I'm not sure what method you are supposed to use to graph this. I would do it the following way although you can also change it to vertex form of the equation and graph it using (h,k) and "a".
The way I'd do it..
vertex (-b/2a , y)
-b/2a = -2
plug in this value for x to find the y coordinate.
-4 +8 -3 = 1
vertex (-2, 1)
the negative in front of the x^2 flips the parabola upside down
So you graph an upside down parabola at (-2,1). It's symmetrical, input easy to do points if needed like x=0 , x=-4
