Problem 50 - Algebra and Trigonometry

Question

Find the period and graph the function. $$ y=\tan \frac{1}{2}\left(x+\frac{\pi}{4}\right) $$

Step-by-Step Solution

Answer

The period is \( 2\pi \), and the graph is a left-shifted tangent with vertical asymptotes every \( 2\pi \).

Step 1: Formula for Period of Tangent Function

The period of the parent tangent function, \( y = \tan(x) \), is \( \pi \). To find the period of any transformed tangent function of the form \( y = \tan(bx) \), the formula for the period is \( \frac{\pi}{|b|} \). In this function, \( b = \frac{1}{2} \).

Step 2: Substitute and Calculate New Period

Substitute \( b = \frac{1}{2} \) into the period formula: \[ \text{Period} = \frac{\pi}{\left|\frac{1}{2}\right|} = \frac{\pi}{0.5} = 2\pi. \] This tells us that the period of the function \( y = \tan \left( \frac{1}{2}\left(x + \frac{\pi}{4}\right) \right) \) is \( 2\pi \).

Step 3: Graph the Function with Correct Period

To graph this function, start by considering the horizontal shift, which is determined by the \( x + \frac{\pi}{4} \) term. This indicates a left shift of \( \frac{\pi}{4} \). The period is \( 2\pi \), so the function repeats every \( 2\pi \) units. Start the graph at \(-\frac{\pi}{4}\), the asymptotes will be at \(-\frac{\pi}{4} + k\cdot2\pi\) for integer \(k\), and the curve will pass through the origin at input \(x = -\frac{\pi}{4} + \pi\) as tangent typically passes through its vertical middle of a period.

Key Concepts

tangent function transformation

Tangent function transformations involve changes to the standard tangent function, altering its period, phase shift, and vertical or horizontal stretching or compressing. The general form of a transformed tangent function is \( y = \tan(bx + c) \). Here, \( b \) and \( c \) are constants, and they significantly affect the shape and position of the graph.
A key transformation is altering the period of the tangent function. The original period of \( y = \tan(x) \) is \( \pi \). When transformed, the period changes based on the coefficient \( b \). Specifically, the new period becomes \( \frac{\pi}{|b|} \).
For example, consider \( y = \tan\left(\frac{1}{2}(x + \frac{\pi}{4})\right) \). Here, \( b = \frac{1}{2} \), thus changing the period to \( 2\pi \), calculated as \( \frac{\pi}{\left|\frac{1}{2}\right|} = 2\pi \). These simpler periods allow for efficient graphing and analysis of trigonometric functions.

• Identify coefficients \( b \) and \( c \) to determine transformations.
• Calculate the altered period using \( \frac{\pi}{|b|} \).
• Understand that transformations can include shifts and scaling of the graph.

This understanding plays a crucial role in understanding function behavior.

graphing trigonometric functions

Graphing trigonometric functions like the tangent requires a good understanding of their properties. The standard tangent graph, \( y = \tan(x) \), exhibits periodicity, vertical asymptotes, and a characteristic wave-like pattern repeating every \( \pi \) units.
When graphing the transformed tangent function \( y = \tan\left(\frac{1}{2}(x + \frac{\pi}{4})\right) \), we incorporate these transformations:

• The new period is \( 2\pi \), so the function repeats every \( 2\pi \) units.
• Vertical asymptotes occur where the tangent function is undefined. For this function, they appear at intervals starting from the horizontal shift, which affects their positions.

The period \( 2\pi \) stretches the distance between asymptotes and thus alters the typical behavior of the tangent function. To graph it:

• Start at the phase shift determined by \( x + \frac{\pi}{4} \), which shifts the graph left by \( \frac{\pi}{4} \).
• Plot key points and asymptotes according to the altered period and phase shift.

Graphing transformed trigonometric functions demands attention to how each parameter modifies the base graph to maintain accurate representations of these functions.

horizontal shift of graphs

The horizontal shift of graphs involves moving the entire graph left or right across the Cartesian plane. This is dictated by the value \( c \) in a function \( y = \tan(bx + c) \). To find the effect of \( c \), we evaluate \( -\frac{c}{b} \), which provides the horizontal displacement.
In the given formula, \( y = \tan\left(\frac{1}{2}(x + \frac{\pi}{4})\right) \), the \( \frac{\pi}{4} \) signifies a leftward shift:

• Calculate \( -\left(\frac{\pi}{4}\right) \times \frac{2}{1} = -\frac{\pi}{2} \), providing a left shift of \( \frac{\pi}{2} \).

This shift affects where key features like asymptotes and zeros occur. Each \( \pi \) unit of the original function adjusts to start at a new location. The new starting point of the graph is important as it shifts the entire cycle of the tangent curve alongside it. Analyzing these shifts is vital:

• Understanding the point from where repetitive parts cycle is key in trigonometric graphing.
• Horizontal shifts impact how we interpret the boundaries and phases of the function.

Horizontal shifts ensure that transformations narrate the correct story of periodic functions, yielding precision in their graphical layout and behavior.