Math Problem Statement
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 over the interval . The red dashed lines represent the asymptotes, which occur where the denominator of , , is equal to zero.
Key points:
- The function has vertical asymptotes where , approximately at , and .
- The function oscillates and repeats in a periodic manner, with alternating positive and negative branches due to the nature of the secant function.
- The graph is restricted to a y-range of for clarity, as tends toward infinity near the asymptotes.
Would you like a step-by-step explanation on how the graph was derived?
Related Questions:
- What is the period of the function ?
- How do you calculate the vertical asymptotes of the secant function?
- What are the key differences between the graphs of and ?
- How does the phase shift affect the graph of the secant function?
- Can the secant function ever have a minimum or maximum value?
Tip: To find the vertical asymptotes of secant, solve for , as 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