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

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Answer 1 · 2014-11-03T05:33:46

#y+1=-2x^2-x#

by factoring out #-2#,

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

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

#=> y+1=-2(x^2+1/2+1/16-1/16)#

by distributing #-2# to #-1/16#,

#y+1=-2(x^2+1/2x+1/16)+1/8#

since #x^2+1/2x+1/16=(x+1/4)^2#,

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

by subtracting #1#,

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

which is in vertex form.

I hope that this was helpful.