MCR3U - Practice Test - Periodic Functions - W2012
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1 Name: Date: May 25, 2012 ID: A MCR3U - Practice Test - Periodic Functions - W One cycle of the graph of a periodic function is shown below. State the period and amplitude. 2. One cycle of the graph of a periodic function is shown below. Extend the graph of the function for one more cycle. 3. For the given periodic relation, state the value of f 45 ( ) assuming the relation continues in the same manner. Page 1 of 3
2 4. Given the function g( x) = 0.5fÎÍ 0.25( x + 540) + 1, with a parent function f( x) = sin( x), (a) list the transformations, in the correct order and using (b) list the key features of the transformed function using appropriate terminology. appropriate terminology. 5. Graph y = 2sinÎÍ 3( x + 45) Determine an equation for a cosine function that has a period of 1440, and amplitude of 2, a vertical shift of 1, and a phase shift of Determine an equation for a cosine function that has a period of 1800, and amplitude of 3, a vertical shift of 3, and a phase shift of Determine a sinusoidal equation that satisfies the given data. x y Page 2 of 3
3 9. For the provided graph, determine 2 different sine equations and 2 different cosine equations. At least one of your answers must include a vertical reflection. 10. Choose an appropriate scale and graph one full period of y = 3cosÎÍ 5( x 18) 3, then state domain and range for the single period of your graph. 11. Determine an equation for the provided graph. Page 3 of 3
4 MCR3U - Practice Test - Periodic Functions - W2012 Answer Section 1. ANS: From the graph we see that the function ranges between y = 4 and y = 3, thus the amplitude is 3 ( 4) = From the graph we see that one cycle starts at x = 1 and ends at x = 4, thus the period is 4?? = 5. [Jan. 03, 3M, A6] 2. ANS: [Jan. 03, 3M, A6] 3. ANS: Clearly the first (second) cycle starts at x = 5 (x = 2) and ends at x = 2 (x = 1) so the period is 2 ( 5) = 3 (1 ( 2) = 3) When you divide 45 by the period, 3, you get 15 with a remainder of 0. Thus f( 45) = f( 0) = 1. [Backup 03, 3M, A2] [Backup 03, 3U, A2] 4. ANS: vertical compression by 2 horizontal stretch by 4 shift left by 540 shift up by 1 amplitude = 0.5 period = 1440 axis of curve: y = 1 phase shift = 540 Page 1 of 3
5 5. ANS: 6. ANS: y = 3cosÎÍ 0.2( x + 225) + 3 Parent: y = cos( x) 7. ANS: y = 2cosÎÍ 0.25( x + 720) 1 Parent: y = cos( x) 8. ANS: y = cosîí 3( x + 15) + 5 Parent: y = cos( x) amplitude = 1 period = 120 axis of curve: y = 5 phase shift = ANS: amplitude = 2 period = 120 axis of curve: y = 4 phase shift = 45 Page 2 of 3
6 10. ANS: Ï D = Ô Ìx x ò,18 x 90 Ô ÓÔ Ô R = ÏÔ Ì ÓÔ y y ò, 6 y 0 Ô Ô amplitude = 3 period = 72 axis of curve: y = 3 phase shift = ANS: Graph y = 4sinÎÍ 3( x 15) + 2 Page 3 of 3
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