Odpowiedź:
y = -1/4(x - 14)² + 43
Szczegółowe wyjaśnienie:
y = a(x - p)² + q - postać kanoniczna
[tex]y = -\frac{1}{4}x^{2}+7x-6\\\\\underline{a = -\frac{1}{4}}, \ b = 7, \ c = -6\\\\p = \frac{-b}{2a} = \frac{-7}{2\cdot(-\frac{1}{4})} = \frac{-7}{-\frac{1}{2}} = -7\cdot(-2) = \underline{14}\\\\q = \frac{-\Delta}{4a} = \frac{-(b^{2}-4ac)}{4a} = \frac{-(7^{2}-4\cdot(-\frac{1}{4})\cdot(-6)}{4\cdot(-\frac{1}{4})} = \frac{-(49-6)}{-1}=49-6 =\underline{ 43}[/tex]
[tex]\boxed{y = -\frac{1}{4}(x-14)^{2}+43} \ - \ postac \ kanoniczna[/tex]