All we need is to put this form in the vertex form f(x) = (ax+b)^2 + c
So we have
f (x)= 3x^2+12x+11 ....
Let's complete the square (if you aware of it)
f(x)= 3x^2+12x+11 = 3(x^2+4x)+11 = 3(x^2+4x+4-4)+11
= 3([x^2+4x+4]-4)+11 = 3[(x+2)^2-4]+11 =3(x+2)^2 - 12 +11 = 3(x+2)^2 -1
so our form would be:

Here is a parabola with vertex of (-2,-1) and with positive slope (concave up)
I hope that helps!