Chapter 10 - Section 10.4 - The Parabola; Identifying Conic Sections - Exercise Set - Page 799: 54

See below.

The vertex of $x=a(y-k)^2+h$ is at $(h,k)$. The vertex of $y=a(x-k)^2+h$ is at $(k,h)$. I modify the original equation: $x=-y^2-6y-10\\x=-(y^2+6y+9)+9-10\\x=-(y+3)^2-1$ a) $y$ is squared, thus the parabola is horizontal b) the coefficient of $y^2$ is negative, thus the parabola opens to the left c) the vertex is at $(-1,-3)$.