Answer:
V(5, -4)
Explanation:
You want the vertex (V) coordinates of a parabola such that the triangle ABV has an area of 12 square units and V is below the x-axis. The coordinates of A and B are A(2, 0) and B(8, 0).
Triangle
The base of the triangle is the length of AB, the difference of their x-coordinates:
AB = 8 -2 = 6
The height of the triangle can be found from the area formula:
A = 1/2bh
12 = 1/2·6·h
4 = h . . . . . . . . divide by the coefficient of h
This tells us the y-coordinate of V is 4 units below the x-axis, at y=-4.
Symmetry
The vertex lies on the line of symmetry of the parabola, which lies halfway between the given x-intercepts. The line of symmetry is located at ...
x = (2 +8)/2 = 10/2 = 5
The vertex coordinates of V are (x, y) = (5, -4).