inflection point, concave up and down question

Please help with the graph analysis

f(x)=x^3-3x^2+3

1.First derivative 3x^2-6x

2. intervals where f is increasing. 3x^2-6x=0, x=0, x=2 Function is increasing from -infinity to 0 and from 2 to infinity

3. decreasing (0,2)

4. local max point (0,3)

5. local min point (2,-1)

6. second derivative 6x-6

I hope everything until now is correct

7. inflection point 6x-6=0, x=1 Is it correct to say that at x=1 function changes from concave down to concave up?

8. what are the intervals where function is concave up and concave down?

Thank you very much

when the second derivative is a positive number

Please explain if I am wrong. Thank you very very much.

Inflection point

F(x) = 3x^2 + 12x + 3

Please correct me if I am wrong

1. We need to find first derivative to find minimum and maximum points

First derivative = 6x +12, 6x + 12 = 0, x = -2

2.Local minimum is (-2, -9)

3. Second derivative = 6 What does it mean? Where is the inflection point? There is no inflection point. Right? Is it correct to say that because the second derivative is a positive number, the graph is always concave up. For this particular function the graph is concave up from negative infinity to positive infinity.