Pomocy daje naj potrzebuje teog jak najszybciej tylko prosze o rozpisanie co z czego wynika

[tex]f(x) = ax^{2}+bx+c \ - \ postac \ ogolna\\\\f(x) = a(x-p)^{2}+q - postac \ kanoniczna[/tex]
1.
[tex]f(x) = 3x^{2}-12x+7\\\\a = 3, \ b = -12, \ c = 7\\\\W = (p,q)\\\\p = \frac{-b}{2a} = \frac{-(-12)}{2\cdot3} = \frac{12}{6} = 2\\\\q = f(p) = f(2) = 3\cdot2^{2}-12\cdot2+7 = 3\cdot4-24+7=12-17 = -5\\\\\boxed{W = (2,-5)} \ - \ wspolrzedne \ wierzcholka\\\\\\\boxed{f(x) = 3(x-2)^{2}-5} \ - \ postac \ kanoniczna[/tex]
3.
[tex]f(x) = \frac{1}{5}(x+5)^{2}-7 = \frac{1}{5}(x^{2}+10x+25)-7 = \frac{1}{5}x^{2}+2x+5-7\\\\\boxed{f(x) = \frac{1}{5}x^{2}+2x-2} \ - \ postac \ ogolna[/tex]
[tex]a = \frac{1}{5}, \ b =2, \ c = -2\\\\\Delta = b^{2}-4ac = 2^{2}-4\cdot\frac{1}{5}\cdot(-2) = 4+1,6 = 5,6\\\\\boxed{\Delta = 5,6}[/tex]
4.
[tex]x = 4\\\\y = 3x^{2}+bx+12\\\\x = p = \frac{-b}{2a}\\\\\frac{-b}{2\cdot3} = 4\\\\\frac{-b}{6} = 4\\\\-b = 24 \ \ /:(-1)\\\\\boxed{b = -24}[/tex]