Features of Graphs of Quadratic Equations Quick Check

Question

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

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

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

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

(1,−17)

Answer 1

To find the point that is symmetrical to \((-1, -17)\) for the graph of the quadratic equation \(y = -x^2 - 16\), we need to determine the axis of symmetry of the graph.

The graph of \(y = -x^2 - 16\) is a downward-opening parabola. The axis of symmetry for a parabola defined by \(y = ax^2 + bx + c\) is the vertical line \(x = -\frac{b}{2a}\). In this case, since there is no \(x\) term (i.e., \(b = 0\)), the axis of symmetry is \(x = 0\) (the y-axis).

The point \((-1, -17)\) is located to the left of the y-axis. To find the symmetric point across the y-axis, we simply change the x-coordinate from \(-1\) to \(1\), while keeping the y-coordinate the same. Therefore, the symmetrical point is:

\[ (1, -17) \]

So the correct answer from the options provided is:

(1, -17).