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Postać kanoniczna funkcji kwadratowej
[tex]y=a(x-p)^2+q\\\\p=\frac{-b}{2a}\\\\q=\frac{-\Delta}{4a} \ (\Delta=b^2-4ac)[/tex]
a)
[tex]f(x)=5x^2-6x\\\\a=5, \ b=-6, \ c=0\\\\\Delta=(-6)^2-4\cdot5\cdot0=36-0=36\\\\p=\frac{-(-6)}{2\cdot5}=\frac{6}{10}=\frac{3}{5}\\\\q=\frac{-36}{4\cdot5}=\frac{-36}{20}=-1,8\\\\\huge\boxed{f(x)=5(x-\frac{3}{5})^2-1,8}[/tex]
b)
[tex]f(x)=-x^2-2x+1\\\\a=-1, \ b=-2, \ c=1\\\\\Delta=(-2)^2-4\cdot(-1)\cdot1=4+4=8\\\\p=\frac{-(-2)}{2\cdot(-1)}=\frac{2}{-2}=-1\\\\q=\frac{-8}{4\cdot(-1)}=\frac{-8}{-4}=2\\\\f(x)=-(x-(-1))^2+2\\\\\huge\boxed{f(x)=-(x+1)^2+2}[/tex]
