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 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).
To find the point that is symmetrical to (-1, -17) for the graph of y = -x^2 - 16, we can use the concept of symmetry.
The graph of the equation y = -x^2 - 16 is a downward-facing parabola with its vertex at the point (0, -16).
Since symmetry occurs with respect to the vertical line passing through the vertex, to find the point symmetric to (-1, -17), we need to find the same x-coordinate on the other side of the vertex.
The x-coordinate of (-1, -17) is -1. Thus, the symmetric x-coordinate will be 1 unit to the right of the vertex, which gives us an x-coordinate of +1.
To find the corresponding y-coordinate, we substitute x = 1 into the equation y = -x^2 - 16:
y = -(1)^2 - 16
y = -1 - 16
y = -17
Therefore, the point that is symmetrical to (-1, -17) for the graph of y = -x^2 - 16 is (1, -17).
To find the point that is symmetrical to a given point on a graph with respect to the equation y = -x^2 - 16, you can follow these steps:
1. Plot the given point (-1, -17) on the coordinate plane. This is the point you want to find the symmetrical point for.
2. Notice that the given equation is a downward-opening parabola. The symmetry of the parabola lies along its axis of symmetry, which is a vertical line passing through the vertex.
3. The equation of the axis of symmetry can be found using the formula x = -b / (2a), where the given equation is in the form y = ax^2 + bx + c. In this case, the equation is already in the form y = -x^2 - 16, so we can identify the coefficients a and b as -1 and 0, respectively. Thus, the axis of symmetry is x = -0 / (2*-1), which simplifies to x = 0.
4. Since the axis of symmetry is a vertical line through x = 0, any point (x, y) on the graph will have its symmetrical point as (-x, y).
5. Apply this concept to the given point. The symmetrical point will be (-(-1), -17), which simplifies to (1, -17).
Therefore, the point symmetrical to (-1, -17) for the graph of y = -x^2 - 16 is (1, -17).