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Original Question
A function is graphed on the coordinate plane. It is a parabola with an open circle at the vertex left-parenthesis 3 comma 0 ri...Asked by freedom
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
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.
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.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
Based on the description provided, the function appears to be a parabola with an open circle at the vertex \( (3, 0) \), indicating that the function is not defined at that point. Additionally, there is a point at \( (3, 3) \) which is not on the parabola.
This configuration suggests that as you approach \( x = 3 \), the function approaches 0 (the value of the parabola) but is not defined at that point (since there is an open circle). The point at \( (3, 3) \) does not suggest continuity either, as the function would jump from being at 0 to being at 3 directly.
Thus, the discontinuity at \( x = 3 \) is removable, because if the function were defined at \( (3, 0) \)—the limit as \( x \) approaches 3 from either side is 0—the function could be made continuous.
Therefore, the correct response would be:
**The function has a removable discontinuity at \( x = 3 \).**
This configuration suggests that as you approach \( x = 3 \), the function approaches 0 (the value of the parabola) but is not defined at that point (since there is an open circle). The point at \( (3, 3) \) does not suggest continuity either, as the function would jump from being at 0 to being at 3 directly.
Thus, the discontinuity at \( x = 3 \) is removable, because if the function were defined at \( (3, 0) \)—the limit as \( x \) approaches 3 from either side is 0—the function could be made continuous.
Therefore, the correct response would be:
**The function has a removable discontinuity at \( x = 3 \).**
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