Math Problem Statement

sketch the graph following the interval [-pi,pi]

Sketch the graph following the interval [-pi, pi].

Solution

sympy import symbols, sec, pi import numpy as np import matplotlib.pyplot as plt

Define the variable and the function

x = symbols('x') y = sec(2*x - pi/2)

Convert the symbolic function to a numerical one for plotting

x_vals = np.linspace(-np.pi, np.pi, 1000) y_vals = 1 / np.cos(2*x_vals - np.pi/2) # sec(x) = 1/cos(x)

Plotting

plt.figure(figsize=(10, 6)) plt.plot(x_vals, y_vals, label=r"$y = \sec(2x - \frac{\pi}{2})$", color='blue')

Set the y-limits to avoid extremely large values

plt.ylim(-10, 10)

Add asymptotes (where cos(2x - pi/2) = 0)

asymptotes = (np.arange(-np.pi/2, np.pi, np.pi/2) + np.pi/4) / 2 for asymptote in asymptotes: plt.axvline(asymptote, color='red', linestyle='--', alpha=0.7, label="Asymptote" if asymptote == asymptotes[0] else "")

Customize the graph

plt.title("Graph of $y = \sec(2x - \pi/2)$ on $[-\pi, \pi]$", fontsize=14) plt.xlabel("x", fontsize=12) plt.ylabel("y", fontsize=12) plt.axhline(0, color='black', linewidth=0.8, linestyle='--') plt.axvline(0, color='black', linewidth=0.8, linestyle='--') plt.xticks(np.arange(-np.pi, np.pi+np.pi/2, np.pi/2), [r"$-\pi$", r"$-\frac{\pi}{2}$", "$0$", r"$\frac{\pi}{2}$", r"$\pi$"]) plt.grid(alpha=0.4) plt.legend() plt.show()Here is the graph of y=sec(2xπ2)y = \sec(2x - \frac{\pi}{2}) over the interval [π,π][-\pi, \pi]. The red dashed lines represent the asymptotes, which occur where the denominator of sec(x)\sec(x), cos(2xπ2)\cos(2x - \frac{\pi}{2}), is equal to zero.

Key points:

  1. The function has vertical asymptotes where cos(2xπ2)=0\cos(2x - \frac{\pi}{2}) = 0, approximately at x=π4,π4,3π4x = -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4}, and 3π4-\frac{3\pi}{4}.
  2. The function oscillates and repeats in a periodic manner, with alternating positive and negative branches due to the nature of the secant function.
  3. The graph is restricted to a y-range of [10,10][-10, 10] for clarity, as sec(x)\sec(x) tends toward infinity near the asymptotes.

Would you like a step-by-step explanation on how the graph was derived?

Related Questions:

  1. What is the period of the function y=sec(2xπ2)y = \sec(2x - \frac{\pi}{2})?
  2. How do you calculate the vertical asymptotes of the secant function?
  3. What are the key differences between the graphs of sec(x)\sec(x) and sec(2xπ2)\sec(2x - \frac{\pi}{2})?
  4. How does the phase shift π2-\frac{\pi}{2} affect the graph of the secant function?
  5. Can the secant function ever have a minimum or maximum value?

Tip: To find the vertical asymptotes of secant, solve for cos(kxϕ)=0\cos(kx - \phi) = 0, as sec(x)\sec(x) is undefined at these points.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Secant Function
Graphing Trigonometric Functions

Formulas

y = sec(kx - phi)
sec(x) = 1/cos(x)

Theorems

Periodic Properties of Trigonometric Functions

Suitable Grade Level

Grades 10-12