Odpowiedź:
[tex]a)\ \ y=-2x^2-8x-3\\\\\Delta=b^2-4ac=(-8)^2-4\cdot(-2)\cdot(-3)=64-24=40\\\\\\p=\frac{-b}{2a}=\frac{-(-8)}{2\cdot(-2)}=\frac{8}{-4}=-2\\\\\\q=\frac{-\Delta}{4a}=\frac{-40}{4\cdot(-2)}=\frac{-40}{-8}=5\\\\\\f(x)=a(x-p)^2+q\\\\f(x)=-2(x-(-2))^2+5\\\\f(x)=-2(x+2)^2+5[/tex]
[tex]b)\ \ y=10x^2-7x+1\\\\\Delta=b^2-4ac=(-7)^2-4\cdot10\cdot1=49-40=9\\\\\\p=\frac{-b}{2a}=\frac{-(-7)}{2\cdot10}=\frac{7}{20}\\\\\\q=\frac{-\Delta}{4a}=\frac{-9}{4\cdot10}=-\frac{9}{40}\\\\\\f(x)=a(x-p)^2+q\\\\f(x)=10(x-\frac{7}{20})^2+(-\frac{9}{40})\\\\f(x)=10(x-\frac{7}{20})^2-\frac{9}{40}[/tex]
[tex]c)\ \ y=2x^2+5x\\\\\Delta=b^2-4ac=5^2-4\cdot2\cdot0=25-0=25\\\\\\p=\frac{-b}{2a}=\frac{-5}{2\cdot2}=-\frac{5}{4}\\\\\\q=\frac{-\Delta}{4a}=\frac{-25}{4\cdot2}=-\frac{25}{8}\\\\\\f(x)=a(x-p)^2+q\\\\f(x)=2(x-(-\frac{5}{4}))^2+(-\frac{25}{8})\\\\f(x)=2(x+\frac{5}{4})^2-\frac{25}{8}[/tex]
[tex]d)\ \ y=x^2+4x-5\\\\\Delta=b^2-4ac=4^2-4\cdot1\cdot(-5)=16+20=36\\\\\\p=\frac{-b}{2a}=\frac{-4}{2\cdot1}=\frac{-4}{2}=-2\\\\\\q=\frac{-\Delta}{4a}=\frac{-36}{4\cdot1}=\frac{-36}{4}=-9\\\\\\f(x)=a(x-p)^2+q\\\\f(x)=1(x-(-2))^2+(-9)\\\\f(x)=(x+2)^2-9[/tex]
[tex]e)\ \ -3x^2+x-\frac{1}{2}\\\\\Delta=b^2-4ac=1^2-4\cdot(-3)\cdot(-\frac{1}{2})=1+12\cdot(-\frac{1}{2})=1-6=-5\\\\\\p=\frac{-b}{2a}=\frac{-1}{2\cdot(-3)}=\frac{-1}{-6}=\frac{1}{6}\\\\\\q=\frac{-\Delta}{4a}=\frac{-(-5)}{4\cdot(-3)}=\frac{5}{-12}=-\frac{5}{12}\\\\\\f(x)=a(x-p)^2+q\\\\f(x)=-3(x-\frac{1}{6})^2+(-\frac{5}{12})\\\\f(x)=-3(x-\frac{1}{6})^2-\frac{5}{12}[/tex]
