Hi, I was looking at a problem from an old test and I got confused about something. Q. Find the slope of the tangent line to the given curve at a given point: y = x(x^2 + 3)^1/2 ; P(1,2) S. First I found the derivative: [(x^2 + 3)^1/2] + [(x^2)(x^2 + 3)^-1/2] Plugged in the values for x and y at the given point: [(4)^1/2] + [(4)^-1/2] The choices for answers were: a) 2 b) -5/2 c) 3/2 d) 5/2 e) none of these Square root of 4 can be + OR - 2, no? That would mean I can have 4 possible solutions: (4/2) + (1/2) = 5/2 (4/2) + (-1/2) = 3/2 (-4/2) + (1/2) = -3/2 (-4/2) + (-1/2) = -5/2 Since 3 of these are listed as choices, how did I pick the right one? (I chose d)5/2 and it was marked correct) Forgive me if this is shockingly stupid...I am trying to review all my calculus and physics to start classes up again in the winter.