Question

已知二次函数y1=ax^2+bx+c(a≠0)的图像经过三点(1,0)(-3,0),(0,-3/2)(1)求二次函数解析式

(2)若反比例函数y2=2/x（x>0）的图像与二次函数y1=ax^2+bx+c(a=/0)的图像在第一象限交于点A（x0,y0）,x0落在两个相邻的正整数之间，请你观察图像，写出这两个相邻的正整数（3）若反比例函数y2=k/x(x>0,k>0)与二次函数y1=ax^2+bx+c(a=/0)的图像在第一象限交于点A,点A的横坐标x0，满足2<x0<3,求实数k的取值范围

Answer 1

(1) 二次函数过(1, 0), (-3, 0), 可以表达为y = a(x -1)(x + 3)
x = 0, y = -3a = -3/2, a = 1/2
y = (x - 1)(x + 3)/2　

(2)
由图可知，1 < x0 < 2

(3)
x0 = 2, y1 = (x0 - 1)(x0 + 3)/2 = 5/2，k = y2*x0 = y1*x0 = (5/2)*2 = 5

x0 = 3, y1 = (x0 - 1)(x0 + 3)/2 = 6, k = y2*x0 = y1*x0 = 6*3 = 18

即 5 < k < 18

Answer 2

检举|今天 09:31唐卫公| 十四级 (1) 二次函数过(1, 0), (-3, 0), 可以表达为y = a(x -1)(x + 3)
x = 0, y = -3a = -3/2, a = 1/2
y = (x - 1)(x + 3)/2　

(2)
由图可知，1 < x0 < 2

(3)
x0 = 2, y1 = (x0 - 1)(x0 + 3)/2 = 5/2，k = y2*x0 = y1*x0 = (5/2)*2 = 5

x0 = 3, y1 = (x0 - 1)(x0 + 3)/2 = 6, k = y2*x0 = y1*x0 = 6*3 = 18

即 5 < k < 18