Question

1. graph the primary period of the function $f(t)=-90\sin(\frac{\pi}{10}t+\frac{3\pi}{5})-120$ and provide the requested information (use interval notation as appropriate).

amplitude:
vertical shift:
average value:
horizontal shift:
maximum value:
endpoints of the primary interval:
minimum value:
domain:
period:
range:

Explanation:

Step1: Identify the general form of the sine function

The general form of a sine function is \(y = A\sin(Bt - C)+D\). For the given function \(f(t)=- 90\sin(\frac{\pi}{10}t+\frac{3\pi}{5})-120\), we can rewrite it as \(f(t)=-90\sin(\frac{\pi}{10}(t + 6))-120\)

Step2: Calculate the amplitude

The amplitude \(|A|\). Here \(A=-90\), so \(|A| = 90\)

Step3: Calculate the period

The period \(T=\frac{2\pi}{|B|}\). Since \(B = \frac{\pi}{10}\), then \(T=\frac{2\pi}{\frac{\pi}{10}}=20\)

Step4: Calculate the vertical shift

The vertical shift is \(D\). Here \(D=-120\)

Step5: Calculate the horizontal shift

The horizontal shift is \(-\frac{C}{B}\). From \(y=-90\sin(\frac{\pi}{10}(t + 6))-120\), the horizontal shift is \(-6\)

Step6: Calculate the average value

The average value is the vertical shift, which is \(-120\)

Step7: Calculate the maximum and minimum values

The maximum value \(=D + |A|=-120 + 90=-30\)
The minimum value \(=D-|A|=-120-90=-210\)

Step8: Find the endpoints of the primary interval

For the sine function \(y = A\sin(Bt - C)+D\), the primary interval for \(\sin u\) (where \(u = Bt - C\)) is \(0\leq u\leq2\pi\).
Set \(\frac{\pi}{10}(t + 6)=0\), then \(t=-6\)
Set \(\frac{\pi}{10}(t + 6)=2\pi\), then \(t = 14\)

Step9: Determine the domain and range

The domain of the primary period is \([-6,14]\)
The range is \([-210,-30]\)

Answer:

Amplitude: \(90\)
Vertical Shift: \(-120\)
Average Value: \(-120\)
Horizontal Shift: \(-6\)
Maximum Value: \(-30\)
Minimum Value: \(-210\)
Period: \(20\)
Endpoints of the Primary Interval: \(-6\) and \(14\)
Domain: \([-6,14]\)
Range: \([-210,-30]\)