the graph to the right is a function of the form y = a sin (bx - c)+d, b>0. the five - quarter points of one…

Explanation:

Step1: Recall period formula for sine - type function

The general form of a sine - type function is (y = A\sin(Bx - C)+D). The period (P) of the function (y = A\sin(Bx - C)+D) is given by (P=\frac{2\pi}{|B|}). If we know the horizontal distance between two key - points (such as two quarter - points) of the sine function, we can find the period. The horizontal distance between two consecutive quarter - points of a sine function is (\frac{P}{4}). Suppose the two consecutive quarter - points are (x_1) and (x_2), then (x_2 - x_1=\frac{P}{4}).

Step2: Calculate the period

Let's assume the two consecutive quarter - points are (x_1 =-\frac{\pi}{6}) and (x_2=\frac{\pi}{6}). Then (\frac{\pi}{6}-\left(-\frac{\pi}{6}\right)=\frac{\pi}{3}), and since (\frac{\pi}{3}=\frac{P}{4}), we can solve for (P) by cross - multiplying. We get (P = \frac{4\pi}{3}).

Step3: Find the value of (B)

Since (P=\frac{2\pi}{|B|}) and (P=\frac{4\pi}{3}), we set (\frac{2\pi}{|B|}=\frac{4\pi}{3}). Cross - multiply: (4\pi|B|=6\pi), then (|B|=\frac{3}{2}). Since (B>0) (given in the problem), (B = \frac{3}{2}).

Step4: Find the phase shift

The phase shift of the function (y = A\sin(Bx - C)+D) is (\frac{C}{B}). We know that for the standard sine function (y=\sin x), the first quarter - point is at (x = 0). For the function (y = A\sin(Bx - C)+D), if the first quarter - point is at (x=-\frac{\pi}{6}) and (B=\frac{3}{2}), and the phase shift formula is (x=\frac{C}{B}), then (-\frac{\pi}{6}=\frac{C}{\frac{3}{2}}). Cross - multiply: (C=-\frac{\pi}{6}\times\frac{3}{2}=-\frac{\pi}{4}).

Answer:

a. The period of this function is (\frac{4\pi}{3}) b. The value of (B) is (\frac{3}{2}) c. The value of the phase shift (using a value at least (-\frac{P}{2}) and less than (\frac{P}{2})) is (-\frac{\pi}{4})