Question

已知函数y等于m的平方加m的和乘x的m的平方减2m加2当m等于多少时，函数是二次函数

Answer 1

解：函察早氏数y = (m2 +m)xm^2 – 2m+2 是二次函数，所以m2 + m≠0 和 m2 – 2m + 2 = 2 => m≠0，m≠-1，败散m2 – 2m = 0 => m = 2 。睁嫌

Answer 2

解：函察早氏数y = (m2 +m)xm^2 – 2m+2 是二次函数，所以m2 + m≠0 和 m2 – 2m + 2 = 2 => m≠0，m≠-1，败散m2 – 2m = 0 => m = 2 。睁嫌