Question

如图,抛物线y=1/2x^2+bx+c与y轴交于点C(0,-4),与x轴交于点A,B,且B点的坐标为(2,0)

Answer 1

过(0,-4),y = c = -4

y = x²/2 + bx - 4

过(2,0):1 + b - 4 = 0,b = 3

y = x²/2 + 3x - 4 = (1/2)(x - 2)(x + 4)

A(-4,0)

令P(p,0),-4 < p < 2

AC的斜率为(-4 - 0)/(0 + 4)= -1

PE的斜率也是-1,PE的方程为y = -(x - p)

BC的方程为:x/2 + y/(-4) = 1

解得E((p+4)/3,(2p - 4)/3)

△PCE的面积 = △PCB的面积 - △PEB的面积

= (1/2)*PB*CO - (1/2)*PB*|E的纵坐标|

= (1/2)(2 - p)*4 - (1/2)(2 - p)(4 - 2p)/3

= (1/3)(2 - p)(p + 4)

此为开口向下的抛物线,对称轴为(2 - 4)/2 = -1

△PCE的面积的最大值为 (1/3)(2 + 1)(-1 +4) = 3