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consider the equation $y = 3\sin(\frac{5\pi}{3}x + \frac{10\pi}{3}) + 7$.
the amplitude is 3.
the period is $\frac{6}{5}$.
the phase shift is units to the left.
the midline is $y = 7$.
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Explanation:

Step1: Recall the general form of sine function

The general form of a sine function is \(y = A\sin(Bx - C)+D\).

Step2: Rewrite the given function in general form

Given \(y = 3\sin(\frac{5\pi}{3}x+\frac{10\pi}{3}) + 7\), we can rewrite it as \(y=3\sin(\frac{5\pi}{3}(x + 2))+7\). Here \(A = 3\), \(B=\frac{5\pi}{3}\), \(C=-\frac{10\pi}{3}\), \(D = 7\).

Step3: Calculate the phase - shift

The formula for phase - shift is \(\frac{C}{B}\). Substituting \(C =-\frac{10\pi}{3}\) and \(B=\frac{5\pi}{3}\), we get \(\frac{-\frac{10\pi}{3}}{\frac{5\pi}{3}}=- 2\).

Answer:

The phase shift is \(2\) units to the left.