Determine the amplitude, period, and phase shift (if any) of the given function. Graph the function. \( y=\frac{3}{5} \sin x \) The amplitude is (Simplify your answer. Type an exact answer, using \( \pi \) as needed.)

To analyze the function \( y = \frac{3}{5} \sin x \), we can determine the amplitude, period, and phase shift as follows: 1. **Amplitude**: The amplitude of a sine function \( y = A \sin(Bx + C) + D \) is given by the absolute value of \( A \). In this case, \( A = \frac{3}{5} \). Therefore, the amplitude is: \[ \text{Amplitude} = \left| \frac{3}{5} \right| = \frac{3}{5} \] 2. **Period**: The period of a sine function is calculated using the formula \( \frac{2\pi}{|B|} \), where \( B \) is the coefficient of \( x \). In this case, since \( B = 1 \), the period is: \[ \text{Period} = \frac{2\pi}{|1|} = 2\pi \] 3. **Phase Shift**: The phase shift is determined by the term \( C \) in the function \( y = A \sin(Bx + C) + D \). In this case, there is no horizontal shift since there is no \( C \) term (i.e., \( C = 0 \)). Therefore, the phase shift is: \[ \text{Phase Shift} = 0 \] ### Summary: - Amplitude: \( \frac{3}{5} \) - Period: \( 2\pi \) - Phase Shift: \( 0 \) ### Graphing the Function: To graph \( y = \frac{3}{5} \sin x \): - The graph will oscillate between \( \frac{3}{5} \) and \( -\frac{3}{5} \). - The sine function starts at \( 0 \) when \( x = 0 \), reaches its maximum at \( x = \frac{\pi}{2} \), returns to \( 0 \) at \( x = \pi \), reaches its minimum at \( x = \frac{3\pi}{2} \), and returns to \( 0 \) at \( x = 2\pi \). You can plot the function using graphing software or a graphing calculator to visualize the sine wave with the specified amplitude and period.