6. sketch a graph of (y = \frac{1}{2}sinleft\frac{2pi}{3}(x - 1)\right+2) for two cycles. be sure to label…

Explanation:

Step1: Identify the general form

The general form of a sinusoidal function is $y = A\sin(B(x - C))+D$. For $y=\frac{1}{2}\sin\left[\frac{2\pi}{3}(x - 1)\right]+2$, we have $A=\frac{1}{2}$, $B = \frac{2\pi}{3}$, $C = 1$, $D=2$. The period $T=\frac{2\pi}{B}=\frac{2\pi}{\frac{2\pi}{3}}=3$.

Step2: Find the key - points for one cycle

For the sine function $y = \sin(u)$, the key - points in one cycle of $u$ from $0$ to $2\pi$ are $u = 0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi$. Set $u=\frac{2\pi}{3}(x - 1)$.

• When $u = 0$: $\frac{2\pi}{3}(x - 1)=0$, then $x - 1=0$, so $x = 1$. And $y=\frac{1}{2}\sin(0)+2=2$.
• When $u=\frac{\pi}{2}$: $\frac{2\pi}{3}(x - 1)=\frac{\pi}{2}$, then $x-1=\frac{3}{4}$, so $x=1 + \frac{3}{4}=\frac{7}{4}$. And $y=\frac{1}{2}\sin\left(\frac{\pi}{2}\right)+2=\frac{1}{2}\times1 + 2=\frac{5}{2}$.
• When $u=\pi$: $\frac{2\pi}{3}(x - 1)=\pi$, then $x - 1=\frac{3}{2}$, so $x=1+\frac{3}{2}=\frac{5}{2}$. And $y=\frac{1}{2}\sin(\pi)+2=2$.
• When $u=\frac{3\pi}{2}$: $\frac{2\pi}{3}(x - 1)=\frac{3\pi}{2}$, then $x - 1=\frac{9}{4}$, so $x=1+\frac{9}{4}=\frac{13}{4}$. And $y=\frac{1}{2}\sin\left(\frac{3\pi}{2}\right)+2=\frac{1}{2}\times(-1)+2=\frac{3}{2}$.
• When $u = 2\pi$: $\frac{2\pi}{3}(x - 1)=2\pi$, then $x - 1=3$, so $x=4$. And $y=\frac{1}{2}\sin(2\pi)+2=2$.

Step3: Find the key - points for two cycles

Since the period is $T = 3$, for the second cycle, we add $3$ to each $x$ - value of the key - points of the first cycle.

• First key - point of second cycle: $x=1 + 3=4$, $y = 2$.
• Second key - point of second cycle: $x=\frac{7}{4}+3=\frac{7 + 12}{4}=\frac{19}{4}$, $y=\frac{5}{2}$.
• Third key - point of second cycle: $x=\frac{5}{2}+3=\frac{5+6}{2}=\frac{11}{2}$, $y = 2$.
• Fourth key - point of second cycle: $x=\frac{13}{4}+3=\frac{13 + 12}{4}=\frac{25}{4}$, $y=\frac{3}{2}$.
• Fifth key - point of second cycle: $x=4 + 3=7$, $y = 2$.

To sketch the graph, plot these ten key - points (five for each cycle) and connect them with a smooth sinusoidal curve.

Answer:

The key - points for the first cycle are $(1,2),\left(\frac{7}{4},\frac{5}{2}\right),\left(\frac{5}{2},2\right),\left(\frac{13}{4},\frac{3}{2}\right),(4,2)$. The key - points for the second cycle are $(4,2),\left(\frac{19}{4},\frac{5}{2}\right),\left(\frac{11}{2},2\right),\left(\frac{25}{4},\frac{3}{2}\right),(7,2)$.