Solved on Sep 26, 2023

Determine the principal cycle, period, vertical asymptotes, center, and halfway points of $y=\tan(\frac{1}{6}x-\pi)$ . Sketch the graph.
The interval of the principal cycle is $[-\pi,\pi]$ . The period is $6\pi$ . The equation of the left vertical asymptote is $x=\frac{7\pi}{3}$ and the right is $x=\frac{11\pi}{3}$ . The coordinates of the center point are $\left(3\pi,0\right)$ . The coordinates of the left-most halfway point are $\left(\frac{\pi}{3},0\right)$ . The coordinates of the right-most halfway point are $\left(\frac{5\pi}{3},0\right)$ .

STEP 1

Assumptions1. The given function is $y=\tan \left(\frac{1}{6} x-\pi\right)$ . We are asked to find the interval for the principal cycle, the period, the equations of the vertical asymptotes, the coordinates of the center point, and the coordinates of the halfway points for the principal cycle.
3. We are also asked to sketch the graph of the function.

STEP 2

The general form of a tangent function is $y = \tan(Bx - C)$ , where $B$ affects the period and $C$ affects the phase shift. The period of the standard tangent function is $\pi$ , and the period of $y = \tan(Bx)$ is $\frac{\pi}{B}$ .
In our function, $B = \frac{1}{6}$ , so we can calculate the period as follows $Period = \frac{\pi}{B}$

STEP 3

Now, plug in the value for $B$ to calculate the period.
$Period = \frac{\pi}{\frac{1}{6}}$

STEP 4

Calculate the period.
$Period = \frac{\pi}{\frac{1}{6}} =6\pi$

STEP 5

The principal cycle of the tangent function is from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ . For $y = \tan(Bx - C)$ , the interval of the principal cycle is from $C - \frac{\pi}{2B}$ to $C + \frac{\pi}{2B}$ .
In our function, $B = \frac{1}{}$ and $C = \pi$ , so we can calculate the interval as follows $Interval = \left[C - \frac{\pi}{2B}, C + \frac{\pi}{2B}\right]$

STEP 6

Now, plug in the values for $B$ and $C$ to calculate the interval.
$Interval = \left[\pi - \frac{\pi}{2 \cdot \frac{1}{6}}, \pi + \frac{\pi}{2 \cdot \frac{1}{6}}\right]$

STEP 7

Calculate the interval.
$Interval = \left[\pi -3\pi, \pi +3\pi\right] = \left[-2\pi,4\pi\right]$

STEP 8

The vertical asymptotes of the tangent function are at the boundaries of each cycle. For $y = \tan(Bx - C)$ , the equations of the vertical asymptotes are $x = C - \frac{\pi}{2B}$ and $x = C + \frac{\pi}{2B}$ .
In our function, $B = \frac{1}{6}$ and $C = \pi$ , so we can calculate the equations of the vertical asymptotes as follows $Left\, asymptote x = C - \frac{\pi}{2B}$ $Right\, asymptote x = C + \frac{\pi}{2B}$

STEP 9

Now, plug in the values for $B$ and $C$ to calculate the equations of the vertical asymptotes.
$Left\, asymptote x = \pi - \frac{\pi}{2 \cdot \frac{}{6}}$ $Right\, asymptote x = \pi + \frac{\pi}{2 \cdot \frac{}{6}}$

STEP 10

Calculate the equations of the vertical asymptotes.
$Left\, asymptote x = \pi -3\pi = -2\pi$ $Right\, asymptote x = \pi +3\pi =4\pi$

STEP 11

The center point of the principal cycle of the tangent function is at the middle of the interval, where $y =0$ . For $y = \tan(Bx - C)$ , the coordinates of the center point are $\left(C,0\right)$ .
In our function, $C = \pi$ , so the coordinates of the center point are $Center\, point \left(\pi,0\right)$

STEP 12

The halfway points of the principal cycle of the tangent function are at the middle of each half of the interval, where $y = \pm$ . For $y = \tan(Bx - C)$ , the coordinates of the halfway points are $\left(C \pm \frac{\pi}{4B}, \pm\right)$ .
In our function, $B = \frac{}{6}$ and $C = \pi$ , so we can calculate the coordinates of the halfway points as follows $Left\, halfway\, point \left(C - \frac{\pi}{4B}, -\right)$ $Right\, halfway\, point \left(C + \frac{\pi}{4B},\right)$

STEP 13

Now, plug in the values for $B$ and $C$ to calculate the coordinates of the halfway points.
$Left\, halfway\, point \left(\pi - \frac{\pi}{ \cdot \frac{}{6}}, -\right)$ $Right\, halfway\, point \left(\pi + \frac{\pi}{ \cdot \frac{}{6}},\right)$

STEP 14

Calculate the coordinates of the halfway points.
$Left\, halfway\, point \left(\pi - \frac{3}{2}\pi, -\right) = \left(-\frac{}{2}\pi, -\right)$ $Right\, halfway\, point \left(\pi + \frac{3}{2}\pi,\right) = \left(\frac{}{2}\pi,\right)$

STEP 15

To sketch the graph of the function, plot the center point, the halfway points, and the vertical asymptotes. Then draw the curve of the tangent function within the interval of the principal cycle.
The graph of the function $y=\tan \left(\frac{}{} x-\pi\right)$ has- Interval of the principal cycle $[-2\pi,4\pi]$ - Period $\pi$ - Equations of the vertical asymptotes $x = -2\pi$ and $x =4\pi$ - Coordinates of the center point $(\pi,0)$ - Coordinates of the halfway points $(-\frac{}{2}\pi, -)$ and $(\frac{5}{2}\pi,)$

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STEP 1

STEP 2

STEP 3

Now, plug in the value for $B$ to calculate the period.
$Period = \frac{\pi}{\frac{1}{6}}$

STEP 4

Calculate the period.
$Period = \frac{\pi}{\frac{1}{6}} =6\pi$

STEP 5

STEP 6

Now, plug in the values for $B$ and $C$ to calculate the interval.
$Interval = \left[\pi - \frac{\pi}{2 \cdot \frac{1}{6}}, \pi + \frac{\pi}{2 \cdot \frac{1}{6}}\right]$

STEP 7

Calculate the interval.
$Interval = \left[\pi -3\pi, \pi +3\pi\right] = \left[-2\pi,4\pi\right]$

STEP 8

STEP 9

Now, plug in the values for $B$ and $C$ to calculate the equations of the vertical asymptotes.
$Left\, asymptote x = \pi - \frac{\pi}{2 \cdot \frac{}{6}}$ $Right\, asymptote x = \pi + \frac{\pi}{2 \cdot \frac{}{6}}$

STEP 10

Calculate the equations of the vertical asymptotes.
$Left\, asymptote x = \pi -3\pi = -2\pi$ $Right\, asymptote x = \pi +3\pi =4\pi$

STEP 11

STEP 12

STEP 13

Now, plug in the values for $B$ and $C$ to calculate the coordinates of the halfway points.
$Left\, halfway\, point \left(\pi - \frac{\pi}{ \cdot \frac{}{6}}, -\right)$ $Right\, halfway\, point \left(\pi + \frac{\pi}{ \cdot \frac{}{6}},\right)$

STEP 14

Calculate the coordinates of the halfway points.
$Left\, halfway\, point \left(\pi - \frac{3}{2}\pi, -\right) = \left(-\frac{}{2}\pi, -\right)$ $Right\, halfway\, point \left(\pi + \frac{3}{2}\pi,\right) = \left(\frac{}{2}\pi,\right)$

STEP 15

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STEP 1

STEP 2

STEP 3

Now, plug in the value for BBB to calculate the period.Period=π16Period = \frac{\pi}{\frac{1}{6}}Period=61​π​

STEP 4

Calculate the period.Period=π16=6πPeriod = \frac{\pi}{\frac{1}{6}} =6\piPeriod=61​π​=6π

STEP 5

STEP 6

Now, plug in the values for BBB and CCC to calculate the interval.Interval=[π−π2⋅16,π+π2⋅16]Interval = \left[\pi - \frac{\pi}{2 \cdot \frac{1}{6}}, \pi + \frac{\pi}{2 \cdot \frac{1}{6}}\right]Interval=[π−2⋅61​π​,π+2⋅61​π​]

STEP 7

Calculate the interval.Interval=[π−3π,π+3π]=[−2π,4π]Interval = \left[\pi -3\pi, \pi +3\pi\right] = \left[-2\pi,4\pi\right]Interval=[π−3π,π+3π]=[−2π,4π]

STEP 8

STEP 9

STEP 10

Calculate the equations of the vertical asymptotes.Left asymptotex=π−3π=−2πLeft\, asymptote x = \pi -3\pi = -2\piLeftasymptotex=π−3π=−2πRight asymptotex=π+3π=4πRight\, asymptote x = \pi +3\pi =4\piRightasymptotex=π+3π=4π

STEP 11

STEP 12

STEP 13

STEP 14

STEP 15

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Now, plug in the value for $B$ to calculate the period.
$Period = \frac{\pi}{\frac{1}{6}}$

Calculate the period.
$Period = \frac{\pi}{\frac{1}{6}} =6\pi$

Now, plug in the values for $B$ and $C$ to calculate the interval.
$Interval = \left[\pi - \frac{\pi}{2 \cdot \frac{1}{6}}, \pi + \frac{\pi}{2 \cdot \frac{1}{6}}\right]$

Calculate the interval.
$Interval = \left[\pi -3\pi, \pi +3\pi\right] = \left[-2\pi,4\pi\right]$

Calculate the equations of the vertical asymptotes.
$Left\, asymptote x = \pi -3\pi = -2\pi$ $Right\, asymptote x = \pi +3\pi =4\pi$