y=x^2+px+q经过两点M(m,m),N(n,n)（m不等于n）：求m+n＝1－p,mn＝q；若点M是抛物线的顶点,

1)将(m,m)(n,n)代入,
m^2+pm+q=m,(1)
n^2+pn+q=n,(2)
两式相减,
(m^2+pm+q)-(n^2+pn+q)=m-n,(1)
m^2-n^2+p(m-n)=m-n,
(m+n)(m-n)+p(m-n)=m-n
因为m,n不相等,
所以m+n+p=1,
即m+n＝1－p,
(1)+(2),
m^2+pm+q+n^2+pn+q=m+n
m^2+n^2+p(m+n)+2q=m+n
(m+n)^2-2mn+p(m+n)+2q=m+n
(1-p)^2-2mn+p(1-p)+2q=1-p
1-2p+p^2-2mn+p-p^2+2q=1-p
所以mn=q
2)n=3代入到m+n=1-p,得,
m+3=1-p,
又点M是抛物线的顶点,所以x=-b/2a=-p/2=m,
解得,
m=2,p=-4,
x=2,p=-4代入抛物线,
2^2-8+q=2,
q=6
所以p=-4,q=6