Odpowiedź:
[tex]\boxed{y = 4(x+\frac{1}{4})(x-5)} \ - \ postac \ iloczynowa[/tex]
Szczegółowe wyjaśnienie:
Postać iloczynowa funkcji kwadratowej:
[tex]y = a(x - x_1)(x - x_2)[/tex]
[tex]y = 4x^{2}-19x - 5\\\\a = 4, \ b = -19, \ c = -5\\\\\Delta = b^{2}-4ac = (-19)^{2}-4\cdot4\cdot(-5) = 361 + 80 = 441\\\\\sqrt{\Delta} = \sqrt{441} = 21\\\\x_1 = \frac{-b-\sqrt{\Delta}}{2a} = \frac{-(-19)-21}{2\cdot4} = \frac{19-21}{8} = \frac{-2}{8}=-\frac{1}{4}\\\\x_2 = \frac{-b+\sqrt{\Delta}}{2a} = \frac{-(-19)+21}{8} = \frac{19+21}{8} = \frac{40}{8} = 5[/tex]
