Odpowiedź:
[tex]Zad.1\\\\a)\\\\2x^2+12=0\ \ |:2\\\\x^2+6=0\\\\x^2=-6\\\\R\'ownanie\ \ nie\ \ ma\ \ rozwiazania\\\\\\b)\\\\x^2-3x+2=0\\\\x^2-x-2x+2=0\\\\x(x-1)-2(x-1)=0\\\\(x-1)(x-2)=0\\\\x-1=0\ \ \ \ \vee\ \ \ \ x-2=0\\\\x=1\ \ \ \ \ \ \ \ \vee\ \ \ \ x=2\\\\\\c)\\\\x^2+64=0\\\\x^2=-64\\\\R\'ownanie\ \ nie\ \ ma\ \ rozwiazania[/tex]
[tex]Zad.2\\\\a)\\\\y=-x^2+3x-2\\\\p=\frac{-b}{2a}=\frac{-3}{2\cdot(-1)}=\frac{-3}{-2}=\frac{3}{2}\\\\q=f(p)=f(\frac{3}{2})=-(\frac{3}{2})^2+3\cdot\frac{3}{2}-2=-\frac{9}{4}+\frac{9}{2}-2=-\frac{9}{4}+\frac{18}{4}-2=\frac{9}{4}-\frac{8}{4}=\frac{1}{4}\\\\W(\frac{3}{2},\frac{1}{4})\\\\\\b)\\\\y=-x^2+3x-2\\\\\Delta=b^2-4ac=3^2-4\cdot(-1)\cdot(-2)=9-8=1\\\\p=\frac{-b}{2a}=\frac{-3}{2\cdot(-1)}=\frac{-3}{-2}=\frac{3}{2}\\\\q=\frac{-\Delta}{4a}=\frac{-1}{4\cdot(-1)}=\frac{-1}{-4}=\frac{1}{4}[/tex]
[tex]f(x)=a(x-p)^2+q\\\\f(x)=-(x-\frac{3}{2})^2+\frac{1}{4}[/tex]
