Problem 66 - Algebra and Trigonometry

Question

If you are given the equation of a tangent function, how do you find a pair of consecutive asymptotes?

Step-by-Step Solution

Answer

The consecutive asymptotes of a tangent function can be found using the formula \(x = \frac{(2n + 1)\pi + 2d}{2k}\), substitute integer values for \(n\) to get the asymptotes.

Step 1: Identify the Tangent Function

The first step is to identify the given tangent function. This function might have a form like \(tan(kx-d)\), where \(k\) indicates the scaling of the horizontal axis and \(d\) shifts the function along the x-axis. If the function is simply stated as \(tan(x)\), then \(k = 1\) and \(d = 0\). In the case that \(k\) and \(d\) are not equal to \(1\) and \(0\) respectively, do take note, as it will affect our calculation for the vertical asymptotes.

Step 2: Compute the Asymptotes

Once you have identified the given tangent function and its parameters \(k\) and \(d\), next step is to compute the asymptotes. The standard tangent function has asymptotes at \(x = (2n + 1)\frac{\pi}{2}\), for any integer \(n\). However, because of the scaling and shifting, the formula changes to \(x = \frac{(2n + 1)\pi + 2d}{2k}\). Substitute integer values for \(n\) to derive asymptotes in pairs.

Step 3: Check the Asymptotes

The last step is to confirm that the derived asymptotes are valid. The tangent function should approach positive or negative infinity at these points. Plug the points extracted for the asymptotes into the tangent function to ensure that the function is undefined.