Stitz-Zeager_College_Algebra_e-book

This transforms the equation into a polynomial in

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Unformatted text preview: ertical asymptotes. y x −π f (x) (x, f (x)) undefined −π 2 2 0 1 (0, 1) π 2 0 π 2,0 π 2 −π, 2 2 undefined 1 −π −π 2 π 2 π x −1 −2 One cycle of y = 1 − tan x 2 . We see that the period is π − (−π ) = 2π . π 2. The ‘quarter marks’ for the fundamental cycle of the cotangent curve are 0, π , π , 34 and π . 42 π π To graph g (x) = 2 cot 2 x + π + 1, we begin by setting 2 x + π equal to each quarter mark and solving for x. a 0 π 4 π 2 3π 4 π π 2x + π = a π 2x + π = 0 π π 2x + π = 4 π π 2x + π = 2 π 3π 2x + π = 4 π 2x + π = π x −2 3 −2 −1 1 −2 0 690 Foundations of Trigonometry We now use these x-values to generate our graph. y x −2 g (x) undefined −3 2 3 3 −2, 3 −1 1 (−1, 1) −1 2 −1 0 3 (x, g (x)) undefined 2 − 1 , −1 2 1 −2 x −1 −1 One cycle of y = 2 cot π x 2 + π + 1. We find the period to be 0 − (−2) = 2. As with the secant and cosecant functions, it is possible to extend the notion of period, phase shift and vertical shift to the tangent and cotangent functions as we did for the cosine and sine functions in Theorem 10.23. Since the number...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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