Problem 45 - Precalculus Essentials

Question

Graph two periods of each function. $$y=2 \tan \left(x-\frac{\pi}{6}\right)+1$$

Step-by-Step Solution

Answer

The graph of the function \(y=2 \tan \left(x-\frac{\pi}{6}\right)+1\) first comprises the graph of \(tan(x)\) stretched vertically by a factor of 2, shifted right by \(\pi/6\) and up by 1 unit. This graph is repeated for every period of length \(\pi\).

Step 1: Identify the transformations

The given function is \(y=2 \tan \left(x-\frac{\pi}{6}\right)+1\). The number 2 is multiplying the tan function which causes a vertical scaling/stretching of the graph by a factor of 2. The expression \(x-\frac{\pi}{6}\) inside the tan function causes a shift of the graph to the right by \(\frac{\pi}{6}\) units. The +1 at the end means there's an upward shift of the graph by 1 unit.

Step 2: Plot the original function

Start by plotting the standard function y = tan(x) without transformations for one period on the interval (-π/2, π/2). Breaks or asymptotes would occur at \(x = \pm \frac{\pi}{2} + k\pi\), where \(k\) is any integer.

Step 3: Apply the horizontal transformation

Next, shift the graph of y = tan(x) to the right by \(\frac{\pi}{6}\) units. This shifts all points and asymptotes to the right by \(\frac{\pi}{6}\) units.

Step 4: Apply the vertical transformations

Now, vertically stretch the graph by a factor of 2 and then shift it upward by 1 unit. All y-coordinates in the original graph will be doubled and then added by 1.

Step 5: Draw the second period

The period of the function y = tan(x) is \(\pi\). With the transformation, this has not changed. Draw the same graph as for the first period on the interval \(\left(\frac{\pi}{2} + \frac{\pi}{6}, \frac{3\pi}{2} + \frac{\pi}{6}\right)\) to complete graphing two periods of the function.