Question
What point is symmetrical to (−1,−17) for the graph of y=−x2−16 ?(1 point)
Responses
(−17,−1)
left parenthesis negative 17 comma negative 1 right parenthesis
(1,17)
left parenthesis 1 comma 17 right parenthesis
(1,−17)
left parenthesis 1 comma negative 17 right parenthesis
(−1,17)
Responses
(−17,−1)
left parenthesis negative 17 comma negative 1 right parenthesis
(1,17)
left parenthesis 1 comma 17 right parenthesis
(1,−17)
left parenthesis 1 comma negative 17 right parenthesis
(−1,17)
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To find the point that is symmetrical to \((-1, -17)\) with respect to the graph of \(y = -x^2 - 16\), we need to understand the concept of symmetry in this context.
The graph of \(y = -x^2 - 16\) is a downward-opening parabola. The line of symmetry for a parabola in standard form \(y = ax^2 + bx + c\) is the vertical line that passes through the vertex.
1. **Find the vertex of the parabola**:
For the parabola \(y = -x^2 - 16\), the vertex occurs at the maximum point since it opens downward. The vertex can be found by noting that there is no linear \(x\) term (the coefficient of \(x\) is 0). Therefore, the x-coordinate of the vertex is 0.
Plugging \(x = 0\) into the equation gives:
\[
y = -0^2 - 16 = -16
\]
Thus, the vertex is at the point \((0, -16)\).
2. **Find the symmetrical point**:
To find a point symmetrical to \((-1, -17)\) with respect to the vertex \((0, -16)\), we can find the midpoint between the two points, which should be equal to the vertex.
Let \((x', y')\) be the coordinates of the symmetrical point. The midpoint between \((-1, -17)\) and \((x', y')\) should equal \((0, -16)\):
\[
\left( \frac{-1 + x'}{2}, \frac{-17 + y'}{2} \right) = (0, -16)
\]
From here, we can set up two equations:
\[
\frac{-1 + x'}{2} = 0 \quad \text{(1)}
\]
\[
\frac{-17 + y'}{2} = -16 \quad \text{(2)}
\]
Solving equation (1):
\[
-1 + x' = 0 \implies x' = 1
\]
Solving equation (2):
\[
-17 + y' = -32 \implies y' = -32 + 17 \implies y' = -15
\]
Thus, the symmetrical point is \((1, -15)\).
Checking against the options, the closest correspondences were provided. However, it seems there is an error since \((1, -15)\) isn't in the provided options. If I narrowed down further based on what you provided:
If we had to choose a point from the options provided based on closest functional behavior or geometry, and since none of the options provide this point effectively based on our findings, please clarify or provide options to closely lead to understanding.
Let's confirm accordingly:
None of these choices were a perfect match for symmetry, if the output was strictly a direct symmetry from the point to vertex. Valid options from the choices you provided were:
- \((−17,−1)\)
- \((1,17)\)
- \((1,−17)\)
- \((−1,17)\)
If no reflection matches on guidelines or correctness, you may adjust symmetrical queries accordingly to guide into clearer outputs. Hope this helps shape clearer outcomes!
The graph of \(y = -x^2 - 16\) is a downward-opening parabola. The line of symmetry for a parabola in standard form \(y = ax^2 + bx + c\) is the vertical line that passes through the vertex.
1. **Find the vertex of the parabola**:
For the parabola \(y = -x^2 - 16\), the vertex occurs at the maximum point since it opens downward. The vertex can be found by noting that there is no linear \(x\) term (the coefficient of \(x\) is 0). Therefore, the x-coordinate of the vertex is 0.
Plugging \(x = 0\) into the equation gives:
\[
y = -0^2 - 16 = -16
\]
Thus, the vertex is at the point \((0, -16)\).
2. **Find the symmetrical point**:
To find a point symmetrical to \((-1, -17)\) with respect to the vertex \((0, -16)\), we can find the midpoint between the two points, which should be equal to the vertex.
Let \((x', y')\) be the coordinates of the symmetrical point. The midpoint between \((-1, -17)\) and \((x', y')\) should equal \((0, -16)\):
\[
\left( \frac{-1 + x'}{2}, \frac{-17 + y'}{2} \right) = (0, -16)
\]
From here, we can set up two equations:
\[
\frac{-1 + x'}{2} = 0 \quad \text{(1)}
\]
\[
\frac{-17 + y'}{2} = -16 \quad \text{(2)}
\]
Solving equation (1):
\[
-1 + x' = 0 \implies x' = 1
\]
Solving equation (2):
\[
-17 + y' = -32 \implies y' = -32 + 17 \implies y' = -15
\]
Thus, the symmetrical point is \((1, -15)\).
Checking against the options, the closest correspondences were provided. However, it seems there is an error since \((1, -15)\) isn't in the provided options. If I narrowed down further based on what you provided:
If we had to choose a point from the options provided based on closest functional behavior or geometry, and since none of the options provide this point effectively based on our findings, please clarify or provide options to closely lead to understanding.
Let's confirm accordingly:
None of these choices were a perfect match for symmetry, if the output was strictly a direct symmetry from the point to vertex. Valid options from the choices you provided were:
- \((−17,−1)\)
- \((1,17)\)
- \((1,−17)\)
- \((−1,17)\)
If no reflection matches on guidelines or correctness, you may adjust symmetrical queries accordingly to guide into clearer outputs. Hope this helps shape clearer outcomes!
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