Odpowiedź:
[tex]Zad\ \ 1\\\\a)\\\\y=-2(x-3)^2+4\\\\y=-2(x^2-6x+9)+4\\\\y=-2x^2+12x-18+4\\\\y=-2x^2+12x-14\\\\\\b)\\\\y=3(x+3)^2-8\\\\y=3(x^2+6x+9)-8\\\\y=3x^2+18x+27-8\\\\y=3x^2+18x+19\\\\\\c)\\\\y=-2(x-3)^2\\\\y=-2(x^2-6x+9)\\\\y=-2x^2+12x-18[/tex]
[tex]Zad\ \ 2\\\\a)\\\\y=-x^2-2x+3\\\\a=-1\ \ ,\ \ b=-2\ \ ,\ \ c=3\\\\\Delta=b^2-4ac\\\\\Delta=(-2)^2-4\cdot(-1)\cdot3=4+12=16\\\\\\p=\frac{-b}{2a}=\frac{-(-2)}{2\cdot(-1)}=\frac{2}{-2}=-1\\\\q=\frac{-\Delta}{4a}=\frac{-16}{4\cdot(-1)}=\frac{-16}{-4}=4\\\\\\y=a(x-p)^2+q\\\\y=-1(x-(-1))^2+4\\\\y=-(x+1)^2+4[/tex]
[tex]b)\\\\y=3x^2-6x+3\\\\a=3\ \ ,\ \ b=-6\ \ ,\ \ c=3\\\\\Delta=b^2-4ac\\\\\Delta=(-6)^2-4\cdot3\cdot3=36-36=0\\\\\\p=\frac{-b}{2a}=\frac{-(-6)}{2\cdot3}=\frac{6}{6}=1\\\\q=\frac{-\Delta}{4a}=\frac{-0}{4\cdot3}=\frac{-0}{12}=0\\\\\\y=a(x-p)^2+q\\\\y=3(x-1)^2+0\\\\y=3(x-1)^2[/tex]
[tex]c)\\\\y=-4x^2+16\\\\a=-4\ \ ,\ \ b=0\ \ ,\ \ c=16\\\\\Delta=b^2-4ac\\\\\Delta=0^2-4\cdot(-4)\cdot16=0+256=256\\\\\\p=\frac{-b}{2a}=\frac{-0}{2\cdot(-4)}=\frac{-0}{-8}=0\\\\q=\frac{-\Delta}{4a}=\frac{-256}{4\cdot(-4)}=\frac{-256}{-16}=16\\\\\\y=a(x-p)^2+q\\\\y=-4(x-0)^2+16[/tex]
