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