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

$\left(9 , - 1\right)$

Explanation:

The given equation:

$x = - {\left(y - 2\right)}^{2} - 6 y + 12$

$x = - {y}^{2} - 4 + 4 y - 6 y + 12$

$x = - {y}^{2} - 2 y + 8$

$x = - \left({y}^{2} + 2 y + 1\right) + 9$

$x = - {\left(y + 1\right)}^{2} + 9$

$x - 9 = - {\left(y + 1\right)}^{2}$

${\left(y + 1\right)}^{2} = - \left(x - 9\right)$

The above quadratic shows a horizontal parabola with arms opening in -ve x-direction: ${Y}^{2} = - 4 A X$ which has the vertex at

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

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

$\setminus \equiv \left(9 , - 1\right)$