Math Problem Statement

Find the period and graph the function. a) y = tan 4x b) y = tan 2(

x + π/ 2) c) y = 3 sec 2 ( x − π/ 2)

Solution

Let's break down each trigonometric function to find the period and then explain how to graph them.

Part a) y=tan(4x)y = \tan(4x)

Period:

The general form of the tangent function is y=tan(kx)y = \tan(kx), where the period TT is given by:

T=πkT = \frac{\pi}{|k|}

For y=tan(4x)y = \tan(4x), k=4k = 4. Therefore, the period is:

T=π4T = \frac{\pi}{4}

Graph:

The graph of y=tan(4x)y = \tan(4x) is a vertical stretch of the standard tangent graph, but the period is compressed to π4\frac{\pi}{4}, meaning the function repeats every π4\frac{\pi}{4}. The vertical asymptotes occur at x=π8+π4nx = \frac{\pi}{8} + \frac{\pi}{4}n, where nZn \in \mathbb{Z}.


Part b) y=tan(2(x+π2))y = \tan\left(2\left(x + \frac{\pi}{2}\right)\right)

Period:

Again, the general form is y=tan(kx)y = \tan(kx), where k=2k = 2. Thus, the period is:

T=π2=π2T = \frac{\pi}{|2|} = \frac{\pi}{2}

The x+π2x + \frac{\pi}{2} inside the function represents a horizontal shift to the left by π2\frac{\pi}{2}, but this does not affect the period itself.

Graph:

The graph of y=tan(2(x+π2))y = \tan\left(2\left(x + \frac{\pi}{2}\right)\right) is shifted to the left by π2\frac{\pi}{2} units compared to y=tan(2x)y = \tan(2x). The vertical asymptotes will now occur at x=π4+π2nx = -\frac{\pi}{4} + \frac{\pi}{2}n.


Part c) y=3sec(2(xπ2))y = 3\sec\left(2\left(x - \frac{\pi}{2}\right)\right)

Period:

The secant function is the reciprocal of the cosine function, and the general form of the cosine function is y=sec(kx)y = \sec(kx), where the period TT is:

T=2πkT = \frac{2\pi}{|k|}

Here, k=2k = 2, so the period is:

T=2π2=πT = \frac{2\pi}{2} = \pi

The xπ2x - \frac{\pi}{2} represents a horizontal shift to the right by π2\frac{\pi}{2}, but this shift does not affect the period.

Graph:

The graph of y=3sec(2(xπ2))y = 3\sec\left(2\left(x - \frac{\pi}{2}\right)\right) is stretched vertically by a factor of 3 and shifted to the right by π2\frac{\pi}{2}. The vertical asymptotes for the secant function will occur at the same places as the zeros of the cosine function x=π4+π2nx = \frac{\pi}{4} + \frac{\pi}{2}n.


Would you like me to generate detailed graphs for these functions or further explanations?

Five Related Questions:

  1. How does the amplitude of y=3sec(2x)y = 3 \sec(2x) compare to y=sec(2x)y = \sec(2x)?
  2. What is the period of y=tan(5xπ3)y = \tan\left(5x - \frac{\pi}{3}\right)?
  3. How do horizontal shifts affect the vertical asymptotes of tangent and secant graphs?
  4. What is the difference between the period of a sine function and a tangent function?
  5. How do vertical stretches and compressions impact the shape of trigonometric graphs?

Tip:

When graphing trigonometric functions, always check for shifts and period changes first, as they significantly affect the shape and position of the graph.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometry
Tangent Function
Secant Function
Periodicity
Graphing Trigonometric Functions

Formulas

Period of tangent: T = π/|k|
Period of secant: T = 2π/|k|

Theorems

Periodicity of Trigonometric Functions
Properties of Tangent and Secant Functions

Suitable Grade Level

Grades 10-12