Finding Asymptotes Of Tangent Function Y=tan(3/4x)

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Introduction

When exploring trigonometric functions, understanding the concept of asymptotes is crucial, especially for functions like the tangent. In this article, we will delve into identifying the asymptotes of the tangent function $y = \tan(\frac{3}{4}x)$. We will explore the properties of tangent functions, how transformations affect their asymptotes, and provide a step-by-step method to find these asymptotes. By the end of this guide, you'll have a solid understanding of how to determine the asymptotes for tangent functions, enabling you to solve related problems effectively.

Understanding Tangent Functions and Asymptotes

The tangent function, denoted as $y = \tan(x)$, is a fundamental trigonometric function. It is defined as the ratio of the sine function to the cosine function, $ \tan(x) = \frac{\sin(x)}{\cos(x)}$. A key characteristic of the tangent function is its periodicity and the presence of vertical asymptotes. Vertical asymptotes occur where the function approaches infinity (or negative infinity) because the denominator (in this case, the cosine function) approaches zero.

For the basic tangent function, $y = \tan(x)$, asymptotes occur at $x = \frac{(2n+1)\pi}{2}$, where n is an integer. This is because the cosine function, which is in the denominator, equals zero at these points. Therefore, the tangent function becomes undefined, leading to vertical asymptotes. Understanding this foundational concept is crucial for analyzing more complex tangent functions.

Now, let’s consider the given function: $y = \tan(\frac{3}{4}x)$. The presence of the coefficient $\frac{3}{4}$ inside the tangent function alters its period and, consequently, the position of its asymptotes. To accurately determine the asymptotes, we need to account for this transformation. The general form of a transformed tangent function is $y = \tan(Bx)$, where B affects the period of the function. The period of $y = \tan(Bx)$ is given by $\frac{\pi}{|B|}$. In our case, $B = \frac{3}{4}$, so the period will be different from the standard tangent function. This change in period directly impacts the location of the asymptotes, which we will calculate in the following sections.

The Impact of Transformations on Tangent Functions

Transformations play a significant role in shaping trigonometric functions, and the tangent function is no exception. Understanding how these transformations affect the graph and asymptotes of a tangent function is crucial. Let’s delve deeper into the specific transformations that influence the function $y = \tan(\frac{3}{4}x)$.

First, consider the general form of a transformed tangent function: $y = A \tan(Bx - C) + D$. Here:

  • A represents the vertical stretch or compression.
  • B affects the period of the function.
  • C introduces a horizontal shift (phase shift).
  • D causes a vertical shift.

In our given function, $y = \tan(\frac{3}{4}x)$, we can identify that A = 1, B = $\frac{3}{4}$, C = 0, and D = 0. The most significant transformation affecting the asymptotes in this case is the value of B, which is $\frac{3}{4}$. B directly influences the period of the tangent function, which in turn affects the spacing and location of the asymptotes.

The period of the transformed tangent function is given by $\frac{\pi}{|B|}$. For $y = \tan(\frac{3}{4}x)$, the period is $\frac{\pi}{\frac{3}{4}} = \frac{4\pi}{3}$. This means the function repeats its pattern every $\frac{4\pi}{3}$ units. The asymptotes, which are normally $\pi$ units apart in the standard tangent function, will now be $\frac{4\pi}{3}$ units apart. This stretching of the period horizontally changes the asymptotes' positions significantly.

The asymptotes of the standard tangent function, $y = \tan(x)$, occur at $x = \frac{(2n+1)\pi}{2}$, where n is an integer. To find the asymptotes of the transformed function, we need to account for the change in the period. The asymptotes of $y = \tan(\frac{3}{4}x)$ will occur when the argument of the tangent function, $\frac{3}{4}x$, is equal to $\frac{(2n+1)\pi}{2}$. This understanding forms the basis for our next step: finding the exact values of the asymptotes for the given function.

Determining the Asymptotes of $y = \tan(\frac{3}{4}x)$

To find the asymptotes of the function $y = \tan(\frac{3}{4}x)$, we need to determine the values of x for which the function is undefined. As discussed earlier, the tangent function is undefined when its argument is an odd multiple of $\frac{\pi}{2}$. Therefore, we set the argument of our tangent function equal to this condition:

\frac{3}{4}x = \frac{(2n+1)\pi}{2}$, where *n* is an integer. Now, we solve for *x* to find the locations of the asymptotes: $x = \frac{4}{3} \cdot \frac{(2n+1)\pi}{2}

x=2(2n+1)Ο€3x = \frac{2(2n+1)\pi}{3}

This general formula gives us the x-values of all the vertical asymptotes for the function. By substituting different integer values for n, we can find specific asymptotes. For example:

  • When n = 0: $x = \frac{2(2(0)+1)\pi}{3} = \frac{2\pi}{3}$
  • When n = 1: $x = \frac{2(2(1)+1)\pi}{3} = \frac{6\pi}{3} = 2\pi$
  • When n = -1: $x = \frac{2(2(-1)+1)\pi}{3} = \frac{-2\pi}{3}$

Thus, we can see that the function has asymptotes at $x = \frac{2\pi}{3}$, $x = 2\pi$, $x = -\frac{2\pi}{3}$, and so on. These asymptotes occur periodically, as determined by the period of the function. Now, let’s evaluate the given options to see which one matches our calculated asymptotes.

Evaluating the Given Options

We have determined the general form of the asymptotes for the function $y = \tan(\frac{3}{4}x)$ as $x = \frac{2(2n+1)\pi}{3}$, where n is an integer. Now, we will evaluate the given options to find the correct asymptote.

A. $x = -\frac{4\pi}{3}$

To check if this is an asymptote, we set $ rac{2(2n+1)\pi}{3} = -\frac{4\pi}{3}$ and solve for n:

2(2n+1)Ο€=βˆ’4Ο€2(2n+1)\pi = -4\pi

2(2n+1)=βˆ’42(2n+1) = -4

2n+1=βˆ’22n+1 = -2

2n=βˆ’32n = -3

n=βˆ’32n = -\frac{3}{2}

Since n is not an integer, $x = -\frac{4\pi}{3}$ is not an asymptote.

B. $x = -\frac{2\pi}{3}$

Similarly, we set $\frac{2(2n+1)\pi}{3} = -\frac{2\pi}{3}$ and solve for n:

2(2n+1)Ο€=βˆ’2Ο€2(2n+1)\pi = -2\pi

2(2n+1)=βˆ’22(2n+1) = -2

2n+1=βˆ’12n+1 = -1

2n=βˆ’22n = -2

n=βˆ’1n = -1

Since n = -1 is an integer, $x = -\frac{2\pi}{3}$ is an asymptote.

C. $x = \frac{3\pi}{4}$

We set $\frac{2(2n+1)\pi}{3} = \frac{3\pi}{4}$ and solve for n:

2(2n+1)Ο€=9Ο€42(2n+1)\pi = \frac{9\pi}{4}

2(2n+1)=942(2n+1) = \frac{9}{4}

2n+1=982n+1 = \frac{9}{8}

2n=182n = \frac{1}{8}

n=116n = \frac{1}{16}

Since n is not an integer, $x = \frac{3\pi}{4}$ is not an asymptote.

D. $x = \frac{3\pi}{2}$

We set $\frac{2(2n+1)\pi}{3} = \frac{3\pi}{2}$ and solve for n:

2(2n+1)Ο€=9Ο€22(2n+1)\pi = \frac{9\pi}{2}

2(2n+1)=922(2n+1) = \frac{9}{2}

2n+1=942n+1 = \frac{9}{4}

2n=542n = \frac{5}{4}

n=58n = \frac{5}{8}

Since n is not an integer, $x = \frac{3\pi}{2}$ is not an asymptote.

Conclusion and Final Answer

After evaluating each option, we found that the only value that satisfies the condition for an asymptote is $x = -\frac{2\pi}{3}$, which corresponds to n = -1. Therefore, the correct answer is option B.

In summary, determining the asymptotes of a transformed tangent function involves understanding the impact of the transformations on the period and applying the general formula for asymptotes. By setting the argument of the tangent function equal to odd multiples of $\frac{\pi}{2}$ and solving for x, we can find the exact locations of the asymptotes. This process is crucial for accurately graphing and analyzing tangent functions.

Final Answer: B. $x = -\frac{2\pi}{3}$