example 4 | a shifted sine curve\nfind the amplitude, period, and phase shift of ( y = 3 sin 2 left( x…

Explanation:

Step1: Find the amplitude

The general form of a sine function is (y = A\sin(B(x - C))+D). For the function (y = 3\sin(2(x-\frac{\pi}{4}))), the amplitude (|A|). Here (A = 3), so the amplitude is (|3|=3).

Step2: Find the period

The formula for the period of a sine function (y = A\sin(B(x - C))+D) is (T=\frac{2\pi}{|B|}). Here (B = 2), so (T=\frac{2\pi}{2}=\pi).

Step3: Find the phase - shift

The formula for the phase - shift of a sine function (y = A\sin(B(x - C))+D) is (C). Here (C=\frac{\pi}{4}), so the phase - shift is (\frac{\pi}{4}) (to the right).

Step4: Graphing

We know that the key points of (y=\sin x) are ((0,0)), ((\frac{\pi}{2},1)), ((\pi,0)), ((\frac{3\pi}{2},- 1)), ((2\pi,0)). For (y = 3\sin(2(x-\frac{\pi}{4}))), we make the substitution (u = 2(x-\frac{\pi}{4})) or (x=\frac{u}{2}+\frac{\pi}{4}). When (u = 0): (x=\frac{0}{2}+\frac{\pi}{4}=\frac{\pi}{4}), (y = 0) When (u=\frac{\pi}{2}): (x=\frac{\pi/2}{2}+\frac{\pi}{4}=\frac{\pi}{2}), (y = 3) When (u=\pi): (x=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}), (y = 0) When (u=\frac{3\pi}{2}): (x=\frac{3\pi/2}{2}+\frac{\pi}{4}=\pi), (y=-3) When (u = 2\pi): (x=\frac{2\pi}{2}+\frac{\pi}{4}=\frac{5\pi}{4}), (y = 0)

Answer:

Amplitude: (3), Period: (\pi), Phase - shift: (\frac{\pi}{4}) (to the right)