Answer
$(8,1)$
Work Step by Step
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 $x=-2y^2+4y+6$
It can be written as:
$x=-2y^2+4y+6$
or, $x=-2(y^2-2y+1)+6+2$
or, $x=-2(y-1)^2+8$
Compare the given equation with the standard form.
Thus, vertex: $(8,1)$