[tex]y=-\frac{1}{2} x^2 -2x + 1\\[/tex]
Postać kanoniczna funkcji kwadratowej to:
[tex]y= a(x-p) + q[/tex]
[tex]p=\frac{x_1+x_2}{2}[/tex]
Δ[tex]=(-2)^2-4*(-\frac{1}{2}) * 1= 4-2=2\\[/tex]
[tex]x_1 = \frac{(-2)^2 + \sqrt{2} }{-\frac{1}{2} * 2} = -4-\sqrt{2}[/tex]
[tex]x_2 = \frac{(-2)^2 - \sqrt{2} }{-\frac{1}{2} * 2} = -4+\sqrt{2}[/tex]
[tex]p=\frac{-4-\sqrt{2} -4+\sqrt{2}}{2}=\frac{-8}{2}=-4[/tex]
q=-Δ/4a
[tex]q=\frac{-2}{4*(-\frac{1}{2})} =\frac{-2}{-2} = 1[/tex]
[tex]y=-\frac{1}{2}(x+4) + 1[/tex]