The equation of a curve is y=(x+3)^2 +5, what are the co-ordinates of the curve's turning point?

Differentiate y with respect to x:dy/dx = 2(x+3) = 2x+6When the above equation is equal to 0, this is where the turning point of the curve is.2x+6 = 02x = -6x = -3Therefore, at x = -3, the curve has a turning point. To find the y co-ordinate, substitute -3 into the original equation and solve for y:y=(-3+3)^2+5=5Therefore, the co-ordinates of the turning point are (-3,5)