name______ period______ 2. determine the features of the following function, then graph at least one period…

Explanation:

Step1: Identify vertical stretch

The general form of a cosecant - type function is $y = A\csc(B(x - C))+D$. For the function $f(x)=2\csc(2(x-\frac{\pi}{2})) + 1$, the value of $A$ gives the vertical stretch. Here, $A = 2$, so the vertical stretch is 2.

Step2: Calculate the period

The period of the cosecant function $y=\csc(Bx)$ is $\frac{2\pi}{|B|}$. For the function $y = \csc(2(x-\frac{\pi}{2}))$, $B = 2$. So the period $T=\frac{2\pi}{2}=\pi$.

Step3: Determine the phase - shift

The phase - shift of the function $y = A\csc(B(x - C))+D$ is given by $C$. For the function $f(x)=2\csc(2(x-\frac{\pi}{2})) + 1$, $C=\frac{\pi}{2}$, so the phase - shift is $\frac{\pi}{2}$ to the right.

Step4: Find the mid - line equation

The mid - line of the function $y = A\csc(B(x - C))+D$ is given by $y = D$. For the function $f(x)=2\csc(2(x-\frac{\pi}{2})) + 1$, $D = 1$. So the mid - line equation is $y = 1$.

Step5: Find the equations of the asymptotes

The asymptotes of the cosecant function $y=\csc(u)$ occur when $u = k\pi$, where $k\in\mathbb{Z}$. For the function $y = 2\csc(2(x-\frac{\pi}{2}))+1$, we set $2(x-\frac{\pi}{2})=k\pi$. Solve for $x$: [ \begin{align*} 2(x-\frac{\pi}{2})&=k\pi\ 2x-\pi&=k\pi\ 2x&=k\pi+\pi\ x&=\frac{k\pi+\pi}{2}=\frac{(k + 1)\pi}{2},k\in\mathbb{Z} \end{align*} ]

Answer:

Vertical stretch: 2 Period: $\pi$ Phase shift: $\frac{\pi}{2}$ to the right Midline equation: $y = 1$ Equation of all asymptotes: $x=\frac{(k + 1)\pi}{2},k\in\mathbb{Z}$