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2. graph one complete primary period of the function $y = 125\\sin(\\fr…

Question

  1. graph one complete primary period of the function $y = 125\sin(\frac{1}{8}t + \frac{\pi}{4}) + 175$ 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: Find the amplitude

For a sine function \(y = A\sin(Bt - C)+D\), the amplitude is \(|A|\). Here \(A = 125\), so the amplitude is \(|125|=125\).

Step2: Find the vertical shift

The vertical shift is \(D\). Here \(D = 175\), so the vertical shift is \(175\).

Step3: Find the average value

The average value of a sine function \(y = A\sin(Bt - C)+D\) is \(D\). So the average value is \(175\).

Step4: Find the horizontal shift

For \(y = A\sin(Bt - C)+D\), the horizontal shift is \(\frac{C}{B}\). Here \(B=\frac{1}{8}\) and \(C =-\frac{\pi}{4}\) (since \(y=125\sin(\frac{1}{8}t+\frac{\pi}{4})+175=125\sin(\frac{1}{8}(t + 2\pi))+175\)), so the horizontal shift is \(- 2\pi\).

Step5: Find the maximum value

The maximum value of \(y = A\sin(Bt - C)+D\) is \(A + D\). So \(125+175 = 300\).

Step6: Find the minimum value

The minimum value of \(y = A\sin(Bt - C)+D\) is \(-A+D\). So \(- 125 + 175=50\).

Step7: Find the period

The period of \(y = A\sin(Bt - C)+D\) is \(T=\frac{2\pi}{|B|}\). Here \(B=\frac{1}{8}\), so \(T=\frac{2\pi}{\frac{1}{8}}=16\pi\).

Step8: Find the endpoints of the primary interval

For \(y=\sin t\), the primary interval is \([0,2\pi]\). For \(y = 125\sin(\frac{1}{8}t+\frac{\pi}{4})+175\), we set \(\frac{1}{8}t+\frac{\pi}{4}=0\) and \(\frac{1}{8}t+\frac{\pi}{4}=2\pi\).
Solving \(\frac{1}{8}t+\frac{\pi}{4}=0\) gives \(t=-2\pi\), and solving \(\frac{1}{8}t+\frac{\pi}{4}=2\pi\) gives \(t = 14\pi\). So the endpoints are \([-2\pi,14\pi]\).

Step9: Find the domain

The domain of a sine function \(y = 125\sin(\frac{1}{8}t+\frac{\pi}{4})+175\) is all real numbers, \((-\infty,\infty)\).

Step10: Find the range

Since the minimum value is \(50\) and the maximum value is \(300\), the range is \([50,300]\).

Answer:

Amplitude: \(125\)
Vertical Shift: \(175\)
Average Value: \(175\)
Horizontal Shift: \(-2\pi\)
Maximum Value: \(300\)
Minimum Value: \(50\)
Period: \(16\pi\)
Endpoints of the Primary Interval: \([-2\pi,14\pi]\)
Domain: \((-\infty,\infty)\)
Range: \([50,300]\)