Question 965700
{{{y-2=(1/2)(x+3)^2}}}

{{{y=(1/2)(x+3)^2+2}}}

 parabola has an axis of symmetry which is the line that runs down its 'center' or vertex

in your case 
{{{y=(1/2)(x+3)^2+2}}} vertex is at ({{{h}}},{{{k}}})=({{{-3}}},{{{2}}})

so, the axis of symmetry is the line {{{x = -3}}}

{{{drawing(600, 600, -10, 10, -10, 15, 
line(-3,15,-3,-10),circle(-3,2,.12),locate(-3,2,V(-3,2)),
 graph(600, 600, -10, 10, -10, 15, (1/2)(x+3)^2+2)) }}}