Question

Question
What point is symmetrical to (−1,−17) for the graph of y=−x2−16 ?(1 point)
Responses

(−1,17)
left parenthesis negative 1 comma 17 right parenthesis

(1,−17)
left parenthesis 1 comma negative 17 right parenthesis

(1,17)
left parenthesis 1 comma 17 right parenthesis

(−17,−1)
left parenthesis negative 17 comma negative 1 right parenthesis
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Answer 1

To find the point that is symmetrical to \((-1, -17)\) for the graph of the function \(y = -x^2 - 16\), we need to first determine the x-coordinate of the point on the curve that corresponds to the x-coordinate of the given point.

1. **Substituting the x-coordinate**: The x-coordinate of the given point is \(-1\). Substitute \(x = -1\) into the equation to find \(y\):

\[
y = -(-1)^2 - 16
\]
\[
y = -1 - 16 = -17
\]

Thus, the point \((-1, -17)\) lies on the graph.

2. **Identifying the axis of symmetry**: The graph of the function \(y = -x^2 - 16\) is a downward-opening parabola, and it is symmetrical about the vertical line \(x = 0\) (the y-axis).

3. **Finding the symmetrical point**: Since the graph is symmetrical about the y-axis, the x-coordinate of the symmetrical point is the negative of the x-coordinate of the original point. Therefore, if the original point is \((-1, -17)\), the symmetrical point is:

\[
(1, -17)
\]

Now, looking at the provided options:

- \((-1, 17)\)
- \((1, -17)\) ✅ (this is our answer)
- \((1, 17)\)
- \((-17, -1)\)

Thus, the symmetrical point to \((-1, -17)\) for the graph of \(y = -x^2 - 16\) is \((1, -17)\).