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Math  /  Trigonometry

QuestionFor the following exercises, graph two full periods of each function and state the amplitude, period, and midline. State the maximum and minimum yy values and their corresponding xx-values on one period for x>0x>0. Round answers to two decimal places if necessary.
6. f(x)=2sin⁡xf(x)=2 \sin x
7. f(x)=23cos⁡xf(x)=\frac{2}{3} \cos x
8. f(x)=−3sin⁡xf(x)=-3 \sin x
9. f(x)=4sin⁡xf(x)=4 \sin x
10. f(x)=2cos⁡xf(x)=2 \cos x
11. f(x)=cos⁡(2x)f(x)=\cos (2 x)
12. f(x)=2sin⁡(12x)f(x)=2 \sin \left(\frac{1}{2} x\right)
13. f(x)=4cos⁡(πx)f(x)=4 \cos (\pi x)
14. f(x)=3cos⁡(65x)f(x)=3 \cos \left(\frac{6}{5} x\right)

Studdy Solution

STEP 1

1. Each function is a trigonometric function involving sine or cosine.
2. The amplitude, period, and midline need to be determined for each function.
3. The maximum and minimum y y values and their corresponding x x -values need to be identified for one period where x>0 x > 0 .
4. Answers should be rounded to two decimal places if necessary.

STEP 2

1. Determine the amplitude, period, and midline for each function.
2. Identify the maximum and minimum y y -values and their corresponding x x -values for one period where x>0 x > 0 .
3. Graph two full periods of each function.

STEP 3

For f(x)=2sin⁡x f(x) = 2 \sin x : - Amplitude: The amplitude is the coefficient of the sine function, which is 2 2 . - Period: The period of sin⁡x \sin x is 2π 2\pi . Since there is no horizontal stretch/compression factor, the period remains 2π 2\pi . - Midline: The midline is y=0 y = 0 since there is no vertical shift.

STEP 4

For f(x)=2sin⁡x f(x) = 2 \sin x : - Maximum y y -value: The maximum value of 2sin⁡x 2 \sin x is 2 2 , occurring at x=π2 x = \frac{\pi}{2} . - Minimum y y -value: The minimum value of 2sin⁡x 2 \sin x is −2 -2 , occurring at x=3π2 x = \frac{3\pi}{2} .

STEP 5

Graph two full periods of f(x)=2sin⁡x f(x) = 2 \sin x : - The graph will start at x=0 x = 0 , reach a maximum at x=π2 x = \frac{\pi}{2} , return to the midline at x=π x = \pi , reach a minimum at x=3π2 x = \frac{3\pi}{2} , and return to the midline at x=2π x = 2\pi . - Repeat the pattern for the second period.

STEP 6

For f(x)=23cos⁡x f(x) = \frac{2}{3} \cos x : - Amplitude: The amplitude is the coefficient of the cosine function, which is 23 \frac{2}{3} . - Period: The period of cos⁡x \cos x is 2π 2\pi . Since there is no horizontal stretch/compression factor, the period remains 2π 2\pi . - Midline: The midline is y=0 y = 0 since there is no vertical shift.

STEP 7

For f(x)=23cos⁡x f(x) = \frac{2}{3} \cos x : - Maximum y y -value: The maximum value of 23cos⁡x \frac{2}{3} \cos x is 23 \frac{2}{3} , occurring at x=0 x = 0 . - Minimum y y -value: The minimum value of 23cos⁡x \frac{2}{3} \cos x is −23 -\frac{2}{3} , occurring at x=π x = \pi .

STEP 8

Graph two full periods of f(x)=23cos⁡x f(x) = \frac{2}{3} \cos x : - The graph will start at x=0 x = 0 , reach a minimum at x=π x = \pi , and return to the maximum at x=2π x = 2\pi . - Repeat the pattern for the second period.
(Continue similarly for each function, following the same steps.)

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