To find the point that is symmetrical to (-1, -17) for the graph of y = -x^2 - 16, we need to find the point that is the same distance from the axis of symmetry as (-1, -17).
The axis of symmetry for a parabola in the form y = ax^2 + bx + c is given by the equation x = -b/(2a).
In this case, the equation is y = -x^2 - 16. Comparing it to the form y = ax^2 + bx + c, we have a = -1, b = 0, and c = -16.
Thus, the axis of symmetry is x = -b/(2a) = -0/(2*(-1)) = 0.
The point (-1, -17) is 1 unit to the left of the axis of symmetry.
Therefore, the point that is symmetric to (-1, -17) is 1 unit to the right of the axis of symmetry. The x-coordinate will still be 0, and the y-coordinate will be the same but with the opposite sign.
Hence, the point that is symmetrical to (-1, -17) for the graph of y = -x^2 - 16 is (1, 17).
The correct answer is C. (1, 17).
What point is symmetrical to (-1, -17) for the graph of y = -x^2 - 16?
A. (-17, -1)
B. (-1, 17)
C. (1, 17)
D. (1, -17)
1 answer