How do you write #x^2+42=12x-3y# into vertex form?

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Answer 1 · 2018-06-22T16:34:51

#y=-1/3(x-6)^2-2#

Convert it to the form #y=a(x-b)^2+c#.

#x^2+42=12x-3y#

Move all terms containing #y# to the LHS.

#-3y=x^2-12x+42#

Divide both sides by #-3#.

#y=-1/3(x^2-12x+42)#

Because #(x+p)^2=x^2+2px+p^2#, the following rule can be used:
#(x+1/2p)^2-p^2=x^2+px#

This gives us:

#y=-1/3((x-6)^2-36+42)#

Simplify to #y=a(x-b)^2+c#.

#y=-1/3((x-6)^2+6)#

#y=-1/3(x-6)^2-2#