What is the vertex of the parabola whose equation is y = (x + 1)^2 + 3?

(-1, -3)
(-1, 3)
(1, 3)

The y-value of the vertex is positive 3, as shown by the +3 on the right hand side of the equation, and the x-value is -1, from the (x+1)^2 (remember, when the number is inside the brackets, flip the sign) The vertex would be (-1, 3) 

If you are looking for a rigorous answer (calculus), we must find the mininum point of the equation: f(x) = (x+1)^2 + 3 f
f'(x) = 2(x+1) = 2x + 2
2x + 2 = 0 
x = -1
f(1) = (-1 + 1)^2 + 3
f(1) = 0 + 3 = 3
(-1, 3)