the graph of which function has an amplitude of 2 and a period of 4π?\n○y = 2 sin 1/2x\n○y = 2 sin 4x\n○y =…

the graph of which function has an amplitude of 2 and a period of 4π?\n○y = 2 sin 1/2x\n○y = 2 sin 4x\n○y = 4 sin 1/2x\n○y = 4 sin 2x\nquestion 13 1 pts\nthe maximum value of 3 sin 1/3θ is\n○1\n○1/3\n○3\n○0\nquestion 14 1 pts\nwhat is the range of the function y = 2 sin 3x?\n○all real numbers\n○ - 1 ≤ y ≤ 1\n○ - 2 ≤ y ≤ 2\n○ - 3 ≤ y ≤ 3

the graph of which function has an amplitude of 2 and a period of 4π?\n○y = 2 sin 1/2x\n○y = 2 sin 4x\n○y = 4 sin 1/2x\n○y = 4 sin 2x\nquestion 13 1 pts\nthe maximum value of 3 sin 1/3θ is\n○1\n○1/3\n○3\n○0\nquestion 14 1 pts\nwhat is the range of the function y = 2 sin 3x?\n○all real numbers\n○ - 1 ≤ y ≤ 1\n○ - 2 ≤ y ≤ 2\n○ - 3 ≤ y ≤ 3

Answer

Explanation:

Step1: Recall amplitude - period formula for 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: Find function with amplitude 2 and period $4\pi$

We want $A = 2$ and $T = 4\pi$. Since $T=\frac{2\pi}{B}=4\pi$, solving for $B$ gives $B=\frac{2\pi}{4\pi}=\frac{1}{2}$. So the function is $y = 2\sin\frac{1}{2}x$.

Step3: Find maximum value of $3\sin\frac{1}{3}\theta$

The range of $\sin t$ is $[- 1,1]$. For the function $y = 3\sin\frac{1}{3}\theta$, when $\sin\frac{1}{3}\theta=1$, $y$ reaches its maximum value. So $y_{max}=3\times1 = 3$.

Step4: Find range of $y = 2\sin3x$

The range of $\sin t$ is $[-1,1]$. For the function $y = 2\sin3x$, when $\sin3x=-1$, $y=-2$; when $\sin3x = 1$, $y = 2$. So the range is $-2\leq y\leq2$.

Answer:

Question 12: A. $y = 2\sin\frac{1}{2}x$ Question 13: C. $3$ Question 14: C. $-2\leq y\leq2$