Question

Determine the period, amplitude, and shifts (both horizontal and vertical) and draw a graph over the interval \(-5 \leq x \leq 5\) for the functions listed. $$ y=3+\sec (x-\pi) $$

Short answer

Period is \( 2\pi \), vertical shift is 3 units up, and horizontal shift is \( \pi \) units right.

Step-by-step solution

Rewrite the Secant Function

The given function is \( y = 3 + \sec(x - \pi) \). To better analyze it, we can express the secant function in terms of cosine: \( y = 3 + \frac{1}{\cos(x-\pi)} \). This doesn't affect the period, amplitude, or shifts but helps in understanding the graph.

Determine the Amplitude

Secant functions, similar to cosine and sine, have no amplitude in the traditional sense because they range from negative to positive infinity. However, we can consider the graph's behavior through vertical and horizontal shifts instead.

Identify the Horizontal Shift

The function \( y = 3 + \sec(x - \pi) \) has a horizontal shift of \( \pi \) units to the right due to the term \( (x - \pi) \). This means the graph of \( \sec(x) \) is shifted \( \pi \) units right.

Identify the Vertical Shift

The graph is shifted vertically by 3 units upwards because of the '+3' in the function \( y = 3 + \sec(x - \pi) \). This means that the middle line, which the secant graph branches from, is \( y = 3 \) rather than \( y = 0 \).

Calculate the Period

For secant functions, the period is the distance over which the function repeats. The functional form \( \sec(bx) \) has a period of \( \frac{2\pi}{b} \). In this problem, \( b = 1 \), so the period remains \( 2\pi \).

Plotting the Graph

To plot \( y = 3 + \sec(x - \pi) \) over \(-5 \leq x \leq 5\), consider the vertical asymptotes where \( \cos(x - \pi) = 0 \), i.e., at \( x = \pi/2, 3\pi/2 \), etc. The graph has branches going to infinity at these points and crosses the horizontal line \( y = 4 \) (maximum) and \( y = 2 \) (minimum due to the negative sections of secant). Include key points within the interval for accuracy.

Key concepts

Secant Function

The secant function, denoted as \( \sec(x) \), is the reciprocal of the cosine function. It equals \( \frac{1}{\cos(x)} \). This means wherever the cosine function is zero, the secant function goes to infinity, causing vertical asymptotes. Understanding the secant as \( y = \frac{1}{\cos(x-\pi)} \) can help visualize these properties as it retains the same critical behavior. It's important to remember that secant doesn't have an amplitude like sine or cosine, since its values range from negative to positive infinity. When graphing a secant function, note the untouchable zones where the cosine function equals zero, as these influence the position of vertical asymptotes, and hence, the secant function's graph.

Amplitude and Period

Unlike sine and cosine, secant functions do not possess a defined amplitude because their range is unbounded. Instead, secant graphs consist of distinct branches that extend to infinity, reflecting its infinite range between positive and negative values. The concept of period, however, applies. A secant function like \( \sec(x) \) repeats itself every \( 2\pi \). For the transformed function \( \sec(x-\pi) \), the period remains \( 2\pi \). This means every \( 2\pi \) interval, the graph will mirror previous sections, creating a predictable and continuous pattern of branches defined by asymptotes and peaks/valleys. These intervals are derived from the period formula \( \frac{2\pi}{b} \), where \( b \) is the coefficient of \( x \).

Graphing Trigonometric Functions

When graphing trigonometric functions like secant, it's essential to understand how they behave over particular x-ranges. For the equation \( y = 3 + \sec(x - \pi) \), the graph features a repetition pattern across each period interval (\( 2\pi \) in this case). The critical aspect of graphing secant includes identifying asymptotes where the cosine part turns zero. Within each period, the graph will stretch and curve away to infinity at these vertical segments. Between the asymptotes, the graph of secant will have upward and downward branches originating from a horizontal midline, adjusted here to \( y = 3 \). This midline shift occurs due to the vertical transformation applied to the basic secant function.

Horizontal and Vertical Shifts

The function \( y = 3 + \sec(x - \pi) \) includes shifts that transform the graph from the basic \( \sec(x) \). The term \( x - \pi \) induces a horizontal shift. Specifically, the graph moves \( \pi \) units to the right. This means, compared to just \( \sec(x) \), every point on the secant curve is relocated to the right. Vertical adjustments are made by the "+3" outside the secant. This shifts the entire graph upwards by 3 units. Such a vertical shift affects where the graph centers, moving the base horizontal line from \( y = 0 \) to \( y = 3 \), changing the vertical extents of secant's upper and lower branches in relation to this new baseline, reflecting how the function scales vertically.