(1) graph the following function. identify amplitude, period, phase shift, vertical shift. show at least two…

Explanation:

Step1: Identify the amplitude

For a sine - function of the form $y = A\sin(Bx - C)+D$, the amplitude is given by $|A|$. Here, $A=\frac{1}{2}$, so the amplitude $|A|=\frac{1}{2}$.

Step2: Calculate the period

The period of the sine - function $y = A\sin(Bx - C)+D$ is $T=\frac{2\pi}{|B|}$. Since $B = \pi$, then $T=\frac{2\pi}{\pi}=2$.

Step3: Determine the phase - shift

The phase - shift is given by $\frac{C}{B}$. In the function $y=\frac{1}{2}\sin(\pi x)+1$, $C = 0$, so the phase - shift is $\frac{0}{\pi}=0$.

Step4: Find the vertical shift

For the function $y = A\sin(Bx - C)+D$, the vertical shift is $D$. Here, $D = 1$.

Step5: Find key points for one - cycle

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

Step6: Determine the domain

The domain of the sine - function $y=\frac{1}{2}\sin(\pi x)+1$ is all real numbers, i.e., $(-\infty,\infty)$.

Step7: Determine the range

The range of $y = A\sin(Bx - C)+D$ is $[D - |A|,D + |A|]$. Here, $D = 1$ and $|A|=\frac{1}{2}$, so the range is $[1-\frac{1}{2},1+\frac{1}{2}]=[\frac{1}{2},\frac{3}{2}]$.

Answer:

Amplitude: $\frac{1}{2}$; Period: $2$; Phase - shift: $0$; Vertical shift: $1$; Domain: $(-\infty,\infty)$; Range: $[\frac{1}{2},\frac{3}{2}]$