The right triangle abc shown below is inscribed inside a parabola. point b is also the maximum point of the parabola (vertex) and point c is the x intercept of the parabola. if the equation of the parabola is given by y = -x2 + 4x + c, find c so that the area of the triangle abc is equal to 32 square units.

H = -b / 2a = 2 : x coordinate of the vertex of the parabola k = -(2)2 + 4(2) + C = 4 + C : y coordinate of vertex x = (2 + √(4 + C)) , x = (2 - √(4 + C)) : the two x intercepts of the parabola. length of BA = k = 4 + C length of AC = 2 + √(4 + C) - 2 = √(4 + C) area = (1/2)BA * AC = (1/2) (4 + C) * √(4 + C) (1/2) (4 + C) * √(4 + C) = 32 : area is equal to 32 C = 12 : solve above for C.

5\16 x 2\5 = 10/80simplify to 1/80