(1)y=x2+x(-1≤x≤1);

(2)y=2/x(-2≤x≤1,且x≠0).

解(1)用描点法可以作出所求函数的图象如图所示.

　　由图可知y=x2+x(-1≤x≤1)的值域为["-" 1/4 "," 2].

　　(2)用描点法可以作出函数的图象如图所示.

　　由图可知y=2/x(-2≤x≤1,且x≠0)的值域为(-∞,-1 ∪[2,+∞).

8.已知f(x)为二次函数,其图象的顶点坐标为(1,3),且过原点,求f(x)的解析:式.

解(方法一)由于函数图象的顶点坐标为(1,3),

　　则设f(x)=a(x-1)2+3(a≠0).

　　∵函数图象过原点(0,0),∴a+3=0,∴a=-3.

　　故f(x)=-3(x-1)2+3.

　　(方法二)设f(x)=ax2+bx+c(a≠0),

依题意得{■("-" b/2a=1"," @(4ac"-" b^2)/4a=3"," @c=0"," )┤即{■(b="-" 2a"," @b^2="-" 12a"," @c=0"." )┤