ 2 2) 2 3 = _ 9 - − _ 12 = 3 period of f(x): _ 4 4 4 The period of f(x) is equal to the period of g(x). π 6 1 6 11. πb = _ → b = _ h 300 h B. tan 60° = _ 300 √3 = _ h 300 300 √3 = h 9A. tan θ = _ ( ) 1 (x) y = tan _ _ 1 6 ― ― y = tan(6x) The correct answer choice is D. 519.6 ≈ h C. 10. Subtract two consecutive asymptotes to find the tan θ = _ 24t 300 300 tan θ = t _ 24 © Houghton Mifflin Harcourt Publishing Company 12.5 tan θ = t Algebra 2 AK155 Journal and Practice Workbook Lesson 16.3 Step It Out Do the Math (Example 1) An equation of the graph resulting from the reflection across y= π is g(x) = -5 sin 4 (x - __ ) + 5. 8 5 The graph of g has the same shape as the graph of f, but one of ( its minimum points is located at A phase shift of ) _ π , 0 . 4 _ π units to the left maps the minimum point on 4 the graph of g to the minimum point on the graph of f. ( ) ( _ π , 0 → 4 ) 0 , 0 Write the equation of a function h(x) whose graph coincides with the graph of f. π - _ h(x) = -5 sin 4 x + _ π + 5 4 8 (( ( = -5 sin 4 x + ) ) ) _ π + 5 8 Do the Math (Example 2) Determine the points where the graph intersects the midline. h(t) = 75 when cos _ π t = 0 10 5π , . . .. The value of the cosine function is 0 for _ π , _ 3π , _ 2 2 2 π t = _ _ π _ 3π π t = _ 10 10 2 2 t=5 t = 15 The graph intersects the midline twice during one cycle, at the points ( 5, 75) and (15, 75). The cycle begins with the point on the observation wheel nearest the ground, so one minimum point is ( 0, 15). A second occurs at the end of the cycle (the period) at the point ( 20, 15). The only maximum point occurs midway through the cycle at ( 10, 135). (0, 15): The observation wheel begins with the point at a height of 15 meters. (5, 75): The point reaches the height of the wheel’s axle, 75 meters, in 5 minutes. (10, 135): At 10 minutes, the point reaches its maximum height of 135 meters at the top of the wheel. Algebra 2 AK156 Journal and Practice Workbook © Houghton Mifflin Harcourt Publishing Company Identify the key parameters of the function. period: 2π • _ 10 π = 20; midline: h(t) = 75 ; amplitude: 60; maximum: 75 + 60 = 135; minimum: 75 − 60 = 15 ( 15, 75): The point comes back down to the height of the axle, 75 meters, in 15 minutes. (20, 15): One trip around the observation wheel takes 20 minutes, ending with the point back at a height of 15 meters. Make a graph of the function. y 120 90 60 30 x 0 10 20 30 Lesson 16.3 More Practice 1. The amplitude is 2, the midline is at 3, 2π the period is __ , and the phase shift is 3 πunits to the left. 4 3. y 2 x 0 π 6 © Houghton Mifflin Harcourt Publishing Company -2 π 3 π 2 2π 3 5π 6 4. -4 2. The amplitude is 4, the midline is at -2, the units to period is 4 π, and the phase shift is π the right. 8 5. y 4 x 0 -4 π 2π 3π 4π 5π y = 0.8 sin(2.5x) − 0.2, y = −0.8 sin(2.5(x + 0.4π)) − 0.2; The period is 0.8π, so b = 0.4, the amplitude is 0.8 because the difference between the maximum and minimum is 1.6 and the midline is at −0.2. A phase shift of 0.4π units to the right or left with a reflection over the midline makes a graph that coincides. ( ) x − 5π y = −6 cos(__ 5x ) + 2, y = 6 cos _____ + 2; The 5 period is 10π, so b = 5, the amplitude is 6 because the difference between the maximum and minimum is 12, and the midline is at 2. A phase shift of 5π units to the right or left with a reflection over the midline makes a graph that coincides. π y = −3 tan(2x) + 6, y = −3 tan 2(x − __ + 6; 2) π 1 __ __ The period is 2 , so b = 2 , a is negative because each period goes down to the right. The point at ( 0, 6) leads to y = a tan(2x) + 6, since the π π asymptotes are at x = ± __ . The point at ( __ , 3 4 8 ) provides a. ( ( )) π + 6 3 = a tan 2_ 8 −3 = a tan _ π 4 −3 = a () -8 A phase shift of __ π2 units to the right or left makes a graph that coincides. Algebra 2 AK157 Journal and Practice Workbook 6. π y = −3 cos(6x) − 2, y = 3 sin 6(x + __ − 2; 4) 9. π , so b = __ 16 , the amplitude is 3 The period is __ 3 because the difference between the maximum and minimum is 6, and the midline is at -2. A π phase shift of __ units to the left with changing 4 from cosine to sine makes a graph that coincides. 7. (( )) π 10. y = 4 tan __ x − __ 12 + 2; The period is 5 π5 , The point at __12 , 2 3 − ( −2) = 5, so b = __ The amplitude is 60, the midline or vertical shift 2π = 10, and the phase shift is 70, the period is ___ π __ ( 5 ) is 1.4 units to the right. The maximum height is 60 + 70 = 130. h(t) ( ) represents the point ( h, k), because it is halfway between two asymptotes. The point __ 47 , 6 provides a. ( ( )) ( ( )) ( ) y = a tan _ π x − _ 1 + k 5 2 6 = a tan _ π _ 7 − _ 1 + 2 5 4 2 () 4 = a tan _ π 4 4=a 120 80 π 11. A phase shift of __ units to the right with a 12 40 reflection over the midline makes a graph that coincides. The correct answer choice is C. t 0 8. π The values are h = __ and k = 5, so each of the 2 five key points of the graph will move 5 units up π and __ units to the right. 2 4 8 12 © Houghton Mifflin Harcourt Publishing Company The amplitude is 33, the midline or vertical shift 2π is 33, the period is __ = __ 12 , and the phase shift 4π is __ 18 unit to the right. The bicycle wheel makes a complete cycle in 0.5 seconds, has a radius of 33 cm, and the wheel stays on the ground, but a point on the wheel goes from 0 to 66 cm and back to 0. h(t) 60 40 20 t 0 Algebra 2 0.5 1 1.5 AK158 Journal and Practice Workbook Lesson 16.4 Step It Out Do the Math (Example 1) The sine function that models the data is: ( 1 m T = -15.8 sin _____ -1.99 3.99 ) + 60.6 . Note: Different regression calculators may provide different parameters that result in equivalent graphs. Check that students' graphs coincide with this graph if their calculator has provided different parameter values. T 60 40 20 m 0 2 4 6 8 10 Do the Math (Example 2) The m-coordinate of the point ( 10.414, 11)means the month is October and the day is 0.414(31) = 12.834 days after the 15th. So the model gives the day October 28 as the first day with no more than 11 hours of daylight in New Orleans. The m-coordinate of the point ( 1.965, 11)means the month is February and the day is 0.035(28) = 0.98 days before the 15th. So, the model gives the day February 14 as the last day with no more than 11 hours of daylight in New Orleans. © Houghton Mifflin Harcourt Publishing Company From about October 28 through approximately February 14, New Orleans, Louisiana gets no more than 11 hours of daylight. Lesson 16.4 More Practice 1. Roughly fit a sine model of the form y = a sin __b1(x - h) + k. The period is 12 months. 2πb = 12 6 b=_ π The horizontal midline is between the highest 94 + 50 and lowest data points which is k = ______ = 72. 2 The closest occurrence of temperature close to 67 is in month 3, so h = 3. The amplitude is half the distance between the highest and lowest data points or 22. Since temperatures increase near the beginning, let a = 22. The rough model is π( x - 3) + 72. y = 22 sin _ 6 Algebra 2 2. Possible answer: 1 (x -22.4) + 71.7 y ≈ -21.7 sin _ 1.95 3. The calculator model seems to go through or very close to all of the points, but the rough model has a phase shift that is not far enough to the right to match the data. 4. The value of R 2for the calculator model is 0.9882, which is very close to 1, so this is a strong correlation to the data. 5. AK159 The m-coordinate of the point ( 5.287, 85) means the month is May and the day is 0.287(31) = 8.897 days after the 15th. So, the model gives the day May 23 as the first day warmer than 85 in Oklahoma City. Journal and Practice Workbook The m-coordinate of the point ( 8.849, 85) means the month is September and the day is 0.151(30) = 4.53 days before the 15th. So, the model gives the day September 11 as the last day above 85 in Oklahoma City. From about December 18 through approximately February 14, Oklahoma City has average lows of 32 degrees or colder. 11. Roughly fit a sine model of the form y = a sin __1(x - h) + k. The period is 12 months. b From about May 23 through approximately September 11, Oklahoma City averages 85 degrees or warmer. 6. 2πb = 12 6 π b=_ Roughly fit a sine model of the form y = a sin __b1(x - h) + k. The period is 12 months. 2πb = 12 The horizontal midline is between 76 and 52, 52 k = ______ 76 +2 = 64. The first occurrence of temperature close to 64 is in month 4, so h = 4. 6 b=_ π The amplitude is half the distance between the maximum and minimum or 12. Since temperatures increase near the beginning, let a = 12. The horizontal midline is between the highest and 29 lowest data points which is k = ______ 72 +2 = 50.5. The closest occurrence of temperature close to 50.5 is in month 4, so h = 4. The amplitude is half the distance between the highest and lowest data points or 21.5. Since temperatures increase near the beginning, let a = 21.5. h 60 The rough model is y = 21.5 sin __π6 (x - 4) + 50.4. 7. 9. 40 20 Possible answer: 1 (x - 28.3) + 50.6 y ≈ 21.6 sin _ -1.93 m 0 The calculator model and the rough model seem to go through or very close to all of the points. The value of R 2for the calculator model is 0.996, which is very close to 1, so this is a very strong correlation to the data. 2 4 6 8 10 12. no; The data are periodic, but there is a second maximum in the winter months that is lower than the maximum in the summer months. 13. The amplitude is half the distance between the max and min or |a| = _______ 14.6 - 9.4 = 2.6. Since 2 daylight increases near the beginning of the year, let a = 2.6. The correct answer choice is A. 10. The m-coordinate of the point (12.096, 32) means the month is December and the day is 0.0967(31) = 2.976 days after the 15th. So, the model gives the day December 18 as the first night colder than 32 in Oklahoma City. The m-coordinate of the point ( 2.026, 32) means the month is February and the day is 0.026(28) = 0.728 days before the 15th. So, the model gives the day February 14 as the last day below 32 in Oklahoma City. Algebra 2 AK160 Journal and Practice Workbook © Houghton Mifflin Harcourt Publishing Company 8. The rough model is y = 12 sin __π6 ( x - 4) + 64. The calculator gives the equation y ≈ 12.61 sin(0.51x - 21.2) + 64.79.
