A function consisting of two rays and an isolated point is graphed on a coordinate plane.A function consisting of two rays and an isolated point is graphed on a coordinate plane. The first ray increases from negative infinity to the point left parenthesis negative 1 comma 2.5 right parenthesis and includes the endpoint. The ray also has a hole at left parenthesis negative 2 comma 2 right parenthesis. A single point is graphed at left parenthesis negative 2 comma 1 right parenthesis. The second ray starts at an open point at left parenthesis negative 1 comma 3 right parenthesis and decreases to infinity.

Question

Question
Use the graph to answer the question.
Describe the continuity or discontinuity of the graphed function.
(4 points)

Answer 1

To describe the continuity or discontinuity of the graphed function, we need to analyze the information given about the rays and isolated points.

1. First Ray: This ray increases and includes the endpoint at \((-1, 2.5)\) but has a hole at \((-2, 2)\). This means that as you approach \((-2, 2)\) from the left on the first ray, the function approaches \(2\), but it is not defined at that point (hole). Therefore, there is a point of discontinuity at \((-2, 2)\).

2. Isolated Point: There is a single point graphed at \((-2, 1)\). This point does not affect the continuity of the function at \((-2, 2)\) since the function was originally discontinuous at that location. The isolated point does not connect with the rays, further emphasizing that the function is not continuous at \(x = -2\).

3. Second Ray: The second ray starts at an open point at \((-1, 3)\) and decreases to infinity. The openness at \((-1, 3)\) indicates another point of discontinuity at \((-1, 3)\) since the function does not actually attain this value at \(x = -1\). It transitions from the first ray to this open point, suggesting a jump discontinuity.

Summary:

• There is a discontinuity at \((-2, 2)\) due to the hole.
• There is another discontinuity at \((-1, 3)\) due to the open point.
• The function is generally discontinuous at these points, and while it is continuous on the intervals outside these points, the specified points cause overall discontinuity in the function.

Overall conclusion: The graphed function has points of discontinuity at \((-2, 2)\) and \((-1, 3)\), making the function discontinuous.