Question

given the equation ( y = 2sinleft(\frac{5pi}{3}x+\frac{20pi}{3}
ight)+5 )
the amplitude is:
the period is:
the horizontal shift is: units to the
the midline is: ( y = )

Explanation:

Step1: Find the amplitude

The general form of a sinusoidal function is \(y = A\sin(Bx - C)+D\). Here \(A\) is the amplitude. For \(y = 2\sin(\frac{5\pi}{3}x+\frac{20\pi}{3})+5\), \(A = 2\).

Step2: Find the period

The period of \(y = A\sin(Bx - C)+D\) is \(T=\frac{2\pi}{|B|}\). Here \(B=\frac{5\pi}{3}\), so \(T=\frac{2\pi}{\frac{5\pi}{3}}=\frac{6}{5} = 1.2\).

Step3: Find the horizontal shift

First, rewrite the function as \(y = 2\sin(\frac{5\pi}{3}(x + 4))+5\). For \(y = A\sin(B(x - h))+D\), the horizontal shift is \(h\). Here \(h=-4\), which means a shift of \(4\) units to the left.

Step4: Find the midline

For \(y = A\sin(Bx - C)+D\), the midline is \(y = D\). Here \(D = 5\).

Answer:

The amplitude is \(2\). The period is \(1.2\). The horizontal shift is \(4\) units to the left. The midline is \(y = 5\).