graph the function y = - 2 sin (\\frac{\\pi}{3}x)+1. show at least two cycles. use the graph to determine…

Explanation:

Step1: Identify the general form of sine - function

The general form of a sine - function is $y = A\sin(Bx - C)+D$. For the function $y=-2\sin(\frac{\pi}{3}x)+1$, we have $A = - 2$, $B=\frac{\pi}{3}$, $C = 0$, and $D = 1$.

Step2: Calculate the period

The period of a sine - function $y = A\sin(Bx - C)+D$ is given by $T=\frac{2\pi}{|B|}$. Substituting $B=\frac{\pi}{3}$ into the formula, we get $T=\frac{2\pi}{\frac{\pi}{3}}=6$.

Step3: Find key points for one - cycle

For $y = \sin x$, the key points in one - cycle are $(0,0),(\frac{\pi}{2},1),(\pi,0),(\frac{3\pi}{2}, - 1),(2\pi,0)$. For $y=-2\sin(\frac{\pi}{3}x)+1$, when $x = 0$, $y=-2\sin(0)+1=1$; when $x=\frac{3}{2}$, $y=-2\sin(\frac{\pi}{2})+1=-2\times1 + 1=-1$; when $x = 3$, $y=-2\sin(\pi)+1=1$; when $x=\frac{9}{2}$, $y=-2\sin(\frac{3\pi}{2})+1=-2\times(-1)+1 = 3$; when $x = 6$, $y=-2\sin(2\pi)+1=1$.

Step4: Graph the function for two - cycles

We can plot the key points for two - cycles ($x$ from $0$ to $12$) and connect them with a smooth curve. The graph of $y=-2\sin(\frac{\pi}{3}x)+1$ oscillates between its maximum and minimum values.

Step5: Determine the domain

The domain of a sine - function $y = A\sin(Bx - C)+D$ is all real numbers. So, the domain of $y=-2\sin(\frac{\pi}{3}x)+1$ is $(-\infty,\infty)$.

Step6: Determine the range

The amplitude of the function is $|A| = 2$. The vertical shift is $D = 1$. The maximum value of $y$ occurs when $\sin(\frac{\pi}{3}x)=-1$, $y=-2\times(-1)+1=3$. The minimum value of $y$ occurs when $\sin(\frac{\pi}{3}x)=1$, $y=-2\times1 + 1=-1$. So, the range is $[-1,3]$.

Answer:

Domain: $(-\infty,\infty)$; Range: $[-1,3]$