[tex]y = \frac{2}{3}x^{2}-5x+8\\\\a = \frac{2}{3}, \ b = -5, \ c = 8\\\\\underline{y = a(x-p)^{2}+q} \ - \ postac \ kanoniczna\\\\\\p = \frac{-b}{2a} = \frac{-(-5)}{2\cdot\frac{2}{3}}=\frac{5}{\frac{4}{3}}=\frac{15}{4} = 3\frac{3}{4}\\\\\\q = f(p) = f(\frac{15}{4}) = \frac{2}{3}\cdot(\frac{15}{4})^{2}-5\cdot\frac{15}{4}+8 = \frac{2}{3}\cdot\frac{225}{16}-\frac{75}{4}+8 = \frac{225}{24}-\frac{450}{8}+\frac{64}{8} =\\\\=-\frac{161}{8}=-20\frac{1}{8}[/tex]
[tex]\boxed{y = \frac{2}{3}(x-3\frac{3}{4})^{2}-20\frac{1}{8}} \ - \ postac \ kanoniczna[/tex]