Question 1005918

Complete the square to get to vertex form,
{{{y=x^2+kx+(k/2)^2+5-(k/2)^2}}}
{{{y=(x+k/2)^2+5-(k/2)^2}}}
The vertex lies on the axis of symmetry so when {{{x=-3}}},
{{{-3+k/2=0}}}
{{{k/2=3}}}
{{{k=6}}}