Question

函数f(x)=ax2+bx+c(a>0)的值域

Answer 1

f(x)=ax2+bx+c

=a(x^2+b/a)^2+c

=a(x+b/2a)^2+c-b^2/4a

=a(x+b/2a)^2+(4ac-b^2)/(4a)

因为a>0

所以当x=-b/2a时,f(x)有最小值(4ac-b^2)/(4a)

即f(x)值域为[(4ac-b^2)/(4a),正无穷)