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)
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