f(x) = a(x - x₁)(x - x₂) - postać iloczynowa funkcji kwadratowej
[tex]a) \ f(x) = -\frac{1}{2}x^{2}-3,5x = -\frac{1}{2}x^{2}-\frac{35}{10}x = -\frac{1}{2}x^{2}-\frac{7}{2}x\\\\\boxed{f(x) = -\frac{1}{2}x(x+7)} \ - \ postac \ iloczynowa[/tex]
[tex]b) \ f(x) = x^{2}+x+1\\\\a = 1, \ b = 1, \ c = 1\\\\\Delta = b^{2}-4ac = 1^{2}-4\cdot1\cdot1 = 1-4 = -3 < 0 \ - \ brak \ postaci \ iloczynowej[/tex]
[tex]c) \ f(x) = -x^{2}+6x-9 = -(x^{2}-6x+9) = -(x-3)^{2}\\\\\boxed{f(x) = -(x-3)(x-3)} \ - \ postac \ iloczynowa[/tex]
[tex]d) \ f(x) = x^{2}+2x-15\\\\a = 1, \ b = 2, \ c = -15\\\\\Delta = b^{2}-4ac = 2^{2}-4\cdot1\cdot(-15) = 4+60 = 64\\\\\sqrt{\Delta} = \sqrt{64} = 8\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-2-8}{2\cdot1} = \frac{-10}{2} = -5\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-2+8}{2} = \frac{6}{2} = 3\\\\f(x) = (x - (-5))(x-3)\\\\\boxed{f(x) = (x+5)(x-3)} \ - \ postac \ iloczynowa[/tex]
