Es ist

${\displaystyle {}{\begin{aligned}f(x)&=x^{2}-3x+{\frac {4}{3}}\\&={\left(x-{\frac {3}{2}}\right)}^{2}-{\frac {9}{4}}+{\frac {4}{3}}\\&={\left(x-{\frac {3}{2}}\right)}^{2}+{\frac {-27+16}{12}}\\&={\left(x-{\frac {3}{2}}\right)}^{2}-{\frac {11}{12}}.\end{aligned}}}$

${\displaystyle {}\geq 0}$

${\displaystyle {}-{\frac {11}{12}}}$