Answer:
Explanation:
I used logic and took the easy way around this as opposed to the long, drawn-out algebraic way. I noticed right off that at x = -3 and x = -1 the y values were the same. In the middle of those two x-values is -2, which is the vertex of the parabola with coordinates (-2, 4). That's the h and k in the formula I'm going to use. Then I picked a point from the table to use as my x and y in the formula I'm going to use. I chose (0, 3) because it's easy. The formula for a quadratic is

and I have everything I need to solve for a. Filling in my h, k, x, and y:
and
and
-1 = 4a so

In work/vertex form the equation for the quadratic is

In standard form it's:
