Determine the amplitude, period, and phase shift of the function. Graph the function. \( y=\frac{1}{4} \sin \left(x-\frac{\pi}{4}\right) \) The amplitude is \( \square \). (Simplify your answer.)

**Step 1. Identify the amplitude** The standard form of a sine function is \[ y = A \sin(Bx - C) \] where: - \( A \) is the amplitude, - \( B \) affects the period, - \( C \) affects the phase shift. Given the function \[ y=\frac{1}{4} \sin \left(x-\frac{\pi}{4}\right), \] we can see that \( A = \frac{1}{4} \). Thus, the amplitude is \[ \boxed{\frac{1}{4}}. \] **Step 2. Determine the period** The period \( P \) of the sine function is given by \[ P = \frac{2\pi}{B}. \] In our function, the coefficient of \( x \) inside the sine function is \( 1 \) (since it appears as \( x-\frac{\pi}{4} \), which is equivalent to \( 1 \cdot x - \frac{\pi}{4} \)). Thus, \( B = 1 \). Hence, the period is \[ P = \frac{2\pi}{1} = 2\pi. \] **Step 3. Determine the phase shift** The phase shift is found by solving \( Bx - C = 0 \), i.e., \[ x - \frac{\pi}{4} = 0. \] Solving for \( x \), we get \[ x = \frac{\pi}{4}. \] Since the sine function is shifted to the right by \( \frac{\pi}{4} \), the phase shift is \[ \boxed{\frac{\pi}{4}}. \] **Step 4. Graph the function** 1. **Plot the key points for one period:** - **Start point:** At \( x = \frac{\pi}{4} \), \( \sin\left(\frac{\pi}{4}-\frac{\pi}{4}\right)=\sin(0)=0 \). Point: \(\left(\frac{\pi}{4}, 0\right)\). - **First quarter period:** At \( x = \frac{\pi}{4} + \frac{1}{4}(2\pi) = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4} \), \(\sin\left(\frac{3\pi}{4}-\frac{\pi}{4}\right)=\sin\left(\frac{\pi}{2}\right)=1\). Adjusting for the amplitude, \( y=\frac{1}{4} \cdot 1 = \frac{1}{4} \). Point: \(\left(\frac{3\pi}{4}, \frac{1}{4}\right)\). - **Midpoint:** At \( x = \frac{\pi}{4} + \frac{1}{2}(2\pi) = \frac{\pi}{4} + \pi = \frac{5\pi}{4} \), \(\sin\left(\frac{5\pi}{4}-\frac{\pi}{4}\right)=\sin(\pi)=0\). Point: \(\left(\frac{5\pi}{4}, 0\right)\). - **Third quarter period:** At \( x = \frac{\pi}{4} + \frac{3}{4}(2\pi) = \frac{\pi}{4} + \frac{3\pi}{2} = \frac{7\pi}{4} \), \(\sin\left(\frac{7\pi}{4}-\frac{\pi}{4}\right)=\sin\left(\frac{3\pi}{2}\right)=-1\). Adjusting for the amplitude, \( y=\frac{1}{4} \cdot (-1) = -\frac{1}{4} \). Point: \(\left(\frac{7\pi}{4}, -\frac{1}{4}\right)\). - **End of one period:** At \( x = \frac{\pi}{4} + 2\pi = \frac{9\pi}{4} \), \(\sin\left(\frac{9\pi}{4}-\frac{\pi}{4}\right)=\sin(2\pi)=0\). Point: \(\left(\frac{9\pi}{4}, 0\right)\). 2. **Sketching the graph:** - Draw the horizontal \( x \)-axis and vertical \( y \)-axis. - Mark the points computed: - \(\left(\frac{\pi}{4}, 0\right)\), - \(\left(\frac{3\pi}{4}, \frac{1}{4}\right)\), - \(\left(\frac{5\pi}{4}, 0\right)\), - \(\left(\frac{7\pi}{4}, -\frac{1}{4}\right)\), - \(\left(\frac{9\pi}{4}, 0\right)\). - Draw a smooth sinusoidal curve passing through these points. - Extend the graph periodically to the left and right. **Summary:** - Amplitude: \(\boxed{\frac{1}{4}}\) - Period: \(\boxed{2\pi}\) - Phase shift: \(\boxed{\frac{\pi}{4}}\) The function \( y=\frac{1}{4} \sin \left(x-\frac{\pi}{4}\right) \) is a sine wave scaled vertically by \(\frac{1}{4}\), with its period unchanged (\(2\pi\)) and shifted right by \(\frac{\pi}{4}\).