1.
[tex]x_1 = 8, \ \ x_2 = -2\\\\ZW =(-\infty;5> \ \ \rightarrow \ \ q = 5\\\\p = \frac{x_1+x_2}{2} = \frac{8-2}{2} = \frac{6}{2} = 3\\\\W = (3,5)\\\\f(x) = a(x-p)^{2}+q \ - \ postac \ kanoniczna\\\\f(x) = a(x-3)^{2}+5\\\\f(x) = a(8-3)^{2}{+5}\\\\0 = a(8-3)^{2} +5\\\\0 = 5^{2}a + 5\\\\25a + 5 = 0\\\\25a = -5 \ \ /:25\\\\a = -\frac{1}{5}\\\\f(x) = -\frac{1}{5}(x-3)^{2}+5 \ - \ postac \ kanoniczna[/tex]
2.
[tex]x_1 = 4, \ \ x_2 = -6\\\\ZW = (-\infty; 8> \ \ \rightarrow \ \ q = 8\\\\p = \frac{x_1+x_2}{2} = \frac{4-6}{2} = \frac{-2}{2} = -1\\\\f(x) = a(x+1)^{2}+8\\\\0 = a(4+1)^{2}+8\\\\5^{2}a + 8 = 0\\\\25a = -8 \ \ /:25\\\\a = -\frac{8}{25}\\\\f(x) = -\frac{8}{25}(x+1)^{2}+8 \ - \ postac \ kanoniczna[/tex]
