Problem 25 Graph one full period of each fu... [FREE SOLUTION]

Chapter 4: Problem 25

Graph one full period of each function. $$y=\sec \left(x+\frac{\pi}{4}\right)$$

Short Answer

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The graph of the function $ y=sec(x + \frac{蟺}{4}) $ for one full period would look like a series of U-shaped curves opening upwards and downwards alternately, starting from $ x = -\frac{蟺}{4} $ to $ x= \frac{7蟺}{4} $ with vertical asymptotes at $ x =\frac{蟺}{4} $, $ x = \frac{5蟺}{4} $ and $ x = \frac{13蟺}{4} $.

Step by step solution

Identify the properties of the secant function

First recognize that the secant function, $ sec(x) $, is equivalent to $ 1/cos(x) $. The secant function has a period of $ 2蟺 $, same as the cosine. However, the secant has vertical asymptotes and hence undefined for $ x = \frac{蟺}{2} + 蟺n $, where $ n $ is an integer, unlike the cosine.

Note the phase shift and draw the cosine function

The given function $ sec(x + \frac{蟺}{4}) $ indicates a phase shift of $ -\frac{蟺}{4} $. So, to graph one full period of this function, let's consider the interval $ -\frac{蟺}{4} $ to $ \frac{7蟺}{4} $. However, it's useful to draw the cosine function over this interval first and identify where the cosine function is undefined.

Graph the secant function

The secant function is not defined at those points the cosine is zero. This means, drawing vertical asymptotes at $ x =\frac{蟺}{2}-\frac{蟺}{4} = \frac{蟺}{4} $, $ x = \frac{3蟺}{2}-\frac{蟺}{4} = \frac{5蟺}{4} $ and $ x = \frac{7蟺}{2}-\frac{蟺}{4} = \frac{13蟺}{4} $. Now, draw the secant curve as the reciprocals of the cosine values. The secant function will be U-shaped curves opening upwards where the cosine function is positive and opening downwards where the cosine function is negative.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Secant Function

The secant function, represented as $ \sec(x) $, is the reciprocal of the cosine function. This means that wherever the cosine function has a value, the secant function takes the reciprocal of that value. Because it is derived from the cosine function, secant shares the same period of $ 2\pi $. This makes it oscillate once it covers this interval, starting to repeat the cycle afterwards.

The secant function can be described using several properties:

Amplitude: Unlike sine and cosine, the secant function does not have a defined amplitude as it can stretch infinitely.
Period: The period is $ 2\pi $, similar to sine and cosine.
Vertical Asymptotes: These occur at points where the cosine function equals zero, because division by zero is undefined. For $ \sec(x) $, these points are $ x = \frac{\pi}{2} + \pi n $, where $ n $ is an integer.
Symmetry: The secant function is even, meaning it is symmetric about the y-axis.

To visualize secant, first sketch the cosine function. Notice where it is undefined and where it reaches its peaks and troughs to guide the secant's curves.

Phase Shift

A phase shift refers to a horizontal shift of a trigonometric function along the x-axis. In the context of trigonometric graphs, this can affect where a function starts its cycle. For the function $ y=\sec(x + \frac{\pi}{4}) $, there is a phase shift to the left by $ \frac{\pi}{4} $. This can be understood as starting the function earlier by this amount compared to the regular, unshifted function $ \sec(x) $.

To understand and draw the graph correctly:

**Identify the shift:** The term $ x + \frac{\pi}{4} $ means the graph shifts to the left by $ \frac{\pi}{4} $.
**Impact on period:** The period remains $ 2\pi $ because the phase shift does not affect the length of a full cycle. So the function completes one cycle in this interval even with the shift.
**Graphing:** Begin plotting points taking into account the phase shift, essentially translating each x-coordinate of the cosine function $ \cos(x) $ by $ -\frac{\pi}{4} $.

This phase shift is crucial for positioning the asymptotes and peaks of the secant curve accurately on the graph.

Vertical Asymptotes

Vertical asymptotes in trigonometric functions, such as secant, are lines where the function heads towards infinity and is undefined. In $ \sec(x) $, these occur where the corresponding $ \cos(x) $ is zero since division by zero leads to undefined values.

For $ \sec(x + \frac{\pi}{4}) $, we apply the phase shift to identify new asymptote locations:

Original Asymptotes: These are at $ x = \frac{\pi}{2} + \pi n $, where $ n $ is an integer.
Shifted Asymptotes: Due to the phase shift $ -\frac{\pi}{4} $, we find new asymptotes at locations like $ x = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{13\pi}{4} $.

Recognizing and correctly plotting these asymptotes is key to sketching the secant curve as they dictate where the function explodes to infinity. These also help differentiate where the U-shaped sections of the secant function open outward.

91影视

Graph one full period of each function. $$y=\sec \left(x+\frac{\pi}{4}\right)$$

Short Answer

Step by step solution

Identify the properties of the secant function

Note the phase shift and draw the cosine function

Graph the secant function

Key Concepts

Secant Function

Phase Shift

Vertical Asymptotes

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