如图,已知二次函数y=ax2+bx+c的图象过(-1,0)和(0,-1)两点,则化简代数式(a?1a)2+4+(a+1a)2?4=____

2024-11-02 18:21:44
推荐回答(1个)
回答1:

把(-1,0)和(0,-1)两点代入y=ax2+bx+c中,得
a-b+c=0,c=-1,
∴b=a+c=a-1,
由图象可知,抛物线对称轴x=-

b
2a
=-
a?1
2a
>0,且a>0,
∴a-1<0,0<a<1,
(a?
1
a
)
2
+4
+
(a+
1
a
)
2
?4

=
(a+
1
a
)
2
+
(a?
1
a
)
2

=|a+
1
a
|+|a-
1
a
|,
=a+
1
a
-a+
1
a

=
2
a

故本题答案为:
2
a