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Answer:
[tex]y=3x^{2} +12x+8\\y=3(x^{2} +4x+\frac{8}{3} )\\y=3[(x^{2} +4x+4)+\frac{8}{3}-4 ]\\y=3[(x+2)^{2} -\frac{4}{3}]\\y=3(x+2)^{2}-4\\So\ the\ vertex\ will\ be\ (-2,-4)\ and\ the\ axis\ of\ symmetry\ will\ be\ x=-2\\Vertex:\ (h,k)\ in\ y=a(x-h)^{2}+k\\Axis\ of\ symmetry:\ x=h[/tex]