the vertex is at -b/(2A). if you use the foil method it will come out to -(x^2+2x+1)+4. then distribute the negative and simplify the equation (add the four) the equation comes out to be -x^2-2x+3. The vertex is at x=2/(2*-1), which is -1. plug -1 into the original equation to find y, (y=-(-1)^2-2*(-1)+3) which is y=4. so the vertex is at (-1,4). then since the parabola opens down (negative a value) and has a maximum at 4, y is always less than or equal to 4. x is all real numbers as it goes on forever. so b is your answer.

[latex]f(x) = a(x-p)^{2} +q \ \ W=(p,q)[/latex] ==================== [latex]f(x) = -(x+1)^{2} +4 \ \ W=(-1,4)[/latex] ------------------------------------------ [latex]D: x in R[/latex] -------------------------------------------- [latex]a extless 0 \ \ R: y leq 4[/latex]