Question

find the amplitude, period, and phase shift of the function. graph the function. be sure to label key points. show at least two periods.

( y = 5 sin ( 4 x - pi ) )

what is the phase shift?

( \frac { pi } { 4 } )
(simplify your answer. type an exact answer, using ( pi ) as needed. use integers or fractions for any numbers in the expression.)

choose the correct graph below.

a.
b.
c.
d.

Explanation:

Step1: Recall the general form of sine function

The general form of a sine function is \(y = A\sin(Bx - C)+D\). For the function \(y = 5\sin(4x-\pi)\), we have \(A = 5\), \(B = 4\), \(C=\pi\), \(D = 0\).

Step2: Calculate the amplitude

The amplitude is given by \(|A|\). So, \(|A|=|5| = 5\).

Step3: Calculate the period

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

Step4: Calculate the phase - shift

The phase - shift is given by \(\frac{C}{B}\). Since \(C=\pi\) and \(B = 4\), the phase - shift is \(\frac{\pi}{4}\).

For graphing, we can find key points. Let \(u=4x-\pi\).
When \(u = 0\) (corresponds to the start of a cycle after phase - shift), \(4x-\pi=0\Rightarrow x=\frac{\pi}{4}\).
When \(u=\frac{\pi}{2}\), \(4x-\pi=\frac{\pi}{2}\Rightarrow x=\frac{3\pi}{8}\), and \(y = 5\).
When \(u=\pi\), \(4x-\pi=\pi\Rightarrow x=\frac{\pi}{2}\), and \(y = 0\).
When \(u=\frac{3\pi}{2}\), \(4x-\pi=\frac{3\pi}{2}\Rightarrow x=\frac{5\pi}{8}\), and \(y=-5\).
When \(u = 2\pi\), \(4x-\pi=2\pi\Rightarrow x=\frac{3\pi}{4}\), and \(y = 0\).

Answer:

Amplitude: \(5\), Period: \(\frac{\pi}{2}\), Phase - shift: \(\frac{\pi}{4}\)