snapkojacky
幼苗
共回答了18个问题采纳率:83.3% 举报
(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= .∴顶点D的坐标为 (3/2 ,-25/8 ).(2)当x = 0时y = -2,∴C(0,-2),OC = 2.当y = 0时,1/2x²- 3/2x-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C点关于X轴对称点C',连接C‘D,设C’D的函数解析式y=kx+b将C‘,D带入此函数解析式求的b=2 k=-41/24,所以m=-24/41
1年前
8