To find the point that is symmetrical to (-1, -17) for the graph of y = -x^2 - 16, we need to determine the reflection of the given point across the axis of symmetry.
For a quadratic function of the form y = ax^2 + bx + c, the axis of symmetry is given by the equation x = -b/2a.
In this case, the equation is y = -x^2 - 16, so we have a = -1 and b = 0. Plugging these values into the equation for the axis of symmetry, we get x = 0/2(-1) = 0.
Since the axis of symmetry is x = 0, any point (a, b) on the graph will have its reflection equal to (-a, b) or (a, -b) across the axis.
Therefore, the point that is symmetrical to (-1, -17) is (1, -17).
what point is symmetrical to (-1, -17) for the graph of y=-x^2-16
1 answer