calculate first the 1rst and 2nd derivative:
The conditions of critical points are: f'(x) = 0 and f''(x) ≠ 0
f(0) = 0 and f''(0) = 0 You've got a point of inflection with a horizontal gradient. Such a point is called in German (literally translated) a "terrace point".
f(3/2) = -27/16 and f''(3/2) = 9 > 0 that means the function has a minimum. From this point the function is continuously increasing.
The condition for the existence of points of inflection is
From 3) you get x = 0 or x = 1
The 2 points are:
The slope in I_1 is f'(0) = 0 therefore the tangent is the x-axis.
The slope in I_2 is f'(1) = -2. Use the point-slope-formula to get the equation of the tangent:
The graph is concave upward if f''(x) > 0 and the graph is concave downward if f''(x) < 0:
I've marked the concave downward part of the graph in red.