Answer
vertex:$(-4,-3)$ and defines at opens to the right.
Domain is: $[-4, \infty)$ and range is: $(-\infty, \infty)$
Work Step by Step
Since, we have $x=y^2+6y+5$
This equation can be written as:
$x=(y^2+6y+9)+5-9$
or, $x=(y-(-3))^2+(-4)$
Standard form of a horizontal parabola is $x=a(y-k)^2+h$
Here, vertex:$(h,k)$
Compare the above equation with the standard form.
Thus, we have vertex:$(-4,-3)$ and defines at opens to the right.
Hence, Domain is: $[-4, \infty)$ and range is: $(-\infty, \infty)$
