A square is formed beneath the parabola y=16-x^2 and the x axis. What is the area of this square.

Answer 1

As the sign of the x^2 term is negative, the parabola opens down. When x = 0, y = 16 which is the vertex. The base of the square lies on the x axis and is between the x intercepts of the parabola. The x intercepts are found by letting y = 0 and solving for x which gives x = +/- 4. So, if the base of the square is on the x axis between x = 4 and x = -4, the length of the base is 8, therefore the area is 64.