D.
1)
f(x) =-2 (x-\frac{2}{3})(x+1) postać iloczynowa
2)
f(x)=(-2x+\frac{4}{3})(x+1)\\\\ f(x)=-2x^2-2x+\frac{4}{3}x+\frac{4}{3}=-2x^2-\frac{6x}{3}+\frac{4x}{3}+\frac{4}{3}
f(x)=-2x^2-\frac{2}{3}x+\frac{4}{3} postać ogólna
a=-2, b= -2/3 , c=4/3
3)
\Delta=(-\frac{2}{3})^2-4\cdot (-2) \cdot \frac{4}{3}=\frac{4}{9}+\frac{32}{3}=\frac{4+96}{9}=\frac{100}{9}\\\\ p=\frac{-(-\frac{2}{3})}{2\cdot (-2)}=\frac{-\not2^1}{3}\cdot \frac{1}{\not4^2}=-\frac{1}{6}\\\\ q=\frac{-\frac{100}{9}}{4\cdot (-2)}=\frac{\not100^{25}}{9}\cdot \frac{1}{\not8^2}=\frac{25}{18}\\\\ f(x)=-2[x-(-\frac{1}{6}]^2+\frac{25}{18}
f(x)=-2(x+\frac{1}{6})^2+\frac{25}{18} postać kanoniczna
E.
1)
f (x)=x^2-2x+3
a=1 , b=-2, c=3
\Delta=(-2)^2-4\cdot 1 \cdot 3=4-12=-8
2)
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 postać kanoniczna
3)
\Delta<0 brak pierwiastków
Brak postaci iloczynowej
F.
f (x)=-2x^2-8x-6 postać ogólna
a=-2, b=-8, c=-6
2)
\Delta=(-8)^2-4\cdot (-2) \cdot (-6)=64-48=16
p=\frac{-(-8)}{2\cdot (-2)}=\frac{8}{-4}=-2 \\\\ q=\frac{-16}{4\cdot (-2)}=\frac{-16}{-8}=2
f(x)=-2[x-(-2)]^2+2
f(x)=-2(x+2)^2+2 postać kanoniczna
3)
f(x)=-2x^2-8x-6\\\\ f(x)=-2(x^2+4x+3)\\\\ f(x)=-2(x^2+x+3x+3)\\\\ f(x)=-2[x(x+1)+3(x+1)]
f(x)=-2(x+1)(x+3) postać iloczynowa