Question 951048
From symmetry,
{{{m-5=5-4}}}
{{{m=6}}}
.
.
{{{f(x)=a(x-5)^2+b}}}
{{{f(4)=a(4-5)^2+b=0}}}
{{{a+b=0}}}
{{{f(6)=a(6-5)^2+b=0}}}
{{{a+b=0}}}
{{{b=-a}}}
Since it has a minimum then {{{a>0}}}
So,
{{{f(x)=a(x-5)^2-a}}}
where {{{a}}} is a non-zero constant.