What point is symmetrical to (-1,-17) for the graph of y = -x^2 - 16
To find the point that is symmetrical to (-1, -17) for the graph of y = -x^2 - 16, we can use the fact that symmetry occurs across the line x = h for a quadratic function of the form y = a(x-h)^2 + k.
In this case, since the original equation is y = -x^2 - 16, the vertex of the parabola is at (h, k) = (0, -16). The line x = h is the vertical line x = 0.
Thus, the point that is symmetrical to (-1, -17) is obtained by reflecting (-1, -17) across the line x = 0. Due to the symmetry property, the x-coordinate of the reflection will be the opposite of -1, which is 1. The y-coordinate remains the same, so the point that is symmetrical to (-1, -17) for the graph of y = -x^2 - 16 is (1, -17).