Odpowiedź:
[tex]c)\ \ y=\frac{1}{2}x^{2}+x\\\\a=\frac{1}{2}\ \ ,\ \ b=1\ \ ,\ \ c=0\\\\\Delta=b^2-4ac=1^2-4\cdot\frac{1}{2}\cdot0=1-0=1\\\\p=\frac{-b}{2a}=\frac{-1}{2\cdot\frac{1}{2}}=\frac{-1}{1}=-1\\\\q=\frac{-\Delta}{4a}=\frac{-1}{4\cdot\frac{1}{2}}=\frac{-1}{2}=-\frac{1}{2}\\\\\\y=a(x-p)^2+q\\\\y=\frac{1}{2}(x-(-1))^2+(-\frac{1}{2})\\\\y=\frac{1}{2}(x+1)^2-\frac{1}{2}[/tex]
[tex]e)\ \ y=-\frac{2}{3}x^2-2x-\frac{3}{2}\\\\a=-\frac{2}{3}\ \ ,\ \ b=-2\ \ ,\ \ c=-\frac{3}{2}\\\\\Delta=b^2-4ac\\\\\Delta=(-2)^2-4\cdot(-\frac{2}{3})\cdot(-\frac{3}{2})=4+\frac{\not8^4}{\not3_{1}}\cdot(-\frac{\not3^1}{\not2_{1}})=4-4=0\\\\\\p=\dfrac{-b}{2a}=\dfrac{-(-2)}{2\cdot(-\frac{2}{3})}=\dfrac{2}{-\frac{4}{3}}=2:(-\frac{4}{3})=-\not2^1\cdot\frac{3}{\not4_{2}}=-\frac{3}{2}\\\\\\q=\dfrac{-\Delta}{4a}=\dfrac{0}{4\cdot(-\frac{2}{3})}=\dfrac{0}{-\frac{8}{3}}=0[/tex]
[tex]y=a(x-p)^2+q\\\\y=-\frac{2}{3}(x-(-\frac{3}{2}))^2+0\\\\y=-\frac{2}{3}(x+\frac{3}{2})^2[/tex]
