Odpowiedź:
[tex]Zad.3\\\\y=\frac{1}{2}x^2+\frac{3}{5}x-\frac{8}{25}\\\\a=\frac{1}{2}\ \ ,\ \ b=\frac{3}{5}\ \ ,\ \ c=-\frac{8}{25}\\\\\Delta=b^2-4ac\\\\\Delta=(\frac{3}{5})^2-4\cdot\frac{1}{2}\cdot(-\frac{8}{25})=\frac{9}{25}-2\cdot(-\frac{8}{25})=\frac{9}{25}+\frac{16}{25}=\frac{25}{25}=1\\\\\\p=\dfrac{-b}{2a}=\dfrac{-\frac{3}{5}}{2\cdot\frac{1}{2}}=\dfrac{-\frac{3}{5} }{1}=-\dfrac{3}{5}\\\\\\q=\dfrac{-\Delta}{4a}=\dfrac{-1}{4\cdot\frac{1}{2}}=\dfrac{-1}{2}=-\dfrac{1}{2}[/tex]
[tex]y=a(x-p)^2+q\\\\y=\frac{1}{2}(x-(-\frac{3}{5}))^2+(-\frac{1}{2})\\\\y=\frac{1}{2}(x+\frac{3}{5})^2-\frac{1}{2}[/tex]
[tex]Zad.4\\\\y=-\frac{1}{4}x^2+5x+75\\\\-\frac{1}{4}x^2+5x+75=0\\\\a=-\frac{1}{4}\ \ ,\ \ b=5\ \ ,\ \ c=75\\\\\Delta=b^2-4ac\\\\\Delta=5^2-\not4\cdot(-\frac{1}{\not4})\cdot75=25+1\cdot75=25+75=100\\\\\sqrt{\Delta}=\sqrt{100}=10\\\\\\x_{1}=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-5-10}{\not2^1\cdot(-\frac{1}{\not4_{2}})}=\dfrac{-15}{-\frac{1}{2}}=15:\frac{1}{2}=15\cdot2=30[/tex]
[tex]x_{2}=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-5+10}{\not2^1\cdot(-\frac{1}{\not4_{2}})}=\dfrac{5}{-\frac{1}{2}}=-5:\frac{1}{2}=-5\cdot2=-10\\\\\\y=a(x-x_{1})(x-x_{2})\\\\y=-\frac{1}{4}(x-30)(x-(-10))\\\\y=-\frac{1}{4}(x-30)(x+10)[/tex]
