What are the vertex, axis of symmetry, maximum or minimum value, and range of y = 3x^2 + 6x − 1?

The vertex is when x=-b/2a, which is also the line of symmetry. 
in this case, b=6, a=3, so x=-6/6=-1
when x=-1. y=3-6-1=-4

another way to do it is to write the equation into vertex form by making a square:
3x²+6x-1
3(x²+2x-1/3)
3(x²+2x+1-1-1/3))
3(x²+2x+1-4/3))
notice the bold part makes a square:
3(x+1)²-4

Either way, 
the vertex is (-1, -4)
the axis of symmetry is x=-1
the coefficient of x² is 3, a positive number, so this parabola opens upward. the function has a minimum value of y=-4
the range is all real number larger than -4

Ive seen this before! the answer is b