Determine the amplitude, period, and phase shift of the function. Graph the function. \( y=\sin (2 x-\pi) \) The amplitude is 1 . (Simplify your answer.) The period is (Type an exact answer, using \( \pi \) as needed. Use integers or fractions for any numbers in the expression.)

**Step 1. Write the function in standard form** We have \[ y = \sin(2x - \pi) \] The standard form of a sine function is \[ y = A \sin(Bx - C), \] where: - \(A\) is the amplitude, - \(\frac{2\pi}{|B|}\) is the period, and - \(\frac{C}{B}\) is the phase shift to the right. **Step 2. Determine the amplitude** The amplitude is given by the absolute value of \(A\). Here, \[ A = 1, \] so the amplitude is \[ 1. \] **Step 3. Determine the period** The period of a sine function is given by \[ \text{Period} = \frac{2\pi}{|B|}. \] In our function, \(B = 2\), so \[ \text{Period} = \frac{2\pi}{2} = \pi. \] **Step 4. Determine the phase shift** The phase shift is calculated by \[ \text{Phase shift} = \frac{C}{B}. \] In our function, the expression inside the sine is \(2x - \pi\), so \(C = \pi\). Thus, \[ \text{Phase shift} = \frac{\pi}{2}. \] This indicates the graph is shifted \(\frac{\pi}{2}\) units to the right. **Summary of the parameters:** - **Amplitude:** \(1\) - **Period:** \(\pi\) - **Phase Shift:** \(\frac{\pi}{2}\) to the right **Step 5. Graph of the function** To graph \(y = \sin(2x-\pi)\): 1. **Start with the basic sine curve** \(y = \sin(x)\). 2. **Horizontal scaling:** Since the period is \(\pi\) (instead of \(2\pi\)), the graph completes one cycle from \(x = 0\) to \(x = \pi\). 3. **Phase shift:** Shift the entire graph to the right by \(\frac{\pi}{2}\). So, instead of starting at \(x = 0\), the cycle now starts at \(x = \frac{\pi}{2}\). 4. **Key points:** - The basic sine function has key points at \(0\), \(\frac{\pi}{2}\), \(\pi\), \( \frac{3\pi}{2}\), and \(2\pi\). - For \(y = \sin(2x-\pi)\) after adjusting: - When \(2x - \pi = 0\), \(2x = \pi\) so \(x = \frac{\pi}{2}\). This is the starting point (\(y = 0\)). - When \(2x-\pi = \frac{\pi}{2}\), \(2x = \frac{3\pi}{2}\) so \(x = \frac{3\pi}{4}\) (\(y = 1\)). - When \(2x-\pi = \pi\), \(2x = 2\pi\) so \(x = \pi\) (\(y = 0\)). - When \(2x-\pi = \frac{3\pi}{2}\), \(2x = \frac{5\pi}{2}\) so \(x = \frac{5\pi}{4}\) (\(y = -1\)). - When \(2x-\pi = 2\pi\), \(2x = 3\pi\) so \(x = \frac{3\pi}{2}\) (\(y = 0\)). 5. **Sketch:** - Draw the \(x\)-axis and mark these key \(x\)-values. - Plot the corresponding \(y\) values for \(y = 0\), \(y = 1\), and \(y = -1\). - Draw a smooth sine curve through these points, starting at \(\frac{\pi}{2}\) and ending at \(\frac{3\pi}{2}\) for one cycle. **Final Answers:** - Amplitude: \(1\) - Period: \(\pi\) - Phase Shift: \(\frac{\pi}{2}\) to the right