Question 1029110
{{{f(3)=0}}}
{{{f(0)=9}}}
Centered about {{{x=3}}}.
{{{f(x)=a(x-3)^2+b}}}
Substituting,
{{{0=a(3-3)^2+b}}}
{{{b=0}}}
.
.
{{{9=a(0-3)^2}}}
{{{9=9a}}}
{{{a=1}}}
.
.
{{{f(x)=(x-3)^2}}}