the graph of $y = \\tan\\left\\frac{1}{4}\\left(x - \\frac{\\pi}{2}\\right)\\right+1$ is shown. what is the…

Explanation:

Step1: Recall tangent - function period formula

The period of the tangent function $y = A\tan(B(x - C))+D$ is given by $T=\frac{\pi}{|B|}$. For the function $y=\tan\left[\frac{1}{4}(x - \frac{\pi}{2})\right]+1$, $B = \frac{1}{4}$.

Step2: Calculate the period

$T=\frac{\pi}{\left|\frac{1}{4}\right|}=4\pi$.

Step3: Recall tangent - function asymptote formula

The asymptotes of the tangent function $y = \tan(x)$ occur at $x=(n+\frac{1}{2})\pi$, where $n\in\mathbb{Z}$. For the function $y=\tan\left[\frac{1}{4}(x - \frac{\pi}{2})\right]+1$, we set $\frac{1}{4}(x - \frac{\pi}{2})=(n+\frac{1}{2})\pi$.

Step4: Solve for $x$

First, multiply both sides of the equation $\frac{1}{4}(x - \frac{\pi}{2})=(n+\frac{1}{2})\pi$ by $4$: $x-\frac{\pi}{2}=4(n + \frac{1}{2})\pi$. Then expand the right - hand side: $x-\frac{\pi}{2}=(4n + 2)\pi$. Add $\frac{\pi}{2}$ to both sides: $x=(4n + 2)\pi+\frac{\pi}{2}=\frac{(8n + 4 + 1)\pi}{2}=\frac{(8n+5)\pi}{2}= \frac{5\pi}{2}+4n\pi$.

Answer:

Period: $4\pi$ Asymptotes: $\frac{5\pi}{2}+4n\pi$