Postać kanoniczna
[tex]y=2x^2-11x+15\\\\a=2\ \ ,\ \ b=-11\ \ ,\ \ c=15\\\\\Delta=b^2-4ac\\\\\Delta=(-11)^2-4\cdot2\cdot15=121-120=1\\\\Obliczamy\ \ warto\'sci\ \ p\ \ i\ \ q\\\\p=\frac{-b}{2a}=\frac{-(-11)}{2\cdot2}=\frac{11}{4}\\\\q=\frac{-\Delta}{4a}=\frac{-1}{4\cdot2}=\frac{-1}{8}=-\frac{1}{8}\\\\Podstawiamy\ \ warto\'sci\ \ p\ \ i\ \ q\ \ do\ \ wzoru\ \ na\ \ posta\'c\ \ kanoniczna\\\\y=a(x-p)^2+q\\\\y=2(x-\frac{11}{4})^2+(-\frac{1}{8})\\\\y=2(x-\frac{11}{4})^2-\frac{1}{8}[/tex]
Postać iloczynowa
[tex]Aby\ \ zapisa\'c\ \ tr\'ojmian\ \ kwadratowy\ \ w\ \ postaci\ \ iloczynowej\ \ obliczamy\\\\delte\ \ i\ \ miejsca\ \ zerowe.\\\\y=2x^2-11x+15\\\\a=2\ \ ,\ \ b=-11\ \ ,\ \ c=15\\\\\Delta=b^2-4ac\\\\\Delta=(-11)^2-4\cdot2\cdot15=121-120=1\\\\\sqrt{\Delta}=\sqrt{1}=1\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-1}{2\cdot2}=\frac{11-1}{4}=\frac{10}{4}=\frac{5}{2}\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+1}{2\cdot2}=\frac{11+1}{4}=\frac{12}{4}=3[/tex]
[tex]Zapisujemy\ \ tr\'ojmian\ \ w\ \ postaci\ \ iloczynowej\\\\y=a(x-x_{1})(x-x_{2})\\\\y=2(x-\frac{5}{2})(x-3)[/tex]
