(1)∵点A(-1,0)在抛物线y=x2 + bx-2上,∴× (-1 )2 + b× (-1)–2 = 0,解得b = ∴抛物线的解析式为y=x2-x-2. y=x2-x-2 = ( x2 -3x- 4 ) =(x-)2-, ∴顶点D的坐标为 (, -). (2)当x = 0时y = -2, ∴C(0,-2),OC = 2。 当y = 0时, x2-x-2 = 0, ∴x1 =" -1," x2 = 4, ∴B (4,0) ∴OA = 1, OB = 4, AB = 5. ∵AB2 = 25, AC2 = OA2 + OC2 = 5, BC2 = OC2 + OB2 = 20, ∴AC2 +BC2 = AB2. ∴△ABC是直角三角形. (3)作出点C关于x轴的对称点C′,则C′(0,2),OC′=2,连接C′D交x轴于点M,根据轴对称性及两点之间线段最短可知,MC + MD的值最小。
解法一:设抛物线的对称轴交x轴于点E. ∵ED∥y轴, ∴∠OC′M=∠EDM,∠C′OM=∠DEM ∴△C′OM∽△DEM. ∴ ∴,∴m =. 解法二:设直线C′D的解析式为y =" kx" + n , 则,解得n =" 2," . ∴ . ∴当y = 0时, , . ∴. |