[tex]Zadanie\ 1\\\\f(x)=a(x-p)^2+q-postac\ kanoniczna\ funkcji\\\\p=\frac{-b}{2a}\\\\q=\frac{-\Delta}{4a}\\\\\Delta=b^2-4ac\\\\f(x)=2x^2+3x-4\\\\a=2\ ;\ \ b=3\ ;\ \ c=-4\\\\\Delta=3^2-4\cdot2\cdot(-4)=9+32=41\\\\p=\frac{-3}{2\cdot2}=-\frac{3}{4}\\\\ q=\frac{-41}{4\cdot2}=-\frac{41}{8}=-5\frac{1}{8}\\\\\\f(x)=2(x+\frac{3}{4})^2- 5\frac{1}{8}[/tex]
