A function is graphed on the coordinate plane. It is a parabola with an open circle at the vertex left-parenthesis 3 comma 0 right-parenthesis. There is a point graphed at left-parenthesis 3 comma 3 right-parenthesis that is not on the parabola.

Question

Question
Use the graph to answer the question.
Describe the continuity of the graphed function.
(1 point)
Responses

The function is continuous.
The function is continuous.

The function has a removable discontinuity at x = 3.
The function has a removable discontinuity at x = 3.

The function has a jump discontinuity at x = 3.
The function has a jump discontinuity at x = 3.

The function has an infinite discontinuity at x = 3.

Answer 1

To analyze the continuity of the graphed function based on your description, let's consider the information provided:

1. The vertex of the parabola is at \((3, 0)\) and it has an open circle at this point, indicating that this point is not included in the function (it is not defined there).
2. There is a point at \((3, 3)\) that is not on the parabola.

Given this information:

• The open circle at \((3, 0)\) means that the function is not defined at \(x = 3\).
• The presence of the point \((3, 3)\) indicates that the function does not approach or equal this value either.

This suggests that there is not a smooth transition through \(x = 3\).

Based on these observations, the most fitting option for the continuity of the function is:

The function has a removable discontinuity at x = 3.

This is because the function is defined everywhere except at a single point, and this point is where the open circle indicates that it could be defined (but isn't).