[tex]\underline{W = (p,q)}[/tex]
[tex]a)\\y = a(x-p)^{2}+q \ - \ postac \ kanoniczna\\\\y = 3(x-2)^{2} \ \ \rightarrow \ \ p = 2, \ q = 0\\\\\boxed{W = (2,0)}[/tex]
[tex]b)\\ y = \frac{1}{2}x^{2}\\\\a = \frac{1}{2}, \ b = 0, \ c = 0\\\\p = \frac{-b}{2a} = \frac{0}{2\cdot\frac{1}{2}} = 0\\\\q = f(p) = f(0) = \frac{1}{2}\cdot0^{2} = 0\\\\\boxed{W = (0,0)}[/tex]
[tex]c)\\y = -5x^{2}-5\\\\a = -5, \ b = 0, \ c = -5\\\\p = \frac{-b}{2a} = \frac{0}{2\cdot(-5)} = 0\\\\q = f(p) = f(0) = -5\cdot0^{2}-5 = -5\\\\\boxed{W = (0,-5)}[/tex]
[tex]d)\\y = (x-p)^{2}+q \ - \ postac \ kanoniczna\\\\y = (x+4)^{2}+8 \ \ \rightarrow \ \ p = -4, \ q = 8\\\\\boxed{W = (-4,8)}[/tex]
