Problem 12
Question
Find the period and sketch the graph of the equation. Show the asymptotes. $$y=\tan \frac{1}{2} x$$
Step-by-Step Solution
Verified Answer
Period is \\(2\pi\\); vertical asymptotes are at \\(x = \pi, 3\pi, 5\pi, \ldots\\).
1Step 1: Understanding the Tangent Function
The basic tangent function, \(y = \tan x\), has a period of \(\pi\) because it completes one full cycle over an interval of length \pi\. We want to understand how transformations affect this period.
2Step 2: Identify the Transformation
The function given is \(y = \tan \frac{1}{2} x\). Here, the input to the tangent function, \(x\), has been replaced by \(\frac{1}{2} x\). This transformation inside the function affects the period.
3Step 3: Calculate the New Period
For a function \(y = \tan(bx)\), the period is given by \(\frac{\pi}{|b|}\). In this problem, \(b = \frac{1}{2}\). Thus, the new period is \(\frac{\pi}{\frac{1}{2}} = 2\pi\). This means the cycle repeats every \(2\pi\).
4Step 4: Determine the Asymptotes
Vertical asymptotes for the tangent function \(y = \tan(bx)\) occur at the points where \(bx = \frac{\pi}{2} + k\pi\), where \(k\) is any integer. Solving \(\frac{1}{2} x = \frac{\pi}{2} + k\pi\) gives \(x = \pi + 2k\pi\). So, the vertical asymptotes are at \(x = \pi, 3\pi, 5\pi, \ldots\).
5Step 5: Sketch the Graph
To sketch the graph of \(y = \tan \frac{1}{2} x\), draw the standard tangent curve between two asymptotes, spaced \(2\pi\) apart. The curve will pass through the origin (0,0) and will have vertical asymptotes at \(x = \pi, 3\pi, 5\pi, \ldots\).
Key Concepts
Period of Trigonometric FunctionsVertical AsymptotesTransformation of Functions
Period of Trigonometric Functions
The period of a trigonometric function is the interval over which it completes one full cycle. For the basic tangent function, \( y = \tan x \), the period is \( \pi \), meaning it repeats every \( \pi \) units. However, transformations can affect this period.
When a function like \( y = \tan(bx) \) is given, the period is calculated by the formula \( \frac{\pi}{|b|} \). In the equation \( y = \tan \frac{1}{2} x \), the value of \( b \) is \( \frac{1}{2} \). Thus, the new period is \[ \frac{\pi}{\frac{1}{2}} = 2\pi \].
This shows that the function completes a cycle over \( 2\pi \) instead of \( \pi \), modifying how often the function repeats. In your graph, this means that every \( 2\pi \), the pattern observed in the tangent function repeats itself.
When a function like \( y = \tan(bx) \) is given, the period is calculated by the formula \( \frac{\pi}{|b|} \). In the equation \( y = \tan \frac{1}{2} x \), the value of \( b \) is \( \frac{1}{2} \). Thus, the new period is \[ \frac{\pi}{\frac{1}{2}} = 2\pi \].
This shows that the function completes a cycle over \( 2\pi \) instead of \( \pi \), modifying how often the function repeats. In your graph, this means that every \( 2\pi \), the pattern observed in the tangent function repeats itself.
Vertical Asymptotes
Vertical asymptotes in trigonometric functions like tangent represent the values where the function becomes undefined and shoots to infinity or negative infinity. For the standard tangent function, \( y = \tan x \), these asymptotes occur at odd multiples of \( \frac{\pi}{2} \), i.e., \( x = \frac{\pi}{2} + k\pi \), where \( k \) is an integer.
In the case of \( y = \tan \frac{1}{2} x \), you solve \( \frac{1}{2} x = \frac{\pi}{2} + k\pi \) to find where these asymptotes lie. Solving gives \( x = \pi + 2k\pi \), indicating that vertical asymptotes appear at \( x = \pi \), \( 3\pi \), \( 5\pi \), and so on. This series is consistent because the period has been adjusted to \( 2\pi \), spacing these asymptotes accordingly.
These lines are crucial when sketching the graph as they guide the placement of the tangent curves, showing where the function diverges toward infinity.
In the case of \( y = \tan \frac{1}{2} x \), you solve \( \frac{1}{2} x = \frac{\pi}{2} + k\pi \) to find where these asymptotes lie. Solving gives \( x = \pi + 2k\pi \), indicating that vertical asymptotes appear at \( x = \pi \), \( 3\pi \), \( 5\pi \), and so on. This series is consistent because the period has been adjusted to \( 2\pi \), spacing these asymptotes accordingly.
These lines are crucial when sketching the graph as they guide the placement of the tangent curves, showing where the function diverges toward infinity.
Transformation of Functions
Transformations alter the typical behaviors of trigonometric functions, such as stretching, compressing, and shifting them along the axes. In the equation \( y = \tan \frac{1}{2} x \), we have an example of horizontal stretching as a result of the change in the input.
This particular transformation, \( \frac{1}{2} x \), stretches the graph horizontally, effectively lengthening the period of the function. A simple way to think about it is: when \( x \) is divided by a fractional number, the graph stretches because it takes longer for the function to complete one cycle.
Transformations like these are vital in understanding the geometry of graphs and can help you predict how the inputs alter their appearance. By mastering transformations, you gain the ability to graph trigonometric functions quickly and accurately, seeing how each change affects the function’s cycle and asymptotes.
This particular transformation, \( \frac{1}{2} x \), stretches the graph horizontally, effectively lengthening the period of the function. A simple way to think about it is: when \( x \) is divided by a fractional number, the graph stretches because it takes longer for the function to complete one cycle.
Transformations like these are vital in understanding the geometry of graphs and can help you predict how the inputs alter their appearance. By mastering transformations, you gain the ability to graph trigonometric functions quickly and accurately, seeing how each change affects the function’s cycle and asymptotes.
Other exercises in this chapter
Problem 12
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