Question Graph the following function: \[ y=\frac{1}{2} \cot \left(3 x+\frac{3 \pi}{4}\right)+1 \] Drag the movable black point to set the left vertical asymptote and shift the function, the red point to set the right vertical asymptote (thereby setting the period of the function), and the blue point at the correct set of coordinates. Note that the two asymptotes can be moved independently of each other and that only one period of the function is shown.

1. Write the function in standard form: \[ y=\frac{1}{2} \cot\left(3x+\frac{3\pi}{4}\right)+1 \] 2. Note that the argument of the cotangent, \(3x+\frac{3\pi}{4}\), can be written in the form \(B(x - C)\) where: - \(B=3\), so the period of \(\cot\) (normally \(\pi\)) becomes \(\frac{\pi}{3}\). - The phase shift is obtained by setting \[ 3x+\frac{3\pi}{4}=3\left(x+\frac{\pi}{4}\right), \] indicating a horizontal shift left by \(\frac{\pi}{4}\). 3. Determine the vertical asymptotes. For the cotangent function, vertical asymptotes occur when its argument equals any integer multiple of \(\pi\): \[ 3x+\frac{3\pi}{4}=k\pi,\quad k\in\mathbb{Z}. \] Choose two consecutive integers to locate the two asymptotes shown on one period. - For the left asymptote, let \(k=0\): \[ 3x+\frac{3\pi}{4}=0 \quad\Longrightarrow\quad 3x=-\frac{3\pi}{4} \quad\Longrightarrow\quad x=-\frac{\pi}{4}. \] Thus, place the black point (left vertical asymptote) at \(x=-\frac{\pi}{4}\). - For the right asymptote, let \(k=1\): \[ 3x+\frac{3\pi}{4}=\pi \quad\Longrightarrow\quad 3x=\pi-\frac{3\pi}{4}=\frac{\pi}{4} \quad\Longrightarrow\quad x=\frac{\pi}{12}. \] Thus, place the red point (right vertical asymptote) at \(x=\frac{\pi}{12}\). 4. Identify the blue point. A convenient reference is the point where the cotangent equals \(0\) (since \(\cot\frac{\pi}{2}=0\)): \[ 3x+\frac{3\pi}{4}=\frac{\pi}{2}\quad\Longrightarrow\quad 3x=\frac{\pi}{2}-\frac{3\pi}{4}=-\frac{\pi}{4}\quad\Longrightarrow\quad x=-\frac{\pi}{12}. \] At this \(x\)-value, compute \(y\): \[ y=\frac{1}{2}\cot\left(\frac{\pi}{2}\right)+1=\frac{1}{2}\times 0+1=1. \] Thus, the blue point is at \(\left(-\frac{\pi}{12},1\right)\). 5. Summary of key points: - **Left vertical asymptote (black point):** \(x=-\frac{\pi}{4}\) - **Right vertical asymptote (red point):** \(x=\frac{\pi}{12}\) - **Midline intersection (blue point):** \(\left(-\frac{\pi}{12},1\right)\) - **Period of the function:** \(\frac{\pi}{3}\) These placements correctly reflect the horizontal shift, period, and vertical translation of the function.