Cześć!
Postać kanoniczna funkcji kwadratowej
[tex]f(x)=a(x-p)^2+q\\\\p=\frac{-b}{2a}\\\\q=\frac{-\Delta}{4a} \ (\Delta=b^2-4ac)[/tex]
Obliczenia
[tex]f(x)=\frac{1}{4}(x+2)(x-10)\\\\f(x)=\frac{1}{4}(x\cdot x-x\cdot10+2\cdot x-2\cdot10)\\\\f(x)=\frac{1}{4}(x^2-10x+2x-20)\\\\f(x)=\frac{1}{4}(x^2-8x-20)\\\\f(x)=\frac{1}{4}\cdot x^2-\frac{1}{4}\cdot8x-\frac{1}{4}\cdot20\\\\f(x)=\frac{1}{4}x^2-2x-5\\\\a=\frac{1}{4}, \ b=-2, \ c=-5\\\\\Delta=(-2)^2-4\cdot\frac{1}{4}\cdot(-5)=4+5=9\\\\p=\frac{-(-2)}{2\cdot\frac{1}{4}}=\frac{2}{\frac{1}{2}}=2\cdot2=4\\\\q=\frac{-9}{4\cdot\frac{1}{4}}=\frac{-9}{1}=-9\\\\\huge\boxed{f(x)=\frac{1}{4}(x-4)^2-9}[/tex]
