The graph of a trigonometric function is shown.

Question

The graph of a trigonometric function is shown. The graph has vertical asymptotes at multiples of pi. The graph alternates between upward-facing and downward-facing curves. The upward-facing curves decrease from infinity to y equals 1, then increase to infinity. The downward-facing curves increase from negative infinity to y equals negative 1, then decrease to negative infinity.
Question
Use the graph to answer the question.
Which function is shown in the graph?
(1 point)
Responses

y = tan θ
y = tan θ

y = sec θ
y = sec θ

y = csc θ
y = csc θ

y = cot θ
y = cot θ

Answer 1

Based on the description provided for the graph, the function that matches the characteristics is y = sec θ.

Here's why:

• The presence of vertical asymptotes at multiples of pi indicates that the function is undefined at those points, which is a characteristic of secant (sec θ).
• The upward-facing curves decrease from infinity to y = 1 and then increase to infinity, while the downward-facing curves increase from negative infinity to y = -1 and then decrease to negative infinity, consistent with the behavior of the secant function, which oscillates between these values.

The other options do not exhibit these characteristics:

• y = tan θ has vertical asymptotes at odd multiples of π/2.
• y = csc θ has vertical asymptotes at integer multiples of π, but the behavior of the curves is different.
• y = cot θ also has vertical asymptotes at integer multiples of π, but its graph behaves differently than described.

Thus, the correct function is: y = sec θ.