Find minimum point of x2 plus 16 equals 0?
This is the long way:x2 + 16 = 0x1= -1/2 + squareroot of -15,75x2 = -1/2 - sqareroot of - 15,75Note: squareroot of 16,25 is roughly ~ 4,03112575Well, so much for the theory, but this question does not have any solution, since you can't draw the squareroot of a negative number like - 15,75.And the short way:x2 + 16 = 0x2 = -16x = squareroot of - 16, which as I declared above, is not possible.Woops, I misunderstood your question; heres the answer to your question:f(x) = x2 + 16derivatef'(x) = 2xf'(0) = 2(0) Put in x = 0f'(0) = 0 When x = 0, the curve turns and you have a minimi point.f(0) = x2 + 16 Put in x = 0 in the first equation to get y.f(0) = 02 + 16f(0) = 16 y = 16minimum point: x = 0, y = 16 --> (0;16)