Question

the graph of a periodic function f is shown below.
a. what is the period of f?
pi
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b. what is the minimum value of f(theta)?
-1
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c what is the maximum value of f(theta)?
5
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d what is the midline for f?
y = (5 + -1)/2
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e what is the amplitude of f?
(5 - -1)/2
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f write a function formula for f (enter \theta\ for theta )
f(theta) = 3 sin(2 theta)
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Explanation:

Step1: Find the period

For a sine function \(y = A\sin(B\theta + C)+D\), the period \(T=\frac{2\pi}{|B|}\).
From the graph, we can see that the function \(y = 3\sin(2\theta)\), here \(B = 2\).
So, \(T=\frac{2\pi}{2}=\pi\).

Step2: Find the minimum value

The general form of a sine function is \(y = A\sin(B\theta + C)+D\). The range of \(\sin(x)\) is \([- 1,1]\).
For \(y = 3\sin(2\theta)\), when \(\sin(2\theta)=-1\), \(y=-3\). But if we consider the mid - line and amplitude, the mid - line \(y = 2\) (since \(\frac{5+( - 1)}{2}=2\)), amplitude \(A = 3\). The minimum value \(y=2 - 3=-1\).

Step3: Find the maximum value

When \(\sin(2\theta)=1\), for \(y = 3\sin(2\theta)\) (with mid - line \(y = 2\)), the maximum value \(y=2 + 3=5\).

Step4: Find the mid - line

The mid - line \(y=\frac{\text{max}+\text{min}}{2}\). Given \(\text{max}=5\) and \(\text{min}=-1\), then \(y=\frac{5+( - 1)}{2}=2\).

Step5: Find the amplitude

The amplitude \(A=\frac{\text{max}-\text{min}}{2}\). Substituting \(\text{max}=5\) and \(\text{min}=-1\), we get \(A=\frac{5-( - 1)}{2}=3\).

Step6: Write the function formula

The general form of a sine function is \(y = A\sin(B\theta)+k\) (where \(k\) is the mid - line). Here \(A = 3\), \(B = 2\), \(k = 2\). So \(y=3\sin(2\theta)+2\).

Answer:

a. The period is \(\pi\).
b. The minimum value is \(-1\).
c. The maximum value is \(5\).
d. The mid - line is \(y = 2\).
e. The amplitude is \(3\).
f. The function formula is \(f(\theta)=3\sin(2\theta)+2\).