Odpowiedź:
[tex]3.40\\\\d)\\\\f(x)=-\frac{1}{2}(x+6)(x-6)\\\\f(x)=-\frac{1}{2}(x^2-36)\\\\f(x)=-\frac{1}{2}x^2+18\\\\a=-\frac{1}{2}\ \ ,\ \ b=0\ \ ,\ \ c=18\\\\\Delta=b^2-4ac=0^2-\not4^2\cdot(-\frac{1}{\not2_{1}})\cdot18=0+2\cdot18=36\\\\\\Obliczamy\ \ warto\'s\'c\ \ p\ \ i\ \ q\\\\p=\frac{-b}{2a}=\frac{0}{2\cdot(-\frac{1}{2})}=\frac{0}{-1}=0\\\\q=\frac{-\Delta}{4a}=\frac{-36}{4\cdot(-\frac{1}{2})}=\frac{36}{-2}=-18[/tex]
[tex]Wyliczone\ \ warto\'sci\ \ p\ \ i\ \ q\ \ podstawiamy\ \ do\ \ wzoru\ \ na\ \ posta\'c\ \ kanoniczna\\\\f(x)=a(x-p)^2+q\\\\f(x)=-\frac{1}{2}(x-0)^2+(-18)\\\\f(x)=-\frac{1}{2}(x-0)^2-18[/tex]
