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\[ y=4 \cot \left(\frac{1}{2} x\right) \] Drag the movable red point to set the vertical asymptotes (dashed red lines) for one period of the given function. Also place the blue point at the correct set of coordinates.

Ask by Gardner Bryant. in the United States
Mar 11,2025

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Set the vertical asymptotes at \( x = 0 \) and \( x = 2\pi \). Place the blue point at \( (\pi, 0) \).

Solución

1. The function is given by \[ y = 4 \cot\left(\frac{1}{2}x\right). \] 2. For the cotangent function, vertical asymptotes occur when its argument is an integer multiple of \(\pi\), i.e. when \[ \cot\theta \quad \text{is undefined for} \quad \theta = k\pi, \quad k\in\mathbb{Z}. \] Here, the argument is \(\frac{1}{2}x\), so set \[ \frac{1}{2}x = k\pi. \] Solving for \(x\) gives \[ x = 2k\pi. \] 3. Choosing one period of the function, select two adjacent vertical asymptotes. For instance, take \[ k=0 \quad \text{and} \quad k=1. \] This places the vertical asymptotes at \[ x=0 \quad \text{and} \quad x=2\pi. \] Drag the movable red points to these \(x\)-values. 4. The cotangent function passes through zero exactly midway between its vertical asymptotes. The midpoint of \(x=0\) and \(x=2\pi\) is \[ x = \frac{0+2\pi}{2} = \pi. \] Substitute \(x=\pi\) into the function: \[ y = 4 \cot\left(\frac{1}{2}\pi\right) = 4\cot\left(\frac{\pi}{2}\right). \] Since \(\cot\left(\frac{\pi}{2}\right)=0\), we have \[ y = 0. \] Place the blue point at the coordinate \[ (\pi, 0). \]

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The function \( y = 4 \cot\left(\frac{1}{2} x\right) \) has vertical asymptotes where the cotangent function is undefined, which occurs at the points where its argument is an integer multiple of \( \pi \). For this particular function, the vertical asymptotes can be found by solving the equation \( \frac{1}{2} x = k\pi \) for integer values of \( k \). This results in \( x = 2k\pi \). For one period of the cotangent function, typically from \( 0 \) to \( \pi \), the vertical asymptotes occur at \( x = 0 \) and \( x = 2\pi \). Move the red point to these coordinates to set the vertical asymptotes at \( x = 0 \) and \( x = 2\pi \). As for placing the blue point, it should be positioned at the point where \( y = 0 \), which occurs at \( x = \pi \) for the given period, yielding the coordinates \( (\pi, 0) \).

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