Question

已知二次函数y=x^2+2x-3,当自变量x=1时，函数y的值为多少，急急急

Answer 1

二次函数y=x^2+2x-3,当自变量x=1时，y=1^2+2-3=3-3=0

Answer 2

x=1时，y=1乘以1+2乘以1-3=0

Answer 3

y=1²+2×1-3=0

Answer 4

答案应该是0

Answer 5

答案应该是0