Odpowiedź:
[tex]a)\\\\g(x)=-2x^2-4x\\\\a=-2\ \ ,\ \ b=-4\ \ ,\ \ c=0\\\\\\Obliczamy\ \ delte\\\\\Delta=b^2-4ac\\\\\Delta=(-4)^2-4\cdot(-2)\cdot0=16+8\cdot0=16+0=16\\\\\\Obliczamy\ \ p\ \ i\ \ q\\\\p=\frac{-b}{2a}=\frac{-(-4)}{2\cdot(-2)}=\frac{4}{-4}=-1\\\\q=\frac{-\Delta}{4a}=\frac{-16}{4\cdot(-2)}=\frac{-16}{-8}=\frac{16}{8}=2\\\\\\Zapisujemy\ \ posta\'c\ \ kanoniczna\\\\g(x)=a(x-p)^2+q\\\\g(x)=-2(x-(-1))^2+2\\\\g(x)=-2(x+1)^2+2[/tex]
[tex]b)\\\\y=-3(x+2)^2+5\\\\y=-3(x^2+4x+4)+5\\\\y=-3x^2-12x-12+5\\\\y=-3x^2-12x-7\\\\\\Wykorzystano\ \ wz\'or\ \ skr\'oconego\ \ mno\.zenia\\\\(a+b)^2=a^2+2ab+b^2[/tex]
