Stitz-Zeager_College_Algebra_e-book

2 x2 1 1 for x 0 2 x2 1 2 1 2 14 x x

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Unformatted text preview: he vertical shift, we 22 2 1 average the endpoints of the range to find B = 1 5 + − 3 = 2 (1) = 1 . Our final answer is 22 2 2 C (x) = 2 cos π x + π + 1 . 3 3 2 2. Most of the work to fit the data to a function of the form S (x) = A sin(ωx + φ) + B is done. The period, amplitude and vertical shift are the same as before with ω = π , A = 2 and 3 1 B = 2 . The trickier part is finding the phase shift. To that end, we imagine extending the graph of the given sinusoid as in the figure below so that we can identify a cycle beginning φ 7 π at 7 , 1 . Taking the phase shift to be 7 , we get − ω = 7 , or φ = − 2 ω = − 7 π = − 76 . 22 2 2 23 π 7π 1 Hence, our answer is S (x) = 2 sin 3 x − 6 + 2 . y 3 5, 5 2 2 1 13 , 1 2 2 7, 1 22 −1 1 2 3 4 5 6 19 , 5 2 2 7 8 9 10 x −1 −2 8, − 3 2 Extending the graph of y = f (x). Note that each of the answers given in Example 10.5.2 is one choice out of many possible answers. 11 For example, when fitting a sine function to the data, we could h...
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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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