Answer
$(4,-1)$
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 $x=3y^2+6y+7$
It can be written as:
$x=3(y^2+2y)+7$
or, $x=3(y^2+2y+1)+7-3$
or, $x=3(y-(-1))^2+4$
Compare the given equation with the standard form.
Thus, vertex: $(4,-1)$
