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. $$ y=\sec \left(\frac{2 \pi}{3} x\right)+2 $$

Short answer

The domain of \(y = \sec \left(\frac{2 \pi}{3} x\right) + 2\) is \(x eq \frac{3}{4} + \frac{3k}{2}\), and its range is \((-\infty, 1] \cup [3, \infty)\).

Step-by-step solution

Understanding the Function

The function to graph is \(y = \sec \left(\frac{2 \pi}{3} x\right) + 2\). Recall that \(\sec x = \frac{1}{\cos x}\). Thus, we need to consider the characteristics of the secant function.

Determine Key Points

Identify points where the cosine function \(\cos \left(\frac{2 \pi}{3} x\right) = 0\), as these will be vertical asymptotes for the secant function since \(\sec x\) becomes undefined. These occur where \(\frac{2 \pi}{3} x = \frac{\pi}{2} + k\pi\) (for integers \(k\)), leading to \(x = \frac{3}{4} + \frac{3k}{2}\).

Plotting the Secant Function

By plotting these asymptotes and finding values between them, draw the secant function. Remember the secant function will be positive where the cosine is positive and negative where the cosine is negative, but shifted up by 2 units due to the \(+2\) in the function.

Cycles and Symmetry

Since the period of \(\sec(\frac{2\pi}{3}x)\) is determined by the coefficient of \(x\), each cycle will span \( \frac{2\pi}{\frac{2\pi}{3}} = 3 \). So, two cycles will span 6 units along the x-axis.

Domain

The domain of \(\sec(\frac{2\pi}{3}x) + 2\) is all real numbers except where the function has vertical asymptotes. This is \(x eq \frac{3}{4} + \frac{3k}{2}\).

Range

The range of the secant function \(\sec \left(\frac{2 \pi}{3} x\right)\) is \((-\infty, -1] \cup [1, \infty)\), and since the entire function is shifted up by 2 units, the range of \(y = \sec \left(\frac{2 \pi}{3} x\right) + 2\) becomes \((-\infty, 1] \cup [3, \infty)\).

Key concepts

secant function

The secant function, denoted as \(\text{sec}(x)\), is the reciprocal of the cosine function. This means \(\text{sec}(x) = \frac{1}{\text{cos}(x)}\). It inherits properties from the cosine function but behaves differently, especially where cosine is zero because division by zero is undefined. Thus, the secant function will have vertical asymptotes at points where cosine is zero. It's crucial to remember that the secant function will have the same period as the cosine function, and it will tend to infinity or negative infinity near these asymptotes. This creates a repetitive pattern of curves going towards positive and negative infinity alternatively, known as branches.

domain and range

The domain of a function refers to all possible input values (x-values) for which the function is defined. For the secant function \(y = \text{sec} \bigg(\frac{2\pi}{3}x\bigg) + 2\), the domain is all real numbers except at the points where the cosine of \( \frac{2 \pi}{3} x \) is zero, because we can't divide by zero. These points occur when \( \frac{2 \pi}{3} x = \frac{\pi}{2} + k\pi\) where k is an integer. Solving for x, we get \( x = \frac{3}{4} + \frac{3k}{2} \). Thus, the domain is all real numbers except \[ x eq \frac{3}{4} + \frac{3k}{2} \].
The range of a function refers to all possible output values (y-values) of the function. The base secant function \( \text{sec} \bigg(\frac{2 \pi}{3} x \bigg)\) has a range of \[(-\infty, -1] \cup [1, \infty)\]. In the case of the given function, \( y = \text{sec} \bigg(\frac{2 \pi}{3} x \bigg) + 2 \), shifting it up by 2 units changes the range to \[ (-\infty, 1] \cup [3, \infty) \]. This means the function will take all values below 1 and all values greater than or equal to 3.

vertical asymptotes

Vertical asymptotes are the lines that the graph of a function approaches but never touches or crosses. For the secant function, these occur at values of x where the cosine value is zero.
In our given function \( y = \text{sec} \bigg(\frac{2 \pi}{3} x \bigg) + 2\), the cosine component \( \frac{2 \pi}{3} x \) is zero at \[ \frac{2 \pi}{3} x = \frac{\pi}{2} + k\pi \]. Solving for x in terms of integers k, we get \[ x = \frac{3}{4} + \frac{3k}{2} \]. Therefore, the vertical asymptotes of this function are at these x-values. These are the lines the secant function will approach but never touch.
This information aids in properly plotting the graph, as it shows where significant behavior change (approaching infinity or negative infinity) occurs.

period of trigonometric functions

The period of a function is the length of one complete cycle of the function. For standard trigonometric functions like sine and cosine, one complete cycle is \( 2 \pi \) radians. In our function, \( y = \text{sec} \bigg( \frac{2 \pi}{3} x \bigg) + 2 \), the coefficient \( \frac{2 \pi}{3} \) in front of x affects the period.
The period is calculated by taking the base period of \( 2 \pi \) and dividing it by the coefficient of x inside the function. So, \[ \text{Period} = \frac{2\pi}{\frac{2 \pi}{3}} = 3 \]. This means every 3 units along the x-axis, the secant function completes one full cycle. Plotting at least two cycles, as the exercise requires, implies drawing the graph from 0 to 6 units on the x-axis.
Understanding the period helps in accurately graphing the function and understanding its repetitive nature.