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trigonometric functions: Given the function \( f(x)=2 \sin (3 x)+1 \), find the amplitude, period, and phase shift of the function.

Ask by Clark Horton. in the United States
Mar 13,2025

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The function \( f(x)=2 \sin (3x)+1 \) has: - **Amplitude**: 2 - **Period**: \( \frac{2\pi}{3} \) - **Phase Shift**: 0

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The amplitude of the function \( f(x) = 2 \sin(3x) + 1 \) is determined by the coefficient of the sine function, which is 2. This means the function oscillates between 3 (1 + 2) and -1 (1 - 2). The period can be found using the formula \( \frac{2\pi}{b} \), where \( b \) is the coefficient of \( x \) inside the sine function. Here, \( b = 3 \), so the period is \( \frac{2\pi}{3} \). Lastly, the phase shift is determined by any horizontal shifts according to the formula \( -\frac{c}{b} \) from \( \sin(bx + c) \). Since there’s no horizontal shift in this function, the phase shift is 0. In summary: Amplitude = 2, Period = \( \frac{2\pi}{3} \), Phase Shift = 0.

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