Question

find the graph of (y = \sec(x)).

Explanation:

⚡ Using what you learned: Graphing Trigonometric Functions

Step 1: Identify key values of the function

The function is \( y = \sec(x) \), which is the reciprocal of the cosine function:

$$ y = \frac{1}{\cos(x)} $$

Let's evaluate \( y \) at key points:

• At \( x = 0 \):

$$ \cos(0) = 1 \implies y = \frac{1}{1} = 1 $$

So, the graph must have a point at \( (0, 1) \).

• At \( x = \pi \):

$$ \cos(\pi) = -1 \implies y = \frac{1}{-1} = -1 $$

So, the graph must have a point at \( (\pi, -1) \).

• At \( x = 2\pi \):

$$ \cos(2\pi) = 1 \implies y = \frac{1}{1} = 1 $$

So, the graph must have a point at \( (2\pi, 1) \).

Step 2: Identify vertical asymptotes

The function \( y = \sec(x) \) is undefined where \( \cos(x) = 0 \). This occurs at:

$$ x = \frac{\pi}{2}, \frac{3\pi}{2} $$

These values correspond to vertical asymptotes on the graph.

Step 3: Match with the correct graph

Looking at the options (oriented sideways in the image):

• First graph (top):
• At \( x = 0 \), the curve opens upwards starting from \( y = 1 \).
• At \( x = \pi \), the curve opens downwards starting from \( y = -1 \).
• At \( x = 2\pi \), the curve opens upwards starting from \( y = 1 \).
• Vertical asymptotes are at \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \).

This matches all the properties of \( y = \sec(x) \).

Answer:

The correct graph is the first graph (the top option in the image), which has a U-shaped curve opening upwards starting at \( (0, 1) \), an upside-down U-shaped curve opening downwards with its vertex at \( (\pi, -1) \), and another U-shaped curve opening upwards starting at \( (2\pi, 1) \).