What are the coordinates of the two turning points of the curve y = x^3+3x^2+3?

To find the turning points we need to find dy/dx and set it equal to zero. From there we can find the x coordinate and substitute it back into the origional equation to find the y coordinate.

First we differentiate y = x^3+3x^2+3

We have dy/dx = 3x^(3-1)+23*x^(2-1)+0 = 3x^2+6x

Now we set dy/dx = 3x^2+6x = 0

Both terms have a common factor of 3x so we can take this outside the brackets so the equation looks like dy/dx = 3x(x+2) = 0

From this we can see the equation equals 0 when x takes the values x = 0 and x = -2

We now substitute these values of x back into the origional equation y = x^3+3x^2+3

For x = 0, y = (0)^3+3(0)^2+3 = 3

For x = -2, y = (-2)^3+3(-2)^2+3 = -8+12 +3 = 7

So the coordinates of the two turning points are (-2,7) and (0,3)