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

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

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

Answer 1

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

The formula for the axis of symmetry for the general quadratic equation \(y = ax^2 + bx + c\) is given by \(x = -\frac{b}{2a}\). In our case, \(a = -1\) and \(b = 0\). Therefore:

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

This indicates that the axis of symmetry is the vertical line \(x = 0\).

To find the symmetric point of \((-1, -17)\), we reflect it across the line \(x = 0\). The x-coordinate of the point \((-1, -17)\) is \(-1\).

The distance from this point to the line \(x = 0\) is 1 unit. Therefore, to find the symmetric point, we move 1 unit to the right of \(x = 0\):

\[
x = 0 + 1 = 1
\]

The y-coordinate remains the same as \(-17\), so the symmetric point is:

\[
(1, -17)
\]

Thus, the correct answer is:

\((1, -17)\) (left parenthesis 1 comma negative 17 right parenthesis)