Chapter 10 - Section 10.4 - The Parabola; Identifying Conic Sections - Exercise Set - Page 799: 16

$(-1,-3)$

Standard form of a horizontal parabola is $x=a(y-k)^2+h$ Here, vertex:$(h,k)$ and $4a$ is latus rectum. Since we are given $x=y^2+6y+8$ It can be written as: $x=(y^2+6y+9)+8-9$ or, $x=(y+3)^2-1$ or, $x=(y-(-3))^2+(-1)$ Compare the given equation with the standard form. Thus, vertex: $(-1,-3)$