graph the function. f(t)=2 sin t

Explanation:

Step1: Recall properties of sine - function

The general form of a sine - function is $y = A\sin(Bt - C)+D$. For the function $f(t)=2\sin t$, we have $A = 2$, $B = 1$, $C = 0$, and $D = 0$.

Step2: Determine amplitude

The amplitude of the function $y = A\sin t$ is given by $|A|$. Here, $|A|=|2| = 2$, which means the graph of $y = 2\sin t$ oscillates between $y=-2$ and $y = 2$.

Step3: Determine period

The period of the function $y=\sin(Bt)$ is $T=\frac{2\pi}{|B|}$. Since $B = 1$, the period $T = 2\pi$.

Step4: Analyze key - points

For $y=\sin t$, key - points in one period $[0,2\pi]$ are $(0,0)$, $(\frac{\pi}{2},1)$, $(\pi,0)$, $(\frac{3\pi}{2}, - 1)$, $(2\pi,0)$. For $y = 2\sin t$, the corresponding key - points are $(0,0)$, $(\frac{\pi}{2},2)$, $(\pi,0)$, $(\frac{3\pi}{2},-2)$, $(2\pi,0)$.

Answer:

We need to find the graph that has an amplitude of 2 (oscillates between $y=-2$ and $y = 2$) and a period of $2\pi$. Without seeing the exact details of each graph, but based on the amplitude and period analysis, the correct graph is the one that has the wave oscillating between $y=-2$ and $y = 2$ with a period of $2\pi$. If we assume the standard orientation and scale of the axes, we look for a sine - wave that reaches up to $y = 2$ and down to $y=-2$ and repeats every $2\pi$ units on the $t$ - axis.