\[ y=4 \tan \left(\frac{1}{2} x+\frac{\pi}{4}\right)-1 \] Drag the movable black point to shift the function, the red points to set the vertical asymptotes, and the blue point at the correct set of coordinates. Note that only one period of the function is shown.

We start with the function \[ y=4 \tan\left(\frac{1}{2}x+\frac{\pi}{4}\right)-1. \] Step 1. Transform inside the tangent. The argument of the tangent is \[ \frac{1}{2}x+\frac{\pi}{4}. \] To find the point on the graph where the “basic” tangent (which normally has its zero‐at the inflection point) occurs, set the inside equal to zero: \[ \frac{1}{2}x+\frac{\pi}{4}=0. \] Solving for \(x\): \[ \frac{1}{2}x=-\frac{\pi}{4} \quad\Longrightarrow\quad x=-\frac{\pi}{2}. \] At \(x=-\frac{\pi}{2}\), the tangent is zero, so \[ y=4\tan(0)-1=0-1=-1. \] Thus the inflection point, which is often used as the “base point” for shifting the function, is \[ \left(-\frac{\pi}{2},\,-1\right). \] This is where the movable black point should be placed to shift the function. Step 2. Locate the vertical asymptotes. For the tangent function, vertical asymptotes occur when the argument is \[ \frac{\pi}{2}+k\pi,\quad k\in\mathbb{Z}. \] So we set \[ \frac{1}{2}x+\frac{\pi}{4}=\frac{\pi}{2}+k\pi. \] Subtract \(\frac{\pi}{4}\) from both sides: \[ \frac{1}{2}x=\frac{\pi}{2}-\frac{\pi}{4}+k\pi=\frac{\pi}{4}+k\pi. \] Multiply by 2: \[ x=\frac{\pi}{2}+2k\pi. \] For one period of the tangent curve (with period \(T=\pi/\frac{1}{2}=2\pi\)), choose two consecutive values of \(k\). For example, taking \(k=0\) and \(k=-1\) gives: - For \(k=0\): \[ x=\frac{\pi}{2}. \] - For \(k=-1\): \[ x=\frac{\pi}{2}-2\pi=-\frac{3\pi}{2}. \] These are the \(x\)-values of the vertical asymptotes (plotted as red points) for the displayed period. Step 3. Identify the blue point. A useful reference point on any tangent curve is the inflection point where the function crosses the midline. We already found this point to be \[ \left(-\frac{\pi}{2},\,-1\right). \] Thus, the blue point should be placed at these coordinates. In summary: - The movable black point (which shifts the function) should be placed at \(\left(-\frac{\pi}{2},-1\right)\). - The red points (the vertical asymptotes) should be set at \(x=-\frac{3\pi}{2}\) and \(x=\frac{\pi}{2}\). - The blue point (marking the central point of the tangent’s period) should be at \(\left(-\frac{\pi}{2},-1\right)\).