Odpowiedź:
[tex]y = - 3 {x}^{2} + 2x + 1 \\ ∆ = {2}^{2} - 4 \times 1 \times ( - 3) = 4 + 12 = 16 \\ \sqrt{∆} = \sqrt{16} = 4 \\ x_{1} = \frac{ - 2 - 4}{ - 6} = \frac{ - 6}{ - 6} = 1 \\ x_{2} = \frac{ - 2 + 4}{ - 6} = \frac{2}{ - 6} = - \frac{1}{3} [/tex]
postać iloczynową
[tex]y = - 3(x - 1)(x + \frac{1}{3} )[/tex]
W(p,q)
[tex]p = - \frac{b}{2a} = - \frac{2}{ - 6} = \frac{2}{6} = \frac{1}{3} \\ q = - \frac{16}{ - 12} = \frac{16}{12} = \frac{4}{3} [/tex]
postać kanoniczna
[tex]y = - 3(x - \frac{1}{3} {)}^{2} - 4[/tex]
