A plane curve is graphed.A graph of the curve is in three sections separated by asymptotes at x equals negative pi and x = pi. The middle section of the graph passes through left-parenthesis negative start fraction pi over 2 end fraction comma 1 right-parenthesis, left-parenthesis 0 comma 2 right parenthesis and left-parenthesis start fraction pi over 2 end fraction comma 3 right-parenthesis.

Question

Question
Use the graph to answer the question.
What is the equation of the graphed trigonometric function?
(1 point)
Responses

y equals tangent left parenthesis one half x right parenthesis plus 2
Image with alt text: y equals tangent left parenthesis one half x right parenthesis plus 2

y equals tangent left parenthesis one half x right parenthesis minus 2
Image with alt text: y equals tangent left parenthesis one half x right parenthesis minus 2

y equals tangent left parenthesis 2 x right parenthesis minus 2
Image with alt text: y equals tangent left parenthesis 2 x right parenthesis minus 2

y equals tangent left-parenthesis 2 x right-parenthesis plus 2
Image with alt text: y equals tangent left-parenthesis 2 x right-parenthesis plus 2

Answer 1

To determine the equation of the graphed trigonometric function, we can analyze the behavior of the tangent function.

1. Identification of Asymptotes: The curve has vertical asymptotes at \( x = -\pi \) and \( x = \pi \). This indicates that the period of the function is \( 2\pi \), which is characteristic of the basic tangent function.

2. Behavior around Asymptotes: The tangent function (\( \tan(x) \)) has vertical asymptotes at odd multiples of \( \frac{\pi}{2} \). Since the asymptotes of the graph are at \( x = -\pi \) and \( x = \pi \), the vertical asymptotes of the standard tangent function need to be scaled.

3. Scaling the function: The function likely has a form of \( \tan(kx) \) where \( k \) determines the frequency of the function. The asymptotes occur at intervals of \( \frac{\pi}{k} \). Since our asymptotes are at \( -\pi \) and \( \pi \), the distance between them is \( 2\pi \). Thus, we need \( k \) such that \( \frac{\pi}{k} = \pi \), which gives \( k = 1 \). This means the tangent function is not simply \( \tan(x) \).

4. Midline transformation: The middle section of the graph passes through the points \( (-\frac{\pi}{2}, 1) \), \( (0, 2) \), and \( (\frac{\pi}{2}, 3) \). The middle point at \( (0, 2) \) suggests that the graph has been shifted vertically, likely by \( +2 \) to match the points.

5. Final equation construction: Since the behavior resembles that of \( \tan(kx) \) and must pass through the vertical shift, we conclude it must shift up by 2. Hence, we have:

The likely candidate is: \[ y = \tan\left(\frac{1}{2}x\right) + 2 \]

However, we also check against the options provided. Given the presence of a coefficient of \( 2 \) which appears final in some forms, we also see that \( y = \tan(2x) + 2 \) could potentially match if the function fits within the standard behavior of tangent.

Conclusion:

Based on the properties inferred, the answer that aligns best is: \[ \text{y equals tangent}\left(2x\right) + 2 \]

That matches the last option: y equals tangent left-parenthesis 2 x right-parenthesis plus 2.