抛物线y=ax^2-2x+c的图像最低点的纵坐标是-2,它与直线y=2x的两个交点的横坐标之差的绝对值为2,求a和c.要详细过程,好的加分.

y=ax^2-2x+c
=a(x-1/a)^2+c-1/a
因此最低点纵坐标是c-1/a
=>c-1/a=-2
=》a=1/(c+2) (A)
它与y=2x的两个交点就是
ax^2-2x+c=2x的两个解
=》ax^2-4x+c=0
x1+x2=4/a
x1*x2=c/a
(x1-x2)^=(x1+x2)^2-4x1x2
=4^2/a^2-4c/a
=(16-4c)/a=2^2=4
=>4+c=a
将（A）代入
=》4-c=1/(c+2)
=>c^2+2c-1=0
=>c1=-1+根号2,a1=根号2-1
c2=-1-根号2,a2=-1-根号2