已知抛物线C1:y1=a(x-1)2+k(a≠0) 急已知抛物线C1:y1=a(x-1)²+k1(a≠0)交x轴玉点(0,0)与点A1(b1,0),抛物线C2:y2=a(x-b1)²+k2交x轴与点(0,0)与点A2(b2,0),抛物线C3:y3=a(x-b2)²+k3交x轴与点(0,0
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![已知抛物线C1:y1=a(x-1)2+k(a≠0) 急已知抛物线C1:y1=a(x-1)²+k1(a≠0)交x轴玉点(0,0)与点A1(b1,0),抛物线C2:y2=a(x-b1)²+k2交x轴与点(0,0)与点A2(b2,0),抛物线C3:y3=a(x-b2)²+k3交x轴与点(0,0](/uploads/image/z/4338945-9-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%8A%9B%E7%89%A9%E7%BA%BFC1%3Ay1%3Da%28x-1%292%2Bk%28a%E2%89%A00%29+%E6%80%A5%E5%B7%B2%E7%9F%A5%E6%8A%9B%E7%89%A9%E7%BA%BFC1%3Ay1%3Da%28x-1%29%26%23178%3B%2Bk1%28a%E2%89%A00%29%E4%BA%A4x%E8%BD%B4%E7%8E%89%E7%82%B9%EF%BC%880%2C0%EF%BC%89%E4%B8%8E%E7%82%B9A1%EF%BC%88b1%2C0%29%2C%E6%8A%9B%E7%89%A9%E7%BA%BFC2%EF%BC%9Ay2%3Da%28x-b1%29%26%23178%3B%2Bk2%E4%BA%A4x%E8%BD%B4%E4%B8%8E%E7%82%B9%EF%BC%880%2C0%EF%BC%89%E4%B8%8E%E7%82%B9A2%EF%BC%88b2%2C0%EF%BC%89%2C%E6%8A%9B%E7%89%A9%E7%BA%BFC3%EF%BC%9Ay3%3Da%28x-b2%29%26%23178%3B%2Bk3%E4%BA%A4x%E8%BD%B4%E4%B8%8E%E7%82%B9%EF%BC%880%2C0)
已知抛物线C1:y1=a(x-1)2+k(a≠0) 急已知抛物线C1:y1=a(x-1)²+k1(a≠0)交x轴玉点(0,0)与点A1(b1,0),抛物线C2:y2=a(x-b1)²+k2交x轴与点(0,0)与点A2(b2,0),抛物线C3:y3=a(x-b2)²+k3交x轴与点(0,0
已知抛物线C1:y1=a(x-1)2+k(a≠0) 急
已知抛物线C1:y1=a(x-1)²+k1(a≠0)交x轴玉点(0,0)与点A1(b1,0),抛物线C2:y2=a(x-b1)²+k2交x轴与点(0,0)与点A2(b2,0),抛物线C3:y3=a(x-b2)²+k3交x轴与点(0,0)与点A3(b3,0)……按此规律,抛物线Cn:yn=a(x-bn-1)²+kn交x轴与点(0,0)与点An(bn,0)(其中n为正整数),我们把抛物线C1,C2,C3……,Cn称为系数为a”关于原点相似“的抛物线族.
(1)市求出b1的值
(2)请用含n的代数式表示线段An-1An的长
(3)探究下列问题
抛物线Cn:yn=a(x-bn-1)²+kn的顶点纵坐标kn与a、n有何数量关系?请说明理由;
若系数为a的”关于原点相似“的抛物线族的顶点坐标记为(T,S),请直接写出S和T所满足的函数关系式.
已知抛物线C1:y1=a(x-1)2+k(a≠0) 急已知抛物线C1:y1=a(x-1)²+k1(a≠0)交x轴玉点(0,0)与点A1(b1,0),抛物线C2:y2=a(x-b1)²+k2交x轴与点(0,0)与点A2(b2,0),抛物线C3:y3=a(x-b2)²+k3交x轴与点(0,0
1.抛物线C1对称轴为X=1,与X轴的交点关于X=1是对称的,所以b1=2
2.由题意可导出,bn=2bn-1,b1=2则bn=2^n,An-1An=bn-bn-1=2^(n-1)条件n大于等于2
3.抛物线过点(0,0),代入方程,a*bn-1^2+kn=0,kn=-a*bn-1^2=-a*2^(2n-2)
S=-aT2