select all the statements about the graph of y = 2 sin(x) that are true.\na. the domain of the function is…

Explanation:

Step1: Analyze domain

The sine - function $y = A\sin(Bx - C)+D$ has a domain of all real numbers. For $y = 2\sin(x)$, the domain is $(-\infty<x<\infty)$. So, statement A is true.

Step2: Check for vertical asymptotes

The sine - function $y=\sin(x)$ and its vertical - stretch $y = 2\sin(x)$ do not have vertical asymptotes. Sine is a continuous function for all real $x$. So, statement B is false.

Step3: Find zeros

Set $y = 2\sin(x)=0$. Then $\sin(x)=0$. The solutions of $\sin(x)=0$ are $x = k\pi$, where $k\in\mathbb{Z}$. When $k = 0$, $x = 0$ and when $k = 2$, $x = 2\pi$. So, statement C is true.

Step4: Analyze increasing and decreasing intervals

The derivative of $y = 2\sin(x)$ is $y'=2\cos(x)$. The function is decreasing when $y'<0$, i.e., $\cos(x)<0$. The interval where $\cos(x)<0$ is $\frac{\pi}{2}+2k\pi<x<\frac{3\pi}{2}+2k\pi,k\in\mathbb{Z}$. For $k = 0$, the function is decreasing when $\frac{\pi}{2}<x<\frac{3\pi}{2}$. So, statement D is true.

Step5: Determine the period

The general form of a sine function is $y = A\sin(Bx - C)+D$, and its period is $T=\frac{2\pi}{|B|}$. For $y = 2\sin(x)$, $B = 1$, so the period $T=\frac{2\pi}{|1|}=2\pi$. So, statement E is true.

Answer:

A. The domain of the function is $(-\infty < x < \infty)$. C. Two of the function's zeros are when $x = 0$ and $x = 2\pi$. D. The function is decreasing when $\frac{\pi}{2}<x<\frac{3\pi}{2}$. E. The period of the function is $2\pi$.