smc

As $f(x)=a{x}^2+bx+c$, we have $f(1)=a\cdot 1^2+b\cdot 1+c=a+b+c$. To compute $f(1)$, set $x=-2$ in the first formula. We get $f(1)=f(-2+3)=3(-2{)}^2+7(-2)+4=12-14+4=\boxed{2}$.