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EXAMPLE 1 Solve a multi-step problem Write a function for the sinusoid shown below. SOLUTION STEP 1 Find the maximum value M and minimum value m. From the graph, M = 5 and m = –1.

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EXAMPLE 1 Solve a multi-step problem STEP 2 Identify the vertical shift, k. The value of k is the mean of the maximum and minimum values. The vertical shift is k M + m 2 = 5 + (–1) 2 = 4 2 = = 2. So, k = 2. STEP 3 Decide whether the graph should be modeled by a sine or cosine function. Because the graph crosses the midline y = 2 on the y-axis, the graph is a sine curve with no horizontal shift. So, h = 0. STEP 4 Find the amplitude and period. The period is π 2 2π b = So, b = 4.

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EXAMPLE 1 Solve a multi-step problem a M – m 2 = 5 – (–1) 2 = 6 2 = The amplitude is = 3. The graph is not a reflection, so a > 0. Therefore, a = 3. ANSWER The function is y = 3 sin 4x + 2.

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EXAMPLE 2 Model circular motion Jump Rope ROPE At a Double Dutch competition, two people swing jump ropes as shown in the diagram below. The highest point of the middle of each rope is 75 inches above the ground, and the lowest point is 3 inches. The rope makes 2 revolutions per second. Write a model for the height h (in feet) of a rope as a function of the time t (in seconds) if the rope is at its lowest point when t = 0.

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EXAMPLE 2 Model circular motion SOLUTION STEP 1 Find the maximum and minimum values of the function. A rope’s maximum height is 75 inches, so M = 75. A rope’s minimum height is 3 inches, so m = 3.

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EXAMPLE 2 Model circular motion STEP 2 Identify the vertical shift. The vertical shift for the model is: k M + m 2 = 75 + 3 2 = = 78 2 = 39 STEP 3 Decide whether the height should be modeled by a sine or cosine function. When t = 0, the height is at its minimum. So, use a cosine function whose graph is a reflection in the x-axis with no horizontal shift (h = 0).

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EXAMPLE 2 Model circular motion STEP 4 Find the amplitude and period. a M – m 2 = 75 – 3 2 = The amplitude is = 36. Because the graph is a reflection, a < 0. So, a = –36. Because a rope is rotating at a rate of 2 revolutions per second, one revolution is completed in 0.5 second. So, the period is 2π b = 0.5, and = 4π. ANSWER A model for the height of a rope is h = –36 cos 4π t + 39.

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GUIDED PRACTICE for Examples 1 and 2 Write a function for the sinusoid. 1. SOLUTION STEP 1 Find the maximum value M and minimum value m. From the graph, M = 2 and m = –2.

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GUIDED PRACTICE for Examples 1 and 2 1. STEP 2 Identify the vertical shift, k. The value of k is the mean of the maximum and minimum values. The vertical shift is k M + m 2 = 2 + (–2) 2 = 2 = = 0. So, k = 0.

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GUIDED PRACTICE for Examples 1 and 2 1. STEP 3 Decide whether the graph should be modeled by a sine or cosine function. Because the graph peaks at y = 2 on the y-axis, the graph is a cos curve with no horizontal shift. So, h = 0. STEP 4 Find the amplitude and period. The period is 2π 3 b = So, b = 3.

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GUIDED PRACTICE for Examples 1 and 2 1. a M – m 2 = 2 – (–2) 2 = 4 2 = The amplitude is = 2. The graph is not a reflection, so a > 0. Therefore, a = 2. ANSWER The function is y = 2 cos 3x.

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GUIDED PRACTICE for Examples 1 and 2 Write a function for the sinusoid. 2. ANSWER y = 2 sin π x – 1

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GUIDED PRACTICE for Examples 1 and 2 3. WHAT IF? Describe how the model in Example 2 would change if the lowest point of a rope is 5 inches above the ground and the highest point is 70 inches above the ground. ANSWER The amplitude changes to 32.5 and the vertical shift becomes 37.5, but the period is not affected

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Translations of Trigonometric Graphs LESSON 12–8.

Translations of Trigonometric Graphs LESSON 12–8.

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