Question

what is the equation of the midline of the sinusoidal function? enter your answer in the box. y = \boxed{}

Explanation:

Step1: Identify midline definition

The midline of a sinusoidal function is the horizontal line that lies exactly halfway between the maximum and minimum values of the function.

Step2: Find max and min values

From the graph, the maximum value (peak) of the sinusoidal function is \( y = 5 \) (wait, no, looking at the graph, the peaks are at \( y = 5 \)? Wait, no, the grid: let's check the y-axis. The peaks are at y=5? Wait, no, the graph has peaks at y=5? Wait, no, looking at the graph, the highest points (peaks) seem to be at y=5? Wait, no, the y-axis has marks at 6,4,2,0,-2,-4,-6,-8. Wait, the first peak after x=-6: let's see the coordinates. Wait, the graph crosses the origin (0,0), and has peaks and troughs. Wait, the midline is the average of the maximum and minimum y-values. Let's find the maximum and minimum. From the graph, the maximum y-value (peak) is \( y = 5 \)? Wait, no, looking at the graph, the peaks are at y=5? Wait, no, the y-axis: the grid lines are at y=6, y=4, y=2, y=0, y=-2, y=-4, y=-6, y=-8. Wait, the peaks: one peak is at (2,5)? Wait, no, the graph at x=2: the peak is at y=5? Wait, no, the y-axis label: 6 is at the top, then 4, 2, 0, -2, -4, -6, -8. Wait, the troughs (minimums) are at y=-5? Wait, no, the troughs: let's see, the first trough (leftmost) is at y=-5? Wait, no, the graph at x=-9 (approx) has a trough at y=-5? Wait, no, maybe I misread. Wait, the midline is the horizontal line that is the average of the maximum and minimum. Let's check the graph: the function oscillates around the x-axis? Wait, no, the graph passes through (0,0), and has peaks above and troughs below. Wait, the maximum y-value (peak) is 5? Wait, no, the y-axis: the distance between the peak and trough. Wait, the peaks are at y=5? No, wait, the graph: when x=0, the function is at (0,0). The peak at x=2: let's see the y-coordinate. The grid: each square is 1 unit? So from y=0, up to y=5? No, the y-axis has 6 at the top, then 4, 2, 0, -2, -4, -6, -8. So the peak is at y=5? Wait, no, the peak is at y=5? Wait, no, the graph shows a peak at y=5? Wait, maybe the maximum is 5 and minimum is -5? Then the midline would be (5 + (-5))/2 = 0? Wait, but the graph passes through (0,0). Wait, let's check the graph again. The sinusoidal function: the midline is the horizontal line that is the average of the maximum and minimum y-values. Let's find the maximum and minimum. Looking at the graph, the highest point (peak) is at y=5? No, wait, the y-axis: the peak is at y=5? Wait, no, the graph has a peak at y=5? Wait, maybe the maximum is 5 and minimum is -5, so midline is (5 + (-5))/2 = 0. Alternatively, maybe the maximum is 5 and minimum is -5, so midline is 0? Wait, but the graph passes through (0,0), which is on the midline. Wait, let's confirm: the midline of a sinusoidal function is the horizontal line \( y = \frac{\text{max} + \text{min}}{2} \). Let's find max and min. From the graph, the maximum y-value (peak) is 5 (looking at the graph, the peaks are at y=5) and the minimum y-value (trough) is -5 (the troughs are at y=-5). So \( \text{midline} = \frac{5 + (-5)}{2} = 0 \). So the equation of the midline is \( y = 0 \). Wait, but let's check: the graph passes through (0,0), and oscillates above and below the x-axis (y=0). So the midline is y=0.

Wait, maybe I made a mistake. Let's re-examine. The graph: it's a sinusoidal function, symmetric about the x-axis (y=0). Because the peaks are above y=0 and troughs are below y=0, and the distance from the midline to peak is equal to the distance from midline to trough. So the midline is y=0.

Step1: Determine max and min y-values

From the graph, the maximum y-value (peak) is \( y = 5 \) (wait, no, maybe the peak is 5? Wait, no, the y-axis: the grid lines are 1 unit apart? So the peak is at y=5, trough at y=-5.

Step2: Calculate midline

The midline is the average of the maximum and minimum: \( \text{midline} = \frac{\text{max} + \text{min}}{2} = \frac{5 + (-5)}{2} = 0 \).

Answer:

\( y = 0 \)