What point must also be on a quadratic function with a vertex of (2, 8) if the point (6, 20) is on the function?

What point must also be on a quadratic function with a vertex of (2, 8) if the point (6, 20) is on the function?

• A. (4, 20)
• B. (0, 20)
• C. (6, 8)
• D. (-2, 20)

Answer: (D)

To determine which point must also be on the quadratic function given the vertex and another point, we can apply the properties of quadratic functions. A quadratic function has a parabolic shape, and it is symmetric around its vertex. In this case, the vertex is at (2, 8). When a point, such as (6, 20), is located on one side of the vertex, there must be a corresponding point on the other side of the vertex that is equidistant from it due to the symmetry of the parabola. The distance from the vertex (2, 8) to the point (6, 20) can be calculated by considering the x-coordinates: 1. The distance in the x-direction from the vertex (2) to the point (6) is 6 - 2 = 4 units to the right. 2. Because of symmetry, there should be a corresponding point that is 4 units to the left of the vertex. This means moving from the x-coordinate of the vertex (2), we subtract 4 to find the corresponding x-coordinate: 2 - 4 = -2. Now, we keep the same y-coordinate as the vertex since we want to find a point at the

To determine which point must also be on the quadratic function given the vertex and another point, we can apply the properties of quadratic functions.

A quadratic function has a parabolic shape, and it is symmetric around its vertex. In this case, the vertex is at (2, 8). When a point, such as (6, 20), is located on one side of the vertex, there must be a corresponding point on the other side of the vertex that is equidistant from it due to the symmetry of the parabola.

The distance from the vertex (2, 8) to the point (6, 20) can be calculated by considering the x-coordinates:

1. The distance in the x-direction from the vertex (2) to the point (6) is 6 - 2 = 4 units to the right.

2. Because of symmetry, there should be a corresponding point that is 4 units to the left of the vertex. This means moving from the x-coordinate of the vertex (2), we subtract 4 to find the corresponding x-coordinate: 2 - 4 = -2.

Now, we keep the same y-coordinate as the vertex since we want to find a point at the