Odpowiedź:
[tex]y=\frac{1}{2}x^2+\frac{3}{5}x-\frac{8}{25}\\\\a=\frac{1}{2}\ \ ,\ \ b=\frac{3}{5}\ \ ,\ \ c=-\frac{8}{25}\\\\\Delta=b^2-4ac\\\\\Delta=(\frac{3}{5})^2-\not4^2\cdot\frac{1}{\not2_{1}}\cdot(-\frac{8}{25})=\frac{9}{25}+2\cdot\frac{8}{25}=\frac{9}{25}+\frac{16}{25}=\frac{25}{25}=1[/tex]
[tex]Obliczamy\ \ warto\'sci\ \ p\ \ i\ \ q\\\\p=\dfrac{-b}{2a}=\dfrac{-\frac{3}{5}}{2\cdot\frac{1}{2}}=\dfrac{-\frac{3}{5}}{1}=-\dfrac{3}{5}\\\\\\q=\dfrac{-\Delta}{4a}=\dfrac{-1}{4\cdot\frac{1}{2}}=\dfrac{-1}{2}=-\dfrac{1}{2}\\\\\\Podstawiamy\ \ obliczone\ \ warto\'sci\ \ p\ \ i\ \ q\ \ do\ \ wzoru\ \ na\ \ posta\'c\ \ kanoniczna\\\\f(x)=a(x-p)^2+q\\\\f(x)=\frac{1}{2}(x-(-\frac{3}{5}))^2+(-\frac{1}{2})\\\\f(x)=\frac{1}{2}(x+\frac{3}{5})^2-\frac{1}{2}[/tex]
