Question

第二个小题，请详细说明过程谢谢

Answer 1

5、说明：^2——表示平方
(2) y=a(x-m)^2-a(x-m)
=a[(x-m)-1/2]^2-a/4
=a[x-(m+1/2)]^2-a/4
C(m+1/2,-a/4)
y=0

a(x-m)^2-a(x-m)=0
a(x-m)(x-m-1)=0
x=m
或者x=m+1
A(m,0)、B(m+1,0)
|AB|=(m+1)-m=1
x=0时
y=a(0-m)^2-a(0-m)
=am^2+am
D(0,am^2+am)
① SΔABC=1/2|AB||-a/4|
=1/2×1×|a|/4
=|a|/8
SΔABC=1
|a|/8=1
|a|=8
a=±8
② SΔABD=1/2|AB||am^2+am|
=1/2×1×|am^2+am|
=|a||m^2+m|/2
SΔABC=SΔABD
|a|/8=|a||m^2+m|/2
|m^2+m|=1/4
m^2+m=±1/4
m^2+m+1/4=±1/4+1/4

(m+1/2)^2=±1/4+1/4
(m+1/2)^2=-1/4+1/4
m+1/2=0
m=-1/2
或者(m+1/2)^2=1/4+1/4
m+1/2=±√2/2
m=(-1±√2)/2