[tex]f(x)=-2(x+1)(x-5)\\f(x)=-2(x^2-5x+x-5)\\f(x)=-2(x^2-4x-5)\\f(x)=-2x^2+8x+10\\\\\Delta=64+4*2*10=64+80=144\\\\p=\frac{-8}{-4}=2\\q=\frac{-144}{-8}=18\\\\\text{Postac kanoniczna: }f(x)=-2(x-2)^2+18\\\text{Os symetrii paraboli: }x=2\\\text{Wykres w zalaczniku}[/tex]