4. on the axes below, graph one cycle of a sine function with amplitude 3 period 4π, midline y = 2, and…

Explanation:

Step1: Recall the general form of sine - function

The general form of a sine function is $y = A\sin(B(x - C))+D$, where $A$ is the amplitude, $B$ is related to the period by $T=\frac{2\pi}{|B|}$, $C$ is the phase - shift, and $D$ is the mid - line. Given $A = 3$, $T = 4\pi$, and $D=2$. Since $T=\frac{2\pi}{|B|}=4\pi$, we can solve for $B$: $$4\pi=\frac{2\pi}{|B|}$$ $$|B|=\frac{2\pi}{4\pi}=\frac{1}{2}$$. Let's assume $B=\frac{1}{2}$ (we can choose the positive value for a basic case). The function is now $y = 3\sin(\frac{1}{2}(x - C))+2$.

Step2: Use the given point to find the phase - shift

The function passes through the point $(0,2)$. Substitute $x = 0$ and $y = 2$ into $y = 3\sin(\frac{1}{2}(x - C))+2$: $$2=3\sin(\frac{1}{2}(0 - C))+2$$ $$0 = 3\sin(-\frac{C}{2})$$ $$\sin(-\frac{C}{2})=0$$ This implies $-\frac{C}{2}=k\pi$, $k\in\mathbb{Z}$. Let $k = 0$, then $C = 0$. So the function is $y = 3\sin(\frac{1}{2}x)+2$.

Step3: Find key points for graphing

For the sine function $y=\sin(u)$, the key points for one period are found when $u = 0,\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi$. For $y = 3\sin(\frac{1}{2}x)+2$, when $\frac{1}{2}x=0$, $x = 0$ and $y=3\sin(0)+2=2$; when $\frac{1}{2}x=\frac{\pi}{2}$, $x=\pi$ and $y=3\sin(\frac{\pi}{2})+2=3 + 2=5$; when $\frac{1}{2}x=\pi$, $x = 2\pi$ and $y=3\sin(\pi)+2=2$; when $\frac{1}{2}x=\frac{3\pi}{2}$, $x = 3\pi$ and $y=3\sin(\frac{3\pi}{2})+2=-3 + 2=-1$; when $\frac{1}{2}x=2\pi$, $x = 4\pi$ and $y=3\sin(2\pi)+2=2$.

Plot the points $(0,2),(\pi,5),(2\pi,2),(3\pi,-1),(4\pi,2)$ and connect them with a smooth curve to graph one cycle of the function $y = 3\sin(\frac{1}{2}x)+2$.

Answer:

Graph the function $y = 3\sin(\frac{1}{2}x)+2$ using the key - points $(0,2),(\pi,5),(2\pi,2),(3\pi,-1),(4\pi,2)$ and connecting them with a smooth curve for one cycle.