What point is symmetrical to (−1,−17)

Question

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 negative 17 right parenthesis

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

(−1,17)

Answer 1

To find the point that is symmetrical to \((-1, -17)\) with respect to the graph of the function \(y = -x^2 - 16\), we need to first find the vertical line of symmetry of the parabola.

The equation \(y = -x^2 - 16\) is a downward-opening parabola. The vertex of this parabola can be found since it is in the form \(y = ax^2 + bx + c\) with \(a = -1\), \(b = 0\), and \(c = -16\). The x-coordinate of the vertex is given by \(x = -\frac{b}{2a}\):

\[ x = -\frac{0}{2(-1)} = 0 \]

So the vertex is at \((0, -16)\). The line of symmetry is the vertical line \(x = 0\).

To find a point symmetrical to \((-1, -17)\), we reflect this point across the line \(x = 0\). The x-coordinate of the point will change sign, while the y-coordinate remains the same.

The original point is \((-1, -17)\). Reflecting it across the line \(x = 0\), we get:

\[ (1, -17) \]

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

(1, -17).

So the correct response is:

(1, -17).