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**Unformatted text preview: **ase shift. Instead of the graph starting at x = π , it ends there. Remember, however, that the
2
graph presented in Example 10.5.1 is only one portion of the graph of y = g (x). Indeed, another
complete cycle begins at x = π , and this is the cycle Theorem 10.23 is detecting. The reason for the
2
discrepancy is that, in order to apply Theorem 10.23, we had to rewrite the formula for g (x) using
the odd property of the sine function. Note that whether we graph y = g (x) using the ‘quarter
marks’ approach or using the Theorem 10.23, we get one complete cycle of the graph, which means
we have completely determined the sinusoid. Example 10.5.2. Below is the graph of one complete cycle of a sinusoid y = f (x).
y 3
−1, 5
2 5, 5
2
2 1
1, 1
22 −1 1 7, 1
22 2 3 4 5 x −1 −2 3
2, − 2 One cycle of y = f (x).
1. Find a cosine function whose graph matches the graph of y = f (x).
2. Find a sine function whose graph matches the graph of y = f (x). 10.5 Graphs of the Trigonometric Functions 679 Solution.
1. We ﬁt the data to a function of the form C (x) = A cos(ωx + φ) + B . Since one cycle is
graphed over the interval [−1, 5], its period is 5 − (−1) = 6. According to Theorem 10.23,
φ
6 = 2π , so that ω = π . Next, we see that the phase shift is −1, so we have − ω = −1, or
ω
3
π
35
φ = ω = 3 . To ﬁnd the amplitude, note that the range of the sinusoid is − 2 , 2 . As a result,
1
the amplitude A = 1 5 − − 3 = 2 (4) = 2. Finally, to determine t...

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