Problem 33 - Precalculus : Building Concepts and Connections

Question

Graph at least two cycles of the given functions. $$r(x)=-\cos (2 \pi x)+2$$

Step-by-Step Solution

Answer

The function \(r(x) = -\sec\left(\frac{2 \pi}{3}x\right) + 2\) has no phase shift, an amplitude of 1, a period of 3 and a vertical shift of 2 units upward. For graphing, divide the period into four equal parts, draw asymptotes where the cosine function equals zero, reflect the secant curve around the x-axis (because of the negative sign), and raise the entire curve two units up. Repeat the cycle to graph multiple periods.

Step 1: Identify the period and phase shift

The general form of a secant function is \(r(x) = a \sec(bx - c) + d\). In this exercise, the function is \(r(x)=-\sec\left(\frac{2\pi}{3}x\right)+2\). First, identify that the function has no phase shift because there is no 'c' present in our function. The period of a regular secant function is \(2\pi\). The frequency or the 'b' in our equation is \(\frac{2\pi}{3}\). To find the period, take the regular period and divide it by the frequency. Therefore, the period here is \(2\pi\div\frac{2\pi}{3}=3\)

Step 2: Identify the amplitude

The absolute value of 'a' gives us the amplitude. In this case, 'a' is given as -1. Even though 'a' is negative, the amplitude, which represents the half distance between the maximum and minimum values of the function, is always positive. Therefore the amplitude is 1. This negative sign means that our secant function is reflected over the x-axis.

Step 3: Identify the vertical shift

The 'd' in our equation represents the vertical shift of the periodic function. In our function, d = 2, so there is a vertical shift of 2 units upward.

Step 4: Sketch the parent function and transformations

To start drawing the graph of the function, divide one period into four equal parts. These are where maximum, minimum and asymptotes occur. Mark these points on the x-axis. Next, draw the graph by sketching the cosine curve, then drawing vertical lines wherever the cosine function equals zero (these are asymptotes). Also, make the necessary reflections and shifts summary of transformations is Vertical Reflection (because of negative sign), Vertical Shift/Translation of 2 units upwards (because of +2), no horizontal shift and a period of 3 units.

Step 5: Draw multiple periods

To graph two cycles of the function, we just duplicate what was graphed previously to the right of the initial period.

Key Concepts

Period of Trigonometric Functions

The period of a trigonometric function indicates how long it takes for the function to complete one full cycle. For the basic secant function, \( \sec(x) \), the period is \( 2\pi \), meaning that it repeats its values every \( 2\pi \) units along the x-axis.

To determine the period of a transformed secant function, such as \( r(x) = -\sec\left(\frac{2\pi}{3}x\right)+2 \), we examine the coefficient \( b \) in the argument of the secant function. The period \( P \) of the transformed function is given by \( P = \frac{2\pi}{|b|} \). Hence, with \( b = \frac{2\pi}{3} \), the period is \( P = 3 \).

This interval \( P \) is the length on the x-axis over which the secant function graph completes its peaks and troughs once before starting the same pattern again. Recognizing the period of the function allows us to predict its behavior and also determines the intervals we use when graphing the function.

Vertical Shift in Trigonometric Graphs

A vertical shift in trigonometric graphs describes the movement of the graph up or down along the y-axis. This transformation doesn't affect the shape of the graph but changes its position.

In the function \( r(x) = -\sec\left(\frac{2\pi}{3}x\right)+2 \), the '+2' at the end signifies a vertical shift. Here, the entire graph of the secant function is raised 2 units above its original position. Graphically, every point \( (x, y) \) on the unshifted graph is moved to \( (x, y+2) \) on the shifted graph.

Identifying Vertical Shift

• Locating the vertical shift is straightforward. It's given by the constant \( d \) added or subtracted at the end of the trigonometric function.
• In \( r(x) = -\sec\left(\frac{2\pi}{3}x\right)+2 \), the vertical shift is '+2', which means every point on the graph moves two units upwards.
• If \( d \) were negative, the graph would shift downwards by that amount.

By considering the vertical shift, we can accurately place the graph on the coordinate system, ensuring it mirrors the function's properties correctly.

Transformations of Trigonometric Graphs

When we talk about transformations of trigonometric graphs, we're referring to changes in the graph's position, shape, or orientation due to modifications in the function's equation. This includes reflections, stretches or compressions, vertical and horizontal shifts, and period changes.

In our example, \( r(x) = -\sec\left(\frac{2\pi}{3}x\right)+2 \), several transformations are at work:

• Reflection: The negative sign in front of the secant indicates a reflection across the x-axis. This means that peaks become troughs and vice versa.
• Vertical Shift: The '+2' at the end of the function indicates a shift upward by 2 units.
• Period Change: The coefficient \( \frac{2\pi}{3} \) in the function modifies the period from its standard \( 2\pi \) to 3 units.

These transformations collectively alter the original \( \sec(x) \) graph into its new form. To successfully sketch the graph of \( r(x) \) by hand, we plot the secant function without transformations, and step by step, add each transformation. This systematic approach ensures a clear understanding of how the final graph is developed from the parent function and helps students visualize and grasp the effects of each transformation.