Question

the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer. which of the following functions is represented by the graph below? \\(\bigcirc\\) \\(y = 2\sin 6x\\) \\(\bigcirc\\) \\(y = 2\sin \left(\frac{1}{3}x\
ight)\\) \\(\bigcirc\\) \\(y = 3\sin 2x\\) \\(\bigcirc\\) \\(y = 2\sin 3x\\)

Explanation:

Step1: Recall the general form of a sine function

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

Step2: Analyze the amplitude

From the graph, the amplitude (the maximum value of \(y\)) is \(2\). So \(A = 2\).

Step3: Analyze the period

Let's assume we count the number of cycles in a certain interval. For \(y=\sin(x)\), the period is \(2\pi\). If we consider the standard - form \(y = A\sin(Bx)\), and we know that the period \(T=\frac{2\pi}{B}\).
If we assume that in the interval from \(x = 0\) to \(x=\pi\), for \(y = 2\sin(6x)\), the period \(T=\frac{2\pi}{6}=\frac{\pi}{3}\). In the interval \(x\in[0, 2\pi]\), the number of cycles \(n=\frac{2\pi}{T}=6\).
For \(y = 2\sin(\frac{1}{3}x)\), the period \(T=\frac{2\pi}{\frac{1}{3}} = 6\pi\).
For \(y=3\sin(2x)\), the amplitude \(A = 3\) (not \(2\)).
For \(y = 2\sin(3x)\), the period \(T=\frac{2\pi}{3}\), and the number of cycles in \(x\in[0,2\pi]\) is \(n = 3\).

Answer:

\(y = 2\sin(6x)\)