SECTION 6.4 Graphs of the Sine and Cosine Functions 435 Check: Graph π ( ) = − y x 2sin 2 using transformations.Which graphing method do you prefer? Now Work PROBLEM 41 USING KEY POINTS If the function to be graphed is of the form ω( ) = + y A x B sin ω [ ] ( ) = + y A x B or cos , first graph ω ω [ ] ( ) ( ) = = y A x y A x sin or cos and then use a vertical shift. Figure 63 x y 4 2 24 (0, 24) (2, 24) (1, 4) ( , 0) 22 (a) 2 1 ( , 0) 1 –– 2 3 –– 2 y 4 –4 –1 (0, 24) (21, 4) (2 , 0) 2 (b) y 5 24 cos (px) 2 1 (2, 24) (1, 4) ( , 0) – 2 1 –– ( , 0) ( , 0) 2 (c) y 5 24 cos (px) 2 2 1 – 3 2 5 – – – (21, 2) Subtract 2; Vertical shift down 2 units x y 2 –6 1 2 –1 (0, 26) (2, 26) (1, 2) (2 , 22) 2 , 22) x – 1 –– ( 1 –– 2 , 22) ( 3 –– 2 Graphing a Sinusoidal Function Using Key Points Graph π( ) = − − y x 4cos 2 using key points. Use the graph to determine the domain and the range of π( ) = − − y x 4cos 2. Solution EXAMPLE 7 Begin by graphing the function π( ) = − y x 4cos . Comparing π( ) = − y x 4cos to ω( ) = y A x cos , note that = − A 4 and ω π = . The amplitude is = − = A 4 4, and the period is π ω π π = = = T 2 2 2. The graph of π( ) = − y x 4cos lies between −4 and 4 on the y-axis. One cycle begins at = x 0 and ends at = x 2. Divide the interval [ ] 0,2 into four subintervals, each of length ÷ = 2 4 1 2 . The x-coordinates of the five key points are + = + = + = + = 0 0 1 2 1 2 1 2 1 2 1 1 1 2 3 2 3 2 1 2 2 1st x-coordinate 2nd x-coordinate 3rd x-coordinate 4th x-coordinate 5th x-coordinate Now evaluate π( ) = − y x 4cos at each of the five x-coordinates listed above. ( ) ( ) ( ) ( ) ( ) − − 0, 4 1 2 , 0 1,4 3 2 , 0 2, 4 Plot these five points, and fill in the graph of the cosine function as shown in Figure 63(a). Extend the graph in each direction to obtain Figure 63(b), the graph of π( ) = − y x 4cos . A vertical shift down 2 units gives the graph of π( ) = − − y x 4cos 2, as shown in Figure 63(c). The domain of π( ) = − − y x 4cos 2 is the set of all real numbers, or ( ) −∞ ∞, . The range of π( ) = − − y x 4cos 2 is { } − ≤ ≤ y y 6 2 , or [ ] −6,2 . Now Work PROBLEM 51
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