Answer :
Answer: y = 2*cos(3pi*x) + 5
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The general template for cosine is
y = A*cos(B(x-C)) + D
where,
|A| = amplitude
B = 2pi/T, with T being the period and T = 1/f, f is the frequency
C = determines horizontal phase shift
D = determines vertical shift, and is midline
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In this case,
midline = 5 means D = 5
amplitude = 2 means A = 2
frequency = 3/2 means the period is T = 2/3, which is the reciprocal of the frequency, so B = 2pi/T = 2pi/(2/3) = 3pi
There is no mention of a phase shift, so C = 0.
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All together we have
A = 2, B = 3pi, C = 0, D = 5
So,
y = A*cos(B(x-C)) + D
y = 2*cos(3pi(x-0)) + 5
y = 2*cos(3pi*x) + 5
The cosine function that has a midline of 5, an amplitude of 2 and a frequency of 3/2 will be y = 2*cos(3pi*x) + 5
Cosine function
- Cos function (or cosine function) in a triangle is the ratio of the adjacent side to that of the hypotenuse
How to solve this problem?
The steps are as follow:
- The general template for cosine is y = A*cos(B(x-C)) + D
- Where,
A = amplitude
B = 2pi/T, with T being the period and T = 1/f, f is the frequency
C = determines horizontal phase shift
D = determines vertical shift, and is midline
- In the given problem,
D = midline = 5
A = amplitude = 2
frequency = 3/2 means the period is T = 2/3, which is the reciprocal of the frequency, so B = 2pi/T = 2pi/(2/3) = 3pi
C = 0.
- By combining all together we get,
A = 2, B = 3pi, C = 0, D = 5
so,
y = A*cos(B(x-C)) + D
y = 2*cos(3pi(x-0)) + 5
y = 2*cos(3pi*x) + 5
So, the cosine function that has a midline of 5, an amplitude of 2 and a frequency of 3/2 will be y = 2*cos(3pi*x) + 5
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