What is the the vertex of #x =-(y-2)^2-6y+12 #?

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Answer 1 · 2018-07-06T00:26:30

#(9, -1)#

The given equation:

#x=-(y-2)^2-6y+12#

#x=-y^2-4+4y-6y+12#

#x=-y^2-2y+8#

#x=-(y^2+2y+1)+9#

#x=-(y+1)^2+9#

#x-9=-(y+1)^2#

#(y+1)^2=-(x-9)#

The above quadratic shows a horizontal parabola with arms opening in -ve x-direction: #Y^2=-4AX# which has the vertex at

#(X=0, Y=0)\equiv (x-9=0, y+1=0)#

#\equiv (x=9, y=-1)#

#\equiv(9, -1)#