Problem 36 - Algebra and Trigonometry with Analytic Geometry

Question

Find the period and sketch the graph of the equation. Show the asymptotes. $$ y=\frac{1}{2} \sec \left(2 x-\frac{\pi}{2}\right) $$

Step-by-Step Solution

Answer

The period is \( \pi \) and asymptotes occur at \( x = \frac{\pi}{2}(n + 1) \).

Step 1: Identify the function type

The given equation is \( y = \frac{1}{2} \sec \left(2x - \frac{\pi}{2}\right) \). This is a secant function, which is related to the cosine function. The secant function is defined as \( \sec(x) = \frac{1}{\cos(x)} \).

Step 2: Find the basic period of the secant function

The basic period of the secant function, \( \sec(x) \), is the same period as the cosine function, \( \cos(x) \), which is \( 2\pi \).

Step 3: Determine the transformed period

The transformed function is \( \sec(2x - \frac{\pi}{2}) \). The formula for finding the period of \( \sec(bx + c) \) is \( \frac{2\pi}{|b|} \). Here \( b = 2 \), so the period is \( \frac{2\pi}{2} = \pi \).

Step 4: Determine phase shift and vertical transformation

The phase shift of \( \sec(bx - c) \) is \( \frac{c}{b} \). For \( \sec(2x - \frac{\pi}{2}) \), this is \( \frac{-\frac{\pi}{2}}{2} = -\frac{\pi}{4} \). There is no vertical shift since the whole expression is not added to or subtracted from a constant, only multiplied by \( \frac{1}{2} \).

Step 5: Identify the asymptotes

Secant functions have vertical asymptotes where the cosine function is zero. For \( \sec(2x - \frac{\pi}{2}) \), find where \( \cos(2x - \frac{\pi}{2}) = 0 \). Solve \( 2x - \frac{\pi}{2} = (2n + 1)\frac{\pi}{2} \), \( x = \frac{\pi}{2}(n + 1) \) for \( n \in \mathbb{Z} \).

Step 6: Sketch the graph

Start with marking the period from \( 0 \) to \( \pi \), including the phase shift. Draw vertical asymptotes at \( x = \frac{\pi}{2}(n + 1) \). The amplitude \( \frac{1}{2} \) indicates that the graph stretches vertically by a factor of \( \frac{1}{2} \). Sketch the \( y = \frac{1}{2} \sec(2x - \frac{\pi}{2}) \) curve within each sub-interval created by these asymptotes.

Key Concepts

Secant Function

The secant function is an important trigonometric function that is related to the cosine function. It is defined as the reciprocal of the cosine function: - \( \sec(x) = \frac{1}{\cos(x)} \).
This means that wherever the cosine function equals zero, the secant function will have a vertical asymptote since division by zero is undefined.
The graph of the secant function, therefore, consists of upward and downward extending curves called branches, with vertical asymptotes occurring at points where the cosine of the angle is zero. A key property of the secant function is that it shares the same period as the cosine function, meaning these functions repeat their values over the same interval. One significant point to remember while sketching secant function graphs is to identify these points of asymptotes and the reciprocal nature of secant values compared to cosine.

Period of Trigonometric Functions

When discussing trigonometric functions, the period refers to the length of one complete cycle of the function values before they start repeating. For the basic secant and cosine functions, this period is \(2\pi\).
If we have a transformed function \(\sec(bx + c)\), like in the equation \(y = \frac{1}{2} \sec\left(2x - \frac{\pi}{2}\right)\), the period changes based on the coefficient \(b\).
The formula used to calculate this new period is: - \(\frac{2\pi}{|b|}\).
For this problem, \(b = 2\), hence the period of the transformed function is \(\frac{2\pi}{2} = \pi\). Understanding the period is crucial when sketching the graph, as it dictates how the repeating pattern of the function aligns along the x-axis. In essence, identifying the period helps in understanding how frequently the secant function will start a new cycle.

Phase Shift

Phase shift occurs when the graph of a trigonometric function is horizontally displaced from its original position. In the expression \(\sec(bx + c)\), phase shift is calculated as: - \(\frac{-c}{b}\).
For our problem, this translates to a phase shift of \(\frac{-\left(-\frac{\pi}{2}\right)}{2} = \frac{\pi}{4}\). This positive phase shift means the entire graph of the secant function moves to the right by \(\frac{\pi}{4}\) units.
Recognizing and applying phase shifts plays an important role in graphing since they define how the starting point of the function is adjusted along the x-axis. Additionally, alongside the period, phase shifts are vital considerations to ensure that the graph accurately represents the function's true behavior.

Vertical Asymptotes

Vertical asymptotes in trigonometric functions like secant occur at specific x-values where the function becomes undefined—in this case, where the cosine of the angle is zero. - For \(\sec(x)\), asymptotes occur at \(x = \frac{\pi}{2} + n\pi\), with \(n \in \mathbb{Z}\).
For the given function \(\frac{1}{2} \sec(2x - \frac{\pi}{2})\), to find the asymptotes, set \(\cos(2x - \frac{\pi}{2}) = 0\).
This results in vertical asymptotes at: - \(x = \frac{\pi}{2}(n+1)\), for \(n \in \mathbb{Z}\). Graphically, asymptotes are shown as dashed lines indicating regions of undefined values approaching infinity. These guide the graph's sketch, as they mark the intervals over which the curves of the secant function extend, illustrating the non-continuity at those points. Remembering the location of these asymptotes aids in accurately portraying the function for any given interval.