Question

find the vertical shift of the sinusoidal function.
simplify any fractions.
vertical shift = \boxed{}

Explanation:

Step1: Recall vertical shift formula

The vertical shift of a sinusoidal function \( y = A\sin(B(x - C)) + D \) (or cosine) is \( D \), which is the midline of the function. The midline is the average of the maximum and minimum values of the function.

Step2: Identify max and min values

From the graph, observe the maximum and minimum \( y \)-values. The graph oscillates around the \( x \)-axis? Wait, no, looking at the graph, the maximum value (peak) and minimum value (trough). Wait, the graph crosses the \( x \)-axis at \( -\frac{\pi}{2} \), \( 0 \) (wait, no, at \( x = 0 \), the function is at \( y = 0 \)? Wait, no, the grid: let's check the midline. Wait, the maximum and minimum: looking at the graph, the highest point (peak) and lowest point (trough). Wait, the graph seems to have a midline at \( y = 0 \)? Wait, no, wait the vertical axis: the graph goes from, let's see, the peak on the left is at \( y = 0 \)? Wait, no, the left peak: when \( x = -\pi \), the \( y \)-value is 0? Wait, no, the grid lines: each square is 1 unit? Wait, the vertical axis has 0, 1, -1, etc. Wait, the function's midline: the average of max and min. Let's find max and min. The maximum \( y \)-value (highest point) and minimum \( y \)-value (lowest point). Looking at the graph, the highest point (peak) is at \( y = 0 \)? No, wait, the left peak: when \( x = -\pi \), the \( y \)-value is 0? Wait, no, the graph crosses the \( x \)-axis at \( -\frac{\pi}{2} \), \( 0 \) (wait, at \( x = 0 \), the function is at \( y = 0 \)? Wait, no, the trough: at \( x = 0 \), the \( y \)-value is 0? Wait, no, maybe I misread. Wait, the vertical shift is the midline, which is the average of the maximum and minimum \( y \)-coordinates. Let's check the graph: the maximum \( y \)-value (peak) and minimum \( y \)-value (trough). Wait, the graph's peaks and troughs: the left peak is at \( y = 0 \)? No, wait, the grid: each horizontal line is 1 unit. The graph oscillates around \( y = 0 \)? Wait, no, the midline is the horizontal line that the graph oscillates around. So if the maximum and minimum are equidistant from the midline. Let's see: the highest point (peak) and lowest point (trough). Wait, the graph's peak on the left: when \( x = -\pi \), \( y = 0 \)? No, wait, the graph at \( x = -\pi \) is at \( y = 0 \)? Wait, no, the left peak: the \( y \)-value is 0? And the trough at \( x = 0 \) is at \( y = 0 \)? No, that can't be. Wait, maybe the midline is 0. Wait, the vertical shift \( D \) is the midline. So if the function is symmetric about the \( x \)-axis ( \( y = 0 \) ), then the vertical shift is 0. Wait, let's confirm: the formula for vertical shift is the midline, which is \( \frac{\text{max} + \text{min}}{2} \). Let's find max and min. From the graph, the maximum \( y \)-value (highest point) is 0? No, wait, the left peak: when \( x = -\pi \), the \( y \)-value is 0? And the right peak: when \( x = \frac{\pi}{2} \), the \( y \)-value is 0? And the trough at \( x = 0 \) is at \( y = 0 \)? No, that would mean the function is \( y = 0 \), but that's not. Wait, maybe I made a mistake. Wait, the graph: let's look at the vertical axis. The function crosses the \( x \)-axis ( \( y = 0 \) ) at \( -\frac{\pi}{2} \), \( 0 \), and another point. The peak on the left: when \( x = -\pi \), the \( y \)-value is 0? No, the left peak is above the \( x \)-axis? Wait, no, the graph is purple: the left peak is at \( y = 0 \)? Wait, no, the grid lines: each square is 1 unit. So the vertical axis: 0 is the middle. The graph's midline is \( y = 0 \), so the vertical shift \( D = 0 \).

Answer:

\( 0 \)