graph the given function over a one - period interval.\ny = csc(\\frac{x}{2}-\frac{3\\pi}{2})

graph the given function over a one - period interval.\ny = csc(\\frac{x}{2}-\frac{3\\pi}{2})

graph the given function over a one - period interval.\ny = csc(\\frac{x}{2}-\frac{3\\pi}{2})

Answer

Explanation:

Step1: Recall the period formula for $y = A\csc(Bx - C)$

The period of $y=\csc(Bx - C)$ is $\frac{2\pi}{|B|}$. For the function $y = \csc\left(\frac{1}{2}x-\frac{3\pi}{2}\right)$, here $B=\frac{1}{2}$.

Step2: Calculate the period

Using the formula, the period $T=\frac{2\pi}{\left|\frac{1}{2}\right|}=4\pi$.

Step3: Find the phase - shift

The phase - shift is given by $\frac{C}{B}$. Here $C = \frac{3\pi}{2}$ and $B=\frac{1}{2}$, so the phase - shift is $\frac{\frac{3\pi}{2}}{\frac{1}{2}}=3\pi$.

Step4: Identify key points

We know that $\csc x=\frac{1}{\sin x}$. First, consider the related sine function $y = \sin\left(\frac{1}{2}x-\frac{3\pi}{2}\right)$. The sine function has a zero at $\frac{1}{2}x-\frac{3\pi}{2}=k\pi$, $k\in\mathbb{Z}$. Solving for $x$ gives $x = 2k\pi+3\pi$. The sine function has a maximum of 1 when $\frac{1}{2}x-\frac{3\pi}{2}=\frac{\pi}{2}+2k\pi$, so $x = 4k\pi + 4\pi$, and a minimum of - 1 when $\frac{1}{2}x-\frac{3\pi}{2}=\frac{3\pi}{2}+2k\pi$, so $x = 4k\pi+6\pi$. The cosecant function has vertical asymptotes where the sine function is zero. For one - period interval, we can choose the interval $[3\pi,7\pi]$. When $x = 3\pi$, $y=\csc\left(\frac{1}{2}(3\pi)-\frac{3\pi}{2}\right)=\csc(0)$ (vertical asymptote). When $x = 4\pi$, $y=\csc\left(\frac{1}{2}(4\pi)-\frac{3\pi}{2}\right)=\csc\left(\frac{\pi}{2}\right)=1$. When $x = 6\pi$, $y=\csc\left(\frac{1}{2}(6\pi)-\frac{3\pi}{2}\right)=\csc\left(\frac{3\pi}{2}\right)= - 1$. When $x = 7\pi$, $y=\csc\left(\frac{1}{2}(7\pi)-\frac{3\pi}{2}\right)=\csc(2\pi)$ (vertical asymptote).

To graph the function:

  1. Plot the vertical asymptotes at $x = 3\pi$ and $x = 7\pi$.
  2. Plot the point $(4\pi,1)$ and $(6\pi, - 1)$.
  3. Sketch the two U - shaped curves of the cosecant function between the asymptotes such that the curve passes through the plotted points.

The one - period interval we can use for graphing is $[3\pi,7\pi]$.

Answer:

One - period interval for graphing is $[3\pi,7\pi]$.