Answer
See below.
Work Step by Step
The vertex of $x=a(y-k)^2+h$ is at $(h,k)$. The vertex of $y=a(x-k)^2+h$ is at $(k,h)$.
a) $x$ is squared, thus the parabola is vertical
b) the coefficient of $x^2$ is negative, thus the parabola opens downwards
c) the vertex is at $(-1,4)$.