Chapter 10 - Section 10.4 - The Parabola; Identifying Conic Sections - Exercise Set - Page 799: 53

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: $x=-y^2+4y+1\\x=-(y^2-4y+4)+4+1\\x=-(y-2)^2+5$ a) $x$ 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 $(5,2)$.