Write the equations in graphing form, then state the vertex of the parabola or the center and radius of the circle.
1. f (x)= 3x^2+12x+11

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!

Re-do the question and enter the picture