如图,已知抛物线y=x2+bx+c交x轴与A(1,0),B(3,0)两点如图,已知抛物线y=x2+bx+c交与x轴与A(1,0),B(3,0)两点交y轴于点C,其顶点为D.(1)求b,c的值并写出抛物线的对称轴;(2) 连接BC,过点O作直线OE⊥BC
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![如图,已知抛物线y=x2+bx+c交x轴与A(1,0),B(3,0)两点如图,已知抛物线y=x2+bx+c交与x轴与A(1,0),B(3,0)两点交y轴于点C,其顶点为D.(1)求b,c的值并写出抛物线的对称轴;(2) 连接BC,过点O作直线OE⊥BC](/uploads/image/z/1671703-7-3.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5%E6%8A%9B%E7%89%A9%E7%BA%BFy%3Dx2%2Bbx%2Bc%E4%BA%A4x%E8%BD%B4%E4%B8%8EA%281%2C0%29%2CB%283%2C0%29%E4%B8%A4%E7%82%B9%E5%A6%82%E5%9B%BE%2C%E5%B7%B2%E7%9F%A5%E6%8A%9B%E7%89%A9%E7%BA%BFy%3Dx2%2Bbx%2Bc%E4%BA%A4%E4%B8%8Ex%E8%BD%B4%E4%B8%8EA%EF%BC%881%2C0%EF%BC%89%2CB%EF%BC%883%2C0%EF%BC%89%E4%B8%A4%E7%82%B9%E4%BA%A4y%E8%BD%B4%E4%BA%8E%E7%82%B9C%2C%E5%85%B6%E9%A1%B6%E7%82%B9%E4%B8%BAD.%EF%BC%881%EF%BC%89%E6%B1%82b%2Cc%E7%9A%84%E5%80%BC%E5%B9%B6%E5%86%99%E5%87%BA%E6%8A%9B%E7%89%A9%E7%BA%BF%E7%9A%84%E5%AF%B9%E7%A7%B0%E8%BD%B4%EF%BC%9B%282%29+%E8%BF%9E%E6%8E%A5BC%2C%E8%BF%87%E7%82%B9O%E4%BD%9C%E7%9B%B4%E7%BA%BFOE%E2%8A%A5BC)
如图,已知抛物线y=x2+bx+c交x轴与A(1,0),B(3,0)两点如图,已知抛物线y=x2+bx+c交与x轴与A(1,0),B(3,0)两点交y轴于点C,其顶点为D.(1)求b,c的值并写出抛物线的对称轴;(2) 连接BC,过点O作直线OE⊥BC
如图,已知抛物线y=x2+bx+c交x轴与A(1,0),B(3,0)两点
如图,已知抛物线y=x2+bx+c交与x轴与A(1,0),B(3,0)两点交y轴于点C,其顶点为D.
(1)求b,c的值并写出抛物线的对称轴;
(2) 连接BC,过点O作直线OE⊥BC交抛物线的对称轴于点E,求证四边形ODBE是等腰梯形;
(3)抛物线上是否存在点Q,使得△OBQ的面积等于四边形ODBE的面积的?若存在,求点Q的坐标;若不存在,请说明理由.
如图,已知抛物线y=x2+bx+c交x轴与A(1,0),B(3,0)两点如图,已知抛物线y=x2+bx+c交与x轴与A(1,0),B(3,0)两点交y轴于点C,其顶点为D.(1)求b,c的值并写出抛物线的对称轴;(2) 连接BC,过点O作直线OE⊥BC
求采纳! 我也很辛苦
(1) 与x轴与A(1,0),B(3,0), 抛物线可表达为y = (x - 1)(x - 3) = x²-4x +3
b = -4, c = 3
对称轴x = (1+ 3)/2 = 2
(2) C(0, 3), D(2, -1)
BC的斜率k = (3 - 0)/(0 - 3) = -1
OE的斜率k' = -1/k = 1
OE的方程...
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(1) 与x轴与A(1,0),B(3,0), 抛物线可表达为y = (x - 1)(x - 3) = x²-4x +3
b = -4, c = 3
对称轴x = (1+ 3)/2 = 2
(2) C(0, 3), D(2, -1)
BC的斜率k = (3 - 0)/(0 - 3) = -1
OE的斜率k' = -1/k = 1
OE的方程y = x, E(2, 2)
BD的斜率k" = (-1-0)/(2 - 3) = 1
BD与OE平行
OD²= (2 - 0)² + (-1-0)² = 5
BE²= (2 - 3)² + (2 - 0)² = 5
OD = BE
四边形ODBE是等腰梯形
(3)
对称轴与x轴交于F(2, 0)
四边形ODBE的面积 = △OBD的面积 + △OBE的面积 = (1/2)OB*DF + (1/2)OB*FE = (1/2)OB(DF + FE) = (1/2)OB*DE
相当于以OB为底, 高为DE = 3的三角形面积, 即Q的纵坐标的绝对值为3,
y = x²-4x +3 = 3
x = 0, Q(0, 3), 即C
x = 4, Q(4, 3)
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