Problem 25 - A Graphical Approach to Precalculus with Limits

Question

Find the (a) period, (b) phase shift (if any), and (c) range of each function. $$y=\frac{1}{2} \sec (2 x+\pi)$$

Step-by-Step Solution

Answer

Period: \(\pi\), Phase Shift: \(\frac{\pi}{2}\) left, Range: \((-\infty, -\frac{1}{2}] \cup [\frac{1}{2}, \infty)\).

Step 1: Identify the standard form

The standard form for a secant function is \( y = A \sec(Bx + C) \). In this equation, \( y = \frac{1}{2} \sec(2x + \pi) \), we can see that \( A = \frac{1}{2} \), \( B = 2 \), and \( C = \pi \).

Step 2: Find the period

The period of the secant function is determined by the formula \( \frac{2\pi}{|B|} \). For our function, \( B = 2 \), so the period is \( \frac{2\pi}{2} = \pi \).

Step 3: Calculate the phase shift

The phase shift is calculated using \( \frac{-C}{B} \). Here, \( C = \pi \) and \( B = 2 \), so the phase shift is \( \frac{-\pi}{2} \). The graph shifts \( \frac{\pi}{2} \) units to the left.

Step 4: Determine the range

The range of the secant function is always unbounded above and below, except for a gap around zero. For the general form \( y = A \sec(Bx + C) \), the range is \( (-\infty, -|A|] \cup [|A|, \infty) \). With \( A = \frac{1}{2} \), the range is \( (-\infty, -\frac{1}{2}] \cup [\frac{1}{2}, \infty) \).

Key Concepts

Period of a Function

In trigonometry, the period of a function is the interval after which the function repeats its behavior. For periodic functions like sine, cosine, and secant, this is a fundamental characteristic.
To find the period of a secant function, use the formula:

• Period = \( \frac{2\pi}{|B|} \)

Here, \(B\) is a coefficient that multiplies the variable \(x\) in the inside of the secant function. This coefficient impacts the frequency of the oscillations.
For example, in the given function \(y = \frac{1}{2} \sec(2x + \pi)\), the coefficient \(B\) is 2. Plugging this into our formula gives:

• Period = \( \frac{2\pi}{2} = \pi \)

This means the secant function repeats every \(\pi\) units along the x-axis, making it half as wide as the parent function \(y = \sec(x)\), which is \(2\pi\).
Understanding the period helps in graphing the function as it provides a clear picture of how frequently the function completes a cycle.

Phase Shift

The phase shift of a trigonometric function determines the horizontal translation of the graph along the x-axis. This can make the curve start at a different point, essentially shifting left or right.
The formula to calculate phase shift is given by:

• Phase Shift = \( \frac{-C}{B} \)

In the given function \(y = \frac{1}{2} \sec(2x + \pi)\), \(C\) is \(\pi\) and \(B\) is 2. Substituting into the formula, we get:

• Phase Shift = \( \frac{-\pi}{2} \)

This means the graph shifts \(\frac{\pi}{2}\) units to the left. Positive phase shift values indicate a shift to the left, whereas negative values indicate a shift to the right.
Understanding this shift is important for accurate graphing because it affects where the function's repeating pattern begins.

Range of a Function

The range of a function defines the set of all possible output values or y-values that the function can take. For a secant function, the range is unbounded except for some values around zero. This is due to the secant function's vertical asymptotes, which lead to large magnitudes of y-values both positive and negative.
For the secant function \(y = A \sec(Bx + C)\), the basic range – not considering amplitude \(A\) – is:

• Range = \((-\infty, -1] \cup [1, \infty)\)

When applying an amplitude \(A\), the range adjusts to:

• Range = \((-\infty, -|A|] \cup [|A|, \infty)\)

In the function \(y = \frac{1}{2} \sec(2x + \pi)\), \(A\) is \(\frac{1}{2}\), so the range becomes:

• Range = \((-\infty, -\frac{1}{2}] \cup [\frac{1}{2}, \infty)\)

Notice the gaps around zero, where zero values cannot be reached. This reflects the function’s undefined nature at certain angles where the vertical asymptotes are located.