已知抛物线C1:y=a(x-1)2+k1(a≠0)交x轴于点(0,0)与点A1(b1,0),抛物线C2:y2=a(x-b1)2+k2交x轴于点(0,0)与点A2(b2,0),抛物线C3:y=a(x-b2)2+k3交x轴于点(0,0)与点A3(b3,0),.按此规律抛物线Cn:y=a(x-bn-1)2+kn交x轴于点(0
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![已知抛物线C1:y=a(x-1)2+k1(a≠0)交x轴于点(0,0)与点A1(b1,0),抛物线C2:y2=a(x-b1)2+k2交x轴于点(0,0)与点A2(b2,0),抛物线C3:y=a(x-b2)2+k3交x轴于点(0,0)与点A3(b3,0),.按此规律抛物线Cn:y=a(x-bn-1)2+kn交x轴于点(0](/uploads/image/z/1144342-46-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%8A%9B%E7%89%A9%E7%BA%BFC1%3Ay%3Da%28x-1%292%2Bk1%28a%E2%89%A00%29%E4%BA%A4x%E8%BD%B4%E4%BA%8E%E7%82%B9%EF%BC%880%2C0%EF%BC%89%E4%B8%8E%E7%82%B9A1%28b1%2C0%29%2C%E6%8A%9B%E7%89%A9%E7%BA%BFC2%EF%BC%9Ay2%3Da%28x-b1%292%2Bk2%E4%BA%A4x%E8%BD%B4%E4%BA%8E%E7%82%B9%280%2C0%29%E4%B8%8E%E7%82%B9A2%28b2%2C0%29%2C%E6%8A%9B%E7%89%A9%E7%BA%BFC3%EF%BC%9Ay%3Da%28x-b2%292%2Bk3%E4%BA%A4x%E8%BD%B4%E4%BA%8E%E7%82%B9%280%2C0%29%E4%B8%8E%E7%82%B9A3%28b3%2C0%29%2C.%E6%8C%89%E6%AD%A4%E8%A7%84%E5%BE%8B%E6%8A%9B%E7%89%A9%E7%BA%BFCn%3Ay%3Da%28x-bn-1%292%2Bkn%E4%BA%A4x%E8%BD%B4%E4%BA%8E%E7%82%B9%280)
已知抛物线C1:y=a(x-1)2+k1(a≠0)交x轴于点(0,0)与点A1(b1,0),抛物线C2:y2=a(x-b1)2+k2交x轴于点(0,0)与点A2(b2,0),抛物线C3:y=a(x-b2)2+k3交x轴于点(0,0)与点A3(b3,0),.按此规律抛物线Cn:y=a(x-bn-1)2+kn交x轴于点(0
已知抛物线C1:y=a(x-1)2+k1(a≠0)交x轴于点(0,0)与点A1(b1,0),抛物线C2:y2=a(x-b1)2+k2交x轴于点(0,0)与点A2(b2,0),抛物线C3:y=a(x-b2)2+k3交x轴于点(0,0)与点A3(b3,0),.按此规律抛物线Cn:y=a(x-bn-1)2+kn交x轴于点(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a的“关于原点位似”的抛物线族.
(1)试求出b1的值.
(2)请用含n的代数式表示线段An-1An的长
(3)探究如下问题:
1.抛物线Cn:y=a(x-bn-1)2+kn的顶点纵坐标kn与a,n有何数量关系?并说明理由
2.若系数为a的“关于原点位似”的抛物线族的各抛物线的顶点坐标记为(T,S),请直接写出S与T所满足的函数关系式
已知抛物线C1:y=a(x-1)2+k1(a≠0)交x轴于点(0,0)与点A1(b1,0),抛物线C2:y2=a(x-b1)2+k2交x轴于点(0,0)与点A2(b2,0),抛物线C3:y=a(x-b2)2+k3交x轴于点(0,0)与点A3(b3,0),.按此规律抛物线Cn:y=a(x-bn-1)2+kn交x轴于点(0
由于C1的对称轴是x=1,所以b1-0=2*1,b1=2
同理
由于C2的对称轴是x=2,所以b2-0=2*2,b2=4
由于C3的对称轴是x=4,所以b3-0=2*4,b3=8
显然b(n)=2*b(n-1)=2^n
线段An-1An的长度=b(n)-b(n-1)=2^n-2^(n-1)=2*2^(n-1)- 2^(n-1)=2^(n-1)
y=a(x-b(n-1))2+k(n)由于都存在共同的解(0,0)
所以 0=a*b(n-1)^2+k(n)
k(n)=-a*b(n-1)^2=-a*[2^(n-1)]^2=-a*2^(2n-2)
T=b(n-1),S=-a*b(n-1)^2=-a*T^2
整理后得 S=-a*T^2
b1=2 2^(n-1) S=-a*T^2