Answer
vertex:$(-6,1)$ and defines at opens to the right.
and Domain is: $[-6, \infty)$ and range is: $(-\infty, \infty)$
Work Step by Step
Since, we have $x=y^2-2y-5$
This equation can be written as:
$x=(y^2-2y+1)-5-1$
or, $x=(y-1)^2+(-6)$
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:$(-6,1)$ and defines at opens to the right.
and Domain is: $[-6, \infty)$ and range is: $(-\infty, \infty)$
