At an ocean depth of 8 meters, a buoy bobs up and then down 5 meters from the ocean's depth. Sixteen seconds pass from the time the buoy is at its highest point to when it is at its lowest point. Assume at x = 0 the buoy is at normal ocean depth.

Use the sine tool to graph the function. The first point must be on the midline and the second point must be a maximum or minimum value on the graph closest to the first point.

At an ocean depth of 8 meters a buoy bobs up and then down 5 meters from the oceans depth Sixteen seconds pass from the time the buoy is at its highest point to class=

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Answer:

  See attached for a graph

Step-by-step explanation:

We're going to plot sea level as y=0 and a depth of 8 meters as y=-8.

The problem statement tells you the initial point (x=0) is at normal ocean depth (y=-8), so the first point you put into your sine tool is ...

  (x, y) = (0, -8)

The buoy takes 16 seconds to go from a high point to a low point, so the time to the first high point is half that, or x=8 seconds. That high point is 5 meters above its average depth, so is at y=-3.

The second point you will put into your sine tool is ...

  (x, y) = (8, -3)

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Answer:

Step-by-step explanation:

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