Given that the graph shows tha the functión at x = 0 is below the y-axis, the constant term of the function has to be negative. This leaves us two possibilities:
y = 8x^2 + 2x - 5 and y = 2x^2 + 8x - 5
To try to discard one of them, let us use the vertex, which is at x = -2.
With y = 8x^2 + 2x - 5, you get y = 8(-2)^2 + 2(-2) - 5 = 32 - 4 - 5 = 23 , which is not the y-coordinate of the vertex of the curve of the graph.
Test the other equation, y = 2x^2 + 8x - 5 = 2(-2)^2 + 8(-2) - 5 = 8 - 16 - 5 = -13, which is exactly the y-coordinate of the function graphed.
