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
F(x) = 3(x-4)^2-38 because finding the perfect square would get you f(x)+38=3(x^2-8x+16) and then finding the squareroot of that and moving the constant on the left back to the right would leave u with f(x)=3(x-4)^2-38
