Odpowiedź:
[tex]1.\\\\f(x)=3x^2-2x+6\\\\a=3\ \ ,\ \ b=-2\ \ ,\ \ c=6\\\\\Delta=b^2-4ac\\\\\Delta=(-2)^2-4\cdot3\cdot6=4-72=-68\\\\\\2.\\\\y=-4x^2+3x+2\\\\\Delta=b^2-4ac\\\\\Delta=3^2-4\cdot(-4)\cdot2=9+32=41\\\\Obliczamy\ \ p\ \ i\ \ q\\\\p=\frac{-b}{2a}=\frac{-3}{2\cdot(-4)}=\frac{-3}{-8}=\frac{3}{8}\\\\q=\frac{-\Delta}{4a}=\frac{-41}{4\cdot(-4)}=\frac{-41}{-16}=2\frac{9}{16}\\\\W(p,q)\\\\W(\frac{3}{8},2\frac{9}{16})[/tex]
