Question 208602
y=x^2 +2x -6
.
(y+6) = x^2 +2x,,,,,,looking  for  form  (y-k) = (x-h)^2,,,,with  vertex(h,k)
.
therefore  complete  the  square, x^2 +2x + 1,,,,,(2/2)^2=1
.
add  +1  to  both  sides  of  eqn
.
y+6+1 = x^2 +2x +1
.
(y+7) = (x+1)^2,,,,,,,,therefore  vertex  at  (-1,-7)
.
positive  coefficient  of  x^2 ,,,therefore  opening  up,,,y  parabola
.
vertex  is  at  a  min,,,as  pointing  up.
.
line  of  symmetry  is  at  ,,,,x=-1
.
Just  check  a  few  points,,(0,-6)  and  (-2,-6),,,,ok