Question

Graph each function. Be sure to label key points and show at least two cycles. Use the graph to determine the domain and the range of each function. $$ y=3 \sec \left(\frac{1}{4} x\right)+1 $$

Short answer

The domain is all real numbers except \(4k\pi + 2\pi\) and the range is \(( -\infty, -2 ] \cup [4, \infty )\).

Step-by-step solution

Understand the given function

The given function is \[y = 3 \, \sec \left(\frac{1}{4}x \right) + 1\]which involves a secant function shifted and scaled. The general form of the secant function is \[y = A \, \sec(Bx - C) + D\].In this particular function, A is 3, B is \(\frac{1}{4}\), C is 0, and D is 1.

Identify the period

The period of the secant function \(\sec(Bx)\) is given by \(\frac{2\pi}{B}\). So for \(\sec \left(\frac{1}{4} x \right)\), the period is \(\frac{2\pi}{\frac{1}{4}} = 8\pi\).

Graph the parent function

Graph the parent function \(y = \sec \left(\frac{1}{4} x \right)\). First, graph \(y = \cos \left(\frac{1}{4} x \right)\), because the secant function is the reciprocal of the cosine function and undefined when the cosine function crosses the x-axis. The vertical asymptotes occur at \(x = 4k \pi + 2 \pi\), where \(k\) is an integer.

Apply vertical stretch and shift

Multiply the secant function by 3 and shift it up by 1. The new function is \(y = 3 \, \sec\left(\frac{1}{4}x\right) + 1\). So, every point on the secant curve will be stretched vertically by a factor of 3 and shifted up by 1 unit. Identify key points and asymptotes.

Label key points and cycles

Label key points, which occur when the original cosine function equals 1 or -1, now these points are (\(0, 4\), (\(8\pi, 4\), (\(4\pi, -2\), (\(12\pi, -2\)). The key asymptotes will be at (\(4\pi + 8k\pi\)), where (k) is an integer. Graph two cycles of the function to show its repetitive nature.

Determine the domain

Since secant is undefined wherever cosine equals zero, vertically asymptotic lines appear at \(x = 4k\pi + 2\pi\), where (k) is any integer. The domain is \(x eq 4k\pi + 2\pi\).

Determine the range

Given the vertical stretch and shift, the range of the function is all real numbers except those between \(-2\) and \(4\). So, the range is \(( -\infty, -2 ] \cup [4, \infty )\).

Key concepts

Secant Function

The secant function, denoted by \(\text{sec}(x)\), is the reciprocal of the cosine function. This means \(\text{sec}(x) = \frac{1}{\text{cos}(x)}\). Because the cosine function can be zero (causing division by zero), the secant function has vertical asymptotes where \text{cos}(x) = 0. Understanding the characteristics of the cosine function helps us to better graph the secant function.

• Secant function has vertical asymptotes where the cosine function crosses the x-axis.
• The secant function has no minimum or maximum values between the vertical asymptotes.
• The peaks and valleys of the secant function mirror the peaks and valleys of the cosine function, but they stretch infinitely.

By understanding these properties, you can graph any transformed secant function more easily.

Period of Trigonometric Functions

The period of a trigonometric function is the length of one complete cycle of the function. For the secant function \(\text{sec}(Bx)\), the period is determined by \(B\). Specifically, the period \(T\) is calculated as \(T = \frac{2\pi}{B}\).
For instance, given \(y = 3 \sec\left(\frac{1}{4} x \right) + 1\), the period is calculated as follows:

• \(B = \frac{1}{4}\)
• \(T = \frac{2\pi}{\frac{1}{4}} = 8\pi\)

This tells us that it takes \(8\pi\) units on the x-axis for the function to complete one full cycle.

Vertical Stretch and Shift

In trigonometric functions, a vertical stretch or compression and a vertical shift transform the graph. For the function \(y = 3 \sec\left(\frac{1}{4} x \right) + 1\), we have two transformations:

• Vertical Stretch: The coefficient 3 stretches the secant function's values by a factor of 3.
• Vertical Shift: The constant 1 shifts the entire function up by 1 unit.

These transformations affect the amplitude and position of the key points and asymptotes. Stretching increases the vertical distances between values and shifting moves all points up or down. As a result, the peaks and valleys become more pronounced, and new key points can be identified at \((0, 4), (8\pi, 4), (4\pi, -2), (12\pi, -2)\).

Identifying Asymptotes

Vertical asymptotes in the secant function occur where the cosine function is zero since the secant function is undefined at these points. For the function \(y = 3 \sec\left(\frac{1}{4} x \right) + 1\), the asymptotes occur where \(\cos\left(\frac{1}{4} x \right) = 0\).
These can be calculated by setting the argument of the cosine function \(\frac{1}{4} x\) equal to \(\frac{\pi}{2} + k\pi\), where \(k\) is any integer. This results in the asymptotes:

• \(x = 4k\pi + 2\pi\)

So, our function has vertical asymptotes at \(x = 4\pi + 2k\pi\).

Domain and Range

The domain and range of a function describe the possible x-values (domain) and y-values (range) that the function can take. For our function \(y = 3 \sec\left(\frac{1}{4} x \right) + 1\), we'll find:

• Domain: Since the secant function is undefined where the cosine function is zero, the domain excludes these points. Thus, the domain can be written as \(x eq 4k\pi + 2\pi\).
• Range: The vertical stretch and shift modifies the range of the function. The function value will never lie between \(-2\) and \(4\); thus, the range is \((-\infty, -2] \cup [4, \infty)\).

By understanding the domain and range, you can identify which x-values the function can accept and which y-values it can produce, giving you comprehensive insight into the behavior of the function.