Problem 52 - Precalculus

Question

In Exercises \(47-56,\) graph the functions over at least one period. $$y=-1+4 \sec (2 x+\pi)$$

Step-by-Step Solution

Answer

Graph the function considering its period, phase shift, transformations, asymptotes, and key points.

Step 1: Analyze the Basic Function

The function is given as \( y = -1 + 4 \sec (2x + \pi) \). The basic function here is \( \sec u \), where \( u = 2x + \pi \). The secant function, \( \sec u \), is the reciprocal of the cosine function and is undefined where the cosine function equals zero.

Step 2: Determine the Phase Shift and Period

The modified function is \( \sec (2x + \pi) \). Here, \( 2x + \pi \) can be rewritten as \( 2(x + \pi/2) \), indicating a phase shift of \( -\pi/2 \). The period of secant is \( \frac{2\pi}{k} \) where \( k = 2 \), so the period is \( \pi \).

Step 3: Identify the Transformations

The expression \( y = -1 + 4 \sec(2x + \pi) \) represents a vertical stretch by a factor of 4 and a vertical shift downward by 1 unit from the form \( y = \sec u \).

Step 4: Find Asymptotes

The secant function has vertical asymptotes where the cosine function is 0. From \( \cos(2x + \pi) = 0 \), solve for asymptotes: \( 2x + \pi = \frac{\pi}{2} + n\pi \). Solving gives \( x = \frac{n\pi}{2} - \frac{\pi}{4} \).

Step 5: Graph Key Points and Asymptotes

For one period from \( x = -\frac{\pi}{4} \) to \( x = \frac{3\pi}{4} \), plot asymptotes at \( x = -\frac{\pi}{4} \), \( x = \frac{\pi}{4} \), and \( x = \frac{3\pi}{4} \). Plot values of \( y \) at key points, such as when the cosine is 1 or -1 corresponding to x-intercepts of parent cosine function.

Step 6: Plot the Graph

Draw the vertical asymptotes and sketch curves approaching these asymptotes, reflecting the signature inverted nature of secant curves, reflecting around lines due to the vertical stretch and shift.

Key Concepts

Secant Function

The secant function, written as \( \sec(u) \), is the reciprocal of the cosine function. This means it is defined as \( \sec(u) = \frac{1}{\cos(u)} \). Because of this relationship, the secant function is undefined wherever the cosine function is equal to zero. This is because division by zero is undefined in mathematics. The secant function exhibits a pattern of peaks that mirror the minima and maxima of the cosine function:

• The maximum points of secant correspond to the minimum points of cosine.
• The minimum points of secant correspond to the maximum points of cosine.

The graph of the secant function has vertical asymptotes, which occur at the x-values where the cosine function crosses the x-axis. This asymptotic behavior is one of the unique features that separates the secant function from other trigonometric functions.

Phase Shift

A phase shift in trigonometric functions involves translating the graph horizontally. In the context of the given function \( y = -1 + 4 \sec(2x + \pi) \), a phase shift occurs because of the transformation \( 2x + \pi \). To identify the phase shift explicitly, rewrite the expression in terms of a recognizable shift:\[ 2\left(x + \frac{\pi}{2}\right) \]This indicates that the graph of the secant function is shifted horizontally to the left by \( \frac{\pi}{2} \) units. Phase shifts are crucial when graphing as they determine the starting point of the periodic function on the x-axis. When analyzing or sketching graphs, acknowledging phase shifts helps in accurately depicting the position of all peaks, valleys, and vertical asymptotes.

Vertical Asymptotes

Vertical asymptotes are fundamental characteristics in functions like secant. They represent x-values where the function heads to infinity or negative infinity. These occur in the secant function because it is the reciprocal of the cosine function, and they specifically occur where the cosine function is zero. In our function \( y = -1 + 4 \sec(2x + \pi) \), the equation for vertical asymptotes comes from solving \( \cos(2x + \pi) = 0 \). This translates to:\[ 2x + \pi = \frac{\pi}{2} + n\pi \]Solving this gives the locations of the asymptotes as \( x = \frac{n\pi}{2} - \frac{\pi}{4} \). These asymptotes repeat every half-period of the parent secant function's period, which is determined by the coefficient of \( x \) in \( 2x + \pi \). Vertical asymptotes are depicted on graphs as dashed lines, showing the boundaries around which the secant function curves.

Graphing Trigonometric Functions

Graphing trigonometric functions like secant involves several steps to ensure accuracy:

• Determine the period and any phase shifts to understand the function's horizontal alterations.
• Identify and label vertical asymptotes with dashed lines. These represent the points where the function is undefined.
• Calculate and plot key points, such as the peaks of secant functions and how they mimic the cosine function's behavior.
• Depict the curve by sketching it near the asymptotes, taking care to show how it approaches these lines without touching them.

When the function is transformed with vertical stretches or shifts, like in \( y = -1 + 4 \sec(2x + \pi) \), these transformations need to be visualized on the graph:

• The coefficient 4 represents a vertical stretch, making the secant arches taller.
• The \(-1\) results in shifting the entire graph 1 unit downward.

This strategy ensures that the graph of the function maintains the distinct shape of the secant function while incorporating the adjustments from transformations.