Odpowiedź :
[tex]y=3x^2-2x+\frac12\\\\p=\frac{-(-3)}{2\cdot3}=\frac36=\frac12\\\\\Delta=(-2)^2-4\cdot3\cdot\frac12=4-6=-2\\\\q=\frac{-(-2)}{4\cdot3}=\frac2{12}=\frac16\\\\\\\underline{\ y=3\left(x-\frac12\right)^2+ \frac16\ }[/tex]
[tex]y=3x^2-2x+\frac12\\\\p=\frac{-(-3)}{2\cdot3}=\frac36=\frac12\\\\\Delta=(-2)^2-4\cdot3\cdot\frac12=4-6=-2\\\\q=\frac{-(-2)}{4\cdot3}=\frac2{12}=\frac16\\\\\\\underline{\ y=3\left(x-\frac12\right)^2+ \frac16\ }[/tex]