初三数学 二次函数 望大神给出详细解答过程 可加悬赏 谢谢
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(1)
A(0, 1), 对称型猛森轴x = -(2m/3)/[2(-1/3)] = m, B(m, 0), C(m, 1)
BE = BC = 1, E(m+1, 0)
ED = AC = m, D(m+1, m)
代入抛物线: m = -(m+1)²/3 + (2/3)m(m+1)+1
m² - 3m + 2 = (m-2)(m - 1)=0, m = 2 (舍去卜亩m = 1)
(2)
E(3, 0), D(3, 2), C(2, 1)
EC: (y - 0)/(1 - 0) = (x - 3)(2 - 3), y = 3 - x
F(0, 3)
AF = 2 = ED
显然ED与AF平行,即该知老四边形为平行四边形。
(3)
显然只有一种可能,即M在第三象限,N在y轴的负半轴。
B(2, 0), AB的斜率为k = (1 - 0)/(0 - 2) = -1/2
BN与AB相互垂直,其斜率为-1/k = 2, BN: y = 2(x - 2)
N(0, -4)
AM: y = 2x + 1(斜截式)
2x + 1 = -x²/3 + 4x/3 + 1
x² + 2x = x(x + 2)=0
x=-2, M(-2, -3)
MN斜率为k' = (-3 + 4)/(-2 - 0) = -1/2 = k
MN与AB平行
A(0, 1), 对称型猛森轴x = -(2m/3)/[2(-1/3)] = m, B(m, 0), C(m, 1)
BE = BC = 1, E(m+1, 0)
ED = AC = m, D(m+1, m)
代入抛物线: m = -(m+1)²/3 + (2/3)m(m+1)+1
m² - 3m + 2 = (m-2)(m - 1)=0, m = 2 (舍去卜亩m = 1)
(2)
E(3, 0), D(3, 2), C(2, 1)
EC: (y - 0)/(1 - 0) = (x - 3)(2 - 3), y = 3 - x
F(0, 3)
AF = 2 = ED
显然ED与AF平行,即该知老四边形为平行四边形。
(3)
显然只有一种可能,即M在第三象限,N在y轴的负半轴。
B(2, 0), AB的斜率为k = (1 - 0)/(0 - 2) = -1/2
BN与AB相互垂直,其斜率为-1/k = 2, BN: y = 2(x - 2)
N(0, -4)
AM: y = 2x + 1(斜截式)
2x + 1 = -x²/3 + 4x/3 + 1
x² + 2x = x(x + 2)=0
x=-2, M(-2, -3)
MN斜率为k' = (-3 + 4)/(-2 - 0) = -1/2 = k
MN与AB平行
追答
提问者不能判断对错,哈哈
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