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

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

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

$\equiv(9, -1)$

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