Question

初三2次函数

已知在直角坐标系内有△ABC，∠C=90°，且点A(1,0)、B(6,0)，点C在x轴下方，求以C为顶点，且过A、B两点的二次函数解析式。

Answer 1

let y = ax^2+bx+c
A(1,0) => 0 = a+b+c (1)
B(6,0) => 0 = 36a+6b +c (2)
(2) - (1)
35a+5b =0
b = -7a (3)
from (1)
a+b+c =0
a-7a+c =0
c = 6a

ie y = ax^2-7ax+6a
y' = 2ax -7a = 0
x = 7/2
f(7/2) = 49a/4 - 49a/2 + 6a
= -25a/4
ie C( 7/2, -25a/4)

slope of AC = (-25a/4)/ (7/2 - 1) = -5a/2
slope of CB = (-25a/4)/(7/2- 6 ) = 5a/2
∠C=90° => slope of AC x slope of CB = -1
=> -25a^2/4 = -1
25a^2 = 4
a = 2/5 or -2/5 ( rejected)
C( 7/2, -5/2)

a = 2/5 , => b = -14/5 , c= 12/5

y = (2/5)x^2 - (14/5)x + 12/5

Answer 2

由题设，C是等腰直角三角形的直角顶点，可得C(7/2,-5/2)
可设此二次函数式为 f(x)=a(x-1)(x-6) 代入点C的坐标，-5/2=a(7/2-1)(7/2-6)
可得 a=2/5. 故所求的二次函数式为 f(x)=(2/5)(x^2-7x+6).