Question

let $y = \sqrt{20} \sin\left(2\pi x + 2e^2\
ight)$. what is the amplitude? what is the period? what is the phase shift? note: if needed, you can enter $\pi$ as pi in your answers.

Explanation:

Step1: Recall the sine function form

The general form of a sine function is \( y = A\sin(Bx + C) \), where \( |A| \) is the amplitude, \( \frac{2\pi}{|B|} \) is the period, and \( -\frac{C}{B} \) is the phase shift.

Step2: Find the amplitude

For the given function \( y = \sqrt{20}\sin(2\pi x + 2e^2) \), compare with the general form. Here, \( A = \sqrt{20} \), which simplifies to \( 2\sqrt{5} \) (since \( \sqrt{20}=\sqrt{4\times5}=2\sqrt{5} \)). So the amplitude is \( \sqrt{20} \) (or \( 2\sqrt{5} \)).

Step3: Find the period

In the function, \( B = 2\pi \). Using the period formula \( \frac{2\pi}{|B|} \), substitute \( B = 2\pi \): \( \frac{2\pi}{|2\pi|}=\frac{2\pi}{2\pi}=1 \). So the period is \( 1 \).

Step4: Find the phase shift

For the phase shift, \( C = 2e^2 \) and \( B = 2\pi \). Using the phase shift formula \( -\frac{C}{B} \), substitute the values: \( -\frac{2e^2}{2\pi}=-\frac{e^2}{\pi} \). So the phase shift is \( -\frac{e^2}{\pi} \) (or \( \frac{-e^2}{\pi} \)).

Answer:

Amplitude: \( \sqrt{20} \) (or \( 2\sqrt{5} \))
Period: \( 1 \)
Phase Shift: \( -\frac{e^2}{\pi} \)