Presentation on theme: "EXAMPLE 1 Solve a multi-step problem"— Presentation transcript:
1EXAMPLE 1Solve a multi-step problemWrite a function for the sinusoid shown below.SOLUTIONSTEP 1Find the maximum value M and minimum value m. From the graph, M = 5 and m = –1.
2EXAMPLE 1Solve a multi-step problemSTEP 2Identify the vertical shift, k. The value of k is the mean of the maximum and minimum values. The vertical shift iskM + m2=5 + (–1)2=42==2.So, k = 2.STEP 3Decide 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 4Find the amplitude and period. The period isπ22πb=So, b = 4.
3EXAMPLE 1Solve a multi-step problemaM – m2=5 – (–1)2=62=The amplitude is= 3.The graph is not a reflection, so a > 0.Therefore,a = 3.ANSWERThe function is y = 3 sin 4x + 2.
4EXAMPLE 2Model circular motionJump RopeROPE 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.
5EXAMPLE 2Model circular motionSOLUTIONSTEP 1Find 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.
6EXAMPLE 2Model circular motionSTEP 2Identify the vertical shift. The vertical shift for the model is:kM + m2=75 + 32==782= 39STEP 3Decide 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).
7EXAMPLE 2Model circular motionSTEP 4Find the amplitude and period.aM – m2=75 – 32=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 is2πb= 0.5, and= 4π.ANSWERA model for the height of a rope is h = –36 cos 4π t + 39.
8GUIDED PRACTICEfor Examples 1 and 2Write a function for the sinusoid.1.SOLUTIONSTEP 1Find the maximum value M and minimum value m. From the graph, M = 2 and m = –2.
9GUIDED PRACTICEfor Examples 1 and 21.STEP 2Identify the vertical shift, k. The value of k is the mean of the maximum and minimum values. The vertical shift iskM + m2=2 + (–2)2=2==0.So, k = 0.
10GUIDED PRACTICEfor Examples 1 and 21.STEP 3Decide 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 4Find the amplitude and period. The period is2π3b=So, b = 3.
11GUIDED PRACTICEfor Examples 1 and 21.aM – m2=2 – (–2)2=42=The amplitude is= 2.The graph is not a reflection, so a > 0.Therefore,a = 2.ANSWERThe function is y = 2 cos 3x.
12GUIDED PRACTICEfor Examples 1 and 2Write a function for the sinusoid.2.ANSWERy = 2 sin π x – 1
13GUIDED PRACTICEfor Examples 1 and 23. WHAT IF? Describe how the model in Example 2would change if the lowest point of a rope is 5inches above the ground and the highest point is 70 inches above the ground.ANSWERThe amplitude changes to 32.5 and the vertical shiftbecomes 37.5, but the period is not affected