f'(x)=dy/dx=3x^2-3; when 3x^2-3=0 there is a turning-point (the gradient is horizontal). So x^2=1 and x=1 and -1.
Second derivative f"(x)=6x. At x=1 f"(1)>0 (minimum) and at x=-1 f"(-1)<0 (maximum).
When x=1 y=1-3+2=0; when x=-1 y=-1+3+2=4.
The coords for min are (1,0) and max (-1,4).
There is no point of inflection.
When x=0, y=2, the y intercept; when y=0, x^3-3x+2=0=(x-1)^2(x+2), so x intercepts are 1 and -2.
By inspection it's clear that x=1 is a solution because 1-3+2=0 as we saw before. Use synthetic division to find the other factor:
1 | 1 0 -3...2
.....1 1 1..-2
.....1 1 -2 | 0 = x^2+x-2=(x-1)(x+2). This gives us the above x intercepts.