Wierzchołek paraboli ma współrzędne....

[tex]wierzcholek\ paraboli:\ \ W=(p,q)=(-2,4)\\\\p=\frac{-b}{2a},\ \ q=\frac{-\Delta }{4a}\\\\\Delta =b^2-4ac\\\\y=ax^2+bx+2\\\\x\neq 0,\ \ c=2\\\\\left \{ {{ \frac{-b}{2a}}=-2 \atop { \frac{-(b^2-4a*2)}{4a}}}=4 \right.[/tex]
[tex]\left \{ {{ -b=-4a \atop { -(b^2-4a*2) =4*4a} \right.\\\\\left \{ {{ b= 4a \atop { -( (4a)^2-8a) = 16a} \right.\\\\ \left \{ {{ b= 4a \atop {-(16a^2-8a)=16a } \right. \\\\\left \{ {{ b= 4a \atop {- 16a^2+8a-16a=0 } \right. \\\\\left \{ {{ b= 4a \atop { -16a^2a-8a=0 } \right. \\\\\left \{ {{ b= 4a \atop { 8a(-2a-1)=0 } \right. \ \ wiemy\ ,ze\ a\neq 0\ ,\ wiec \\\\\left \{ {{ b= 4a \atop { -2a-1 =0 } \right. \\\\\left \{ {{ b= 4a \atop { -2a=1 } \right. \ \[/tex]
[tex]\left \{ {{ b= 4*(-\frac{1}{2}) \atop { a=-\frac{1}{2} } \right. \\\\\left \{ {{ b= -2 \atop { a=-\frac{1}{2} } \right.\\\\y=-\frac{1}{2}x^2-2x+2[/tex]