How do you write y+1=-2x^2-xy+1=2x2x in the vertex form?

1 Answer
Nov 3, 2014

y+1=-2x^2-xy+1=2x2x

by factoring out -22,

=> y+1=-2(x^2+1/2x)y+1=2(x2+12x)

by adding and subtracting 1/16116 in the parentheses on the right,
(Note: (1/2 divide 2)^2=1/16(12÷2)2=116)

=> y+1=-2(x^2+1/2+1/16-1/16)y+1=2(x2+12+116116)

by distributing -22 to -1/16116,

y+1=-2(x^2+1/2x+1/16)+1/8y+1=2(x2+12x+116)+18

since x^2+1/2x+1/16=(x+1/4)^2x2+12x+116=(x+14)2,

y+1=-2(x+1/4)^2+1/8y+1=2(x+14)2+18

by subtracting 11,

y=-2(x+1/4)^2-7/8y=2(x+14)278,

which is in vertex form.


I hope that this was helpful.