The intersection of the image of the quadratic function y = - x ^ 2 / 2 + X + 4 and the X axis is a and B from left to right, the intersection with the Y axis is C, and the vertex is d 1. Finding the area of quadrilateral abdc 2. Find a point d 'on the parabola in the first quadrant to maximize the area of quad abd'c

The intersection of the image of the quadratic function y = - x ^ 2 / 2 + X + 4 and the X axis is a and B from left to right, the intersection with the Y axis is C, and the vertex is d 1. Finding the area of quadrilateral abdc 2. Find a point d 'on the parabola in the first quadrant to maximize the area of quad abd'c

（1）
-x^2/2+x+4=0
X1=-2,X2=4
A(-2,0),B(4,0)
AB=6
When x = 0, y = 4
So, C (0,4)
Let D (x, y)
X=-1/[(-1/2)*2]=1,Y=(-8-1)/(-2)=4.5
So, D (1,4.5)
Area of triangle ABC = 1 / 2 * 6 * 4 = 12
Area of triangle abd = 1 / 2 * 6*
Area of quadrilateral abdc = 1 / 2 * 2 * 4 + 1 / 2 * (4 + 4.5) * 1 + 1 / 2 * 3 * 4.5
=4+4.25+6.75
=15
(2) Let d '(x, - x ^ 2 / 2 + X + 4) (x > 0, - x ^ 2 / 2 + X + 4 > 0)
S quadrilateral abd'c = 1 / 2 * 2 * 4 + 1 / 2 * (4-x ^ 2 / 2 + X + 4) * x + 1 / 2 (- x ^ 2 / 2 + X + 4) * (4-x)
=4+2X-X^3/4+X^2/2+2X-X^2+2X+8+X^3/4-X^2/2-2X
=-X^2+4X+12
When x = 4 / 2 = 2, abd'c of s quadrilateral has a maximum
-x^2/2+x+4=-4/2+2+4=4
So, d '(2,4)