如图抛物线y=ac^2+bx经过点A(2,0),B(1,1)

抛物线y=ax²+bx经过点A(2,0),B(1,1)；(1)求抛物线方程；(2)直线y=-(1/2)x与抛物线交于C点,请求出C点坐标；(3)P是直线OC上方的抛物线上一动点(不与点O,C重合),过P做x轴的垂线交OC于D
①设P横坐标m,用含m的代数式表示PD的长；②PD是否存在最大值,说明理由
(1).将A,B的坐标代入抛物线方程得 4a+2b=0.(1)；a+b=1.(2)；
(1)(2)联立求解,得a=-1,b=2；故抛物线方程为y=-x²+2x.
(2).由-x²+2x=-(1/2)x,得2x²-5x=x(2x-5)=0,解得x=5/2,y=-5/4,即C点的坐标为(5/2,-5/4)；
(3).①∣PD∣=-m²+2m+(1/2)m=-m²+(5/2)m；
②∣PD∣=-m²+(5/2)m=-[m²-(5/2)m]=-[(m-5/4)²-25/16]=-(m-5/4)²+25/16≦25/16；即当m=5/4
时∣PD∣获得最大值25/16.

(1)
4a + 2b = 0
a + b = 1
a = -1, b= 2
y = -x^2 + 2x
(2) y = -x/2 = -x^2 + 2x
x(x - 5/2) = 0
x = 5/2, y = -5/4
C(5/2, -5/4)
(3)
P(m, 2m - m^2)
x = m, y = -x/2 = -m/2
PD = 2m - m^2 - (-m/2)
= 5m/2 - m^2
= -(m -5/4)^2 + 25/16
m = 5/4, PD最大