有三位学生分别说出了某二次函数图象的一些特点：甲：对称轴直线x=4甲:对称轴是直线x=4乙:与x轴交

有三位学生分别说出了某二次函数图象的一些特点：甲：对称轴直线x=4甲:对称轴是直线x=4乙:与x轴交

显然,此二次函数与x轴的交点A, B(A在B左侧)关于x = 4对称. 设A与对称轴的距离为b (b>0, 且是整数), 则A(4 -b, 0), B(4 + b, 0).于是二次函数可表达为y = a[x - (4-b)][x - (4 + b)]x = 0, 与y轴交点C的纵坐标 c = a(16-b²)AB = (4+b) - (4-b) = 2b三角形ABC的面积S = (1/2)AB*|C的纵坐标| = (1/2)*2b|a(16 -b²)| = b|a(16 -b²)| = 3因为b为正整数,现在分别讨论：(1) b = 1|a| = 1/5. a = ±1/5y = ±(x-3)(x-5)/5(2) b = 2|a| = 1/8. a = ±1/8y = ±(x-2)(x-6)/8(3) b = 3|a| = 1/7. a = ±1/7y = ±(x-1)(x-7)/7(4) b = 4此时A,C重合,无解(5) b = 5|a| = 1/15. a = ±1/15y = ±(x+1)(x-9)/15依此类推,有无穷多解

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