446 CHAPTER 6 Trigonometric Functions Solution Figure 71 shows the steps using transformations. Graphing a Function of the Form ω( ) = + y A x B cot Graph ( ) = y x 3cot 2 . Use the graph to determine the domain and the range of ( ) = y x 3cot 2 . Solution EXAMPLE 2 Figure 72 shows the steps using transformations. Figure 71 y y x 1 p p–– 2 p–– 2 2 3p ––– 2 ( , 1) (0, 0) (p, 0) (0, 0) (p, 0) p –– 4 (2 , 21) p –– 4 (a) y 5 tan x (b) y 5 2 tan x x 1 2 1 2 21 2 22 p p–– 2 p–– 2 2 3p ––– 2 ( , 2) p –– 4 (2 , 22) p –– 4 Multiply by 2; Vertical stretch by a factor of 2 Subtract 1; Vertically shift down 1 unit x 5 x 5 x 5 x 5 x 5 x 5 y (c) y 5 2 tan x 21 x 21 (p, 21) p–– 2 p–– 2 2 3p ––– 2 ( , 1) (0, 21) p –– 4 (2 , 23) p –– 4 x 5 x 5 x 5 The domain of = − y x 2tan 1 is π { } ≠ x x k k 2 , isanoddinteger , and the range is the set of all real numbers, or ( ) −∞ ∞, . Check: Graph = Y 2 1 tan −x 1 to verify the graph shown in Figure 71(c). Now Work PROBLEM 21 The graph of ω( ) = + y A x B cot has characteristics similar to those of the tangent function. The cotangent function ω( ) = + y A x B cot has period π ω . The cotangent function has no amplitude. The role of A is to provide the magnitude of the vertical stretch; the presence of B indicates a vertical shift. Figure 72 x y 2 ( , 3) p –– 4 ( , 0) p –– 2 ( , 3) 5p ––– 4 ( , 0) 3p ––– 2 ( , 23) 3p ––– 4 ( , 23) 7p ––– 4 x 5 0 x 5 p x 5 2p y x 2 ( , 1) p –– 4 ( , 0) p –– 2 ( , 1) 5p ––– 4 ( , 0) 3p ––– 2 ( , 21) 3p ––– 4 ( , 21) 7p ––– 4 x 5 0 x 5 p x 5 2p x y 2 ( , 3) p –– 8 ( , 0) p –– 4 ( , 3) 5p ––– 8 ( , 0) 3p ––– 4 ( , 23) 3p ––– 8 ( , 23) 7p ––– 8 x 5 0 x 5 p p–– 2 x 5 Multiply by 3; Vertical stretch by a factor of 3 Replace x by 2x ; Horizontal compression by a factor of (a) y 5 cot x (c) y 5 3 cot (2x ) (b) y 5 3 cot x 1 –– 2 The domain of ( ) = y x 3cot2 is π { } ≠ x x k k 2 , isaninteger , and the range is the set of all real numbers, or ( ) −∞ ∞, . Check: Graph ( ) = Y x 3cot 2 1 to verify the graph in Figure 72(c).
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