Problem 29 - Precalculus: Functions and Graphs

Question

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

Step-by-Step Solution

Answer

The period is \(2\pi\), with vertical asymptotes at \(x = \pi + k\pi\).

Step 1: Understand the Function

The function given is \( y = \sec\left(x - \frac{\pi}{2}\right) \). Recall that the secant function, \(\sec(x)\), is the reciprocal of the cosine function, \(\cos(x)\). Thus, \(\sec(x) = \frac{1}{\cos(x)}\). This means \(\sec(x)\) will have asymptotes wherever \(\cos(x)\) is zero.

Step 2: Identify Horizontal Shift

The function \( y = \sec\left(x - \frac{\pi}{2}\right) \) indicates a horizontal shift of \(\frac{\pi}{2}\) units to the right. This transformation shifts the standard secant graph, resulting in a phase shift.

Step 3: Determine the Period

The period of the standard \(\sec(x)\) is \(2\pi\), since it is based on the periodicity of the cosine function. The function \(\sec\left(x - \frac{\pi}{2}\right)\) has not been vertically stretched or horizontally scaled, so its period remains \(2\pi\).

Step 4: Determine Asymptotes

For \( y = \sec(x) \), vertical asymptotes occur at \( x = \frac{\pi}{2} + k\pi \), where \( k \) is an integer because \(\cos(x) = 0\) at these points. Considering the horizontal shift, the asymptotes for \( y = \sec(x - \frac{\pi}{2}) \) now occur at \( x = \frac{\pi}{2} + \frac{\pi}{2} + k\pi = \pi + k\pi \). So, the equations of the asymptotes are \( x = \pi + k\pi \).

Step 5: Sketch the Graph

Sketch the graph by plotting multiple periods between \(-2\pi\) and \(2\pi\). Begin with drawing vertical asymptotes using \( x = \pi + k\pi \). In each interval defined by two consecutive asymptotes, draw the secant curve, repeating its standard \"U-shaped\" behavior between \(-\infty\) and \(+\infty\), in accordance with the horizontal shift.

Key Concepts

Secant Function

The secant function, denoted as \(\sec(x)\), is a core concept in trigonometry and often arises in analysis of waveforms and oscillations. It is the reciprocal of the cosine function, expressed as \(\sec(x) = \frac{1}{\cos(x)}\).

One key point to keep in mind about \(\sec(x)\) is its behavior at points where \(\cos(x) = 0\), such as \(\frac{\pi}{2}\), \(\frac{3\pi}{2}\), etc. At these values, the cosine function becomes zero, making the reciprocal undefined.

• The graph of the secant function has vertical "gaps" at these points, known as vertical asymptotes.
• The typical "U" shape of the secant graph corresponds to the peaks and troughs of the cosine function completed over an interval of \(2\pi\).

Understanding this reciprocal relationship and how it affects graph shape and asymptotes helps in plotting and analyzing secant functions under various transformations.

Graph Transformation

In trigonometry, graph transformations involve shifting, stretching, or compressing a function's graph. For the equation \(y = \sec\left(x - \frac{\pi}{2}\right)\), the main transformation involved is a horizontal shift.

This transformation represents a phase shift of the secant graph, specifically a rightward shift by \(\frac{\pi}{2}\) units.

• Phase Shift: By altering the input as \(x - \frac{\pi}{2}\), the entire graph is moved horizontally, aligning its phase with that of the modified function.
• No Vertical Scaling: There's no vertical scaling or reflection because there are no coefficients affecting \(y\) or modifications to the secant function itself.

This type of transformation alters the positioning of asymptotes and affects where the "U" shapes of the secant graph appear along the x-axis.

Periodicity

Periodicity in trigonometric functions refers to the repeating nature of the function's graph over regular intervals. The secant function has a periodic nature because it is directly tied to its reciprocal, the cosine function, which repeats every \(2\pi\).

For \(y = \sec(x)\), the period is \(2\pi\), just like \(y = \cos(x)\).

• No Altered Period: Since the equation \(y = \sec(x - \frac{\pi}{2})\) does not involve any changes to the coefficient of the \(x\), the period remains \(2\pi\).
• Understanding the Period: Recognizing the period allows us to accurately predict the graph's behavior across intervals of \(x\), ensuring we capture the full repetitive nature of the secant function.

Knowing how transformations affect periodicity is crucial. They can shift the entire wave along the x-axis but the interval remains the same.

Asymptotes

Asymptotes are key features of the graph of \(\sec(x)\), occurring wherever \(\cos(x)\) is zero, causing \(\sec(x)\) to be undefined.

For the standard secant function, these vertical asymptotes appear at intervals of \(x = \frac{\pi}{2} + k\pi\), where \(k\) is an integer.

• Influence of Transformations: With a function like \(y = \sec\left(x - \frac{\pi}{2}\right)\), horizontal shifts need adjustment in the asymptote formula. The asymptotes shift right by \(\frac{\pi}{2}\), resulting in \(x = \pi + k\pi\).
• Graphical Importance: These asymptotes are depicted as vertical lines where the secant curve approaches but never touches or crosses, leading to break the graph into intervals.

Accounting for these shifts ensures accuracy in sketching and interpreting the graph of transformed secant functions.