Question

assignment 8.1: graphs of the sine and cosine functions
score: 13/21 answered: 5/10
question 6
given the equation ( y = 3 sin left( \frac { 5 pi } { 4 } x - \frac { 35 pi } { 4 }
ight) + 8 )
the amplitude is:
the period is:
the horizontal shift is: units to the select an answer
the midline is: ( y = )
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Explanation:

Step1: Find the amplitude

For a sine function \(y = A\sin(Bx - C)+D\), the amplitude is \(|A|\).
In \(y = 3\sin(\frac{5\pi}{4}x-\frac{35\pi}{4}) + 8\), \(A = 3\), so the amplitude is \(|3|=3\).

Step2: Find the period

The period of a sine function \(y = A\sin(Bx - C)+D\) is \(T=\frac{2\pi}{|B|}\).
Here \(B=\frac{5\pi}{4}\), so \(T=\frac{2\pi}{\frac{5\pi}{4}}=\frac{2\pi\times4}{5\pi}=\frac{8}{5}\).

Step3: Find the horizontal shift

The horizontal shift of a sine function \(y = A\sin(Bx - C)+D\) is \(\frac{C}{B}\).
Since \(B=\frac{5\pi}{4}\) and \(C = \frac{35\pi}{4}\), then \(\frac{C}{B}=\frac{\frac{35\pi}{4}}{\frac{5\pi}{4}} = 7\).
Because the form is \(y=A\sin(B(x - h))+D=A\sin(Bx - Bh)+D\) (here \(h=\frac{C}{B}\)), and the function is \(y = 3\sin(\frac{5\pi}{4}(x - 7))+8\), the shift is \(7\) units to the right.

Step4: Find the mid - line

For a sine function \(y = A\sin(Bx - C)+D\), the mid - line is \(y = D\).
In \(y = 3\sin(\frac{5\pi}{4}x-\frac{35\pi}{4}) + 8\), \(D = 8\), so the mid - line is \(y = 8\).

Answer:

The amplitude is: \(3\)
The period is: \(\frac{8}{5}\)
The horizontal shift is: \(7\) units to the right
The midline is: \(y = 8\)