To find the point symmetrical to \((-1, -17)\) for the graph of \(y = -x^2 - 16\), we first need to determine the axis of symmetry of the parabola described by the equation.
The parabola given by \(y = -x^2 - 16\) opens downward and has its vertex at the point where \(x = 0\). The vertex can be found by evaluating \(y\) at \(x = 0\):
\[ y = -0^2 - 16 = -16 \]
Thus, the vertex of the parabola is at \((0, -16)\).
The axis of symmetry for this parabola is the vertical line \(x = 0\). To find the point that is symmetrical to \((-1, -17)\) across this axis (which is the y-axis), we reflect the x-coordinate of \((-1, -17)\).
The symmetrical point will have the same y-coordinate but the opposite x-coordinate:
\[ (-1, -17) \rightarrow (1, -17) \]
Therefore, the point symmetrical to \((-1, -17)\) for the graph of \(y = -x^2 - 16\) is \((1, -17)\).
So the answer is:
\((1,−17)\)
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