Question 1001656
 x^4 + 2x^3 + 1 
derivative =0 is 4x^3+6x^2=0.  Your critical points are correct.  I would use open brackets, because at the critical point for a minimum, the graph is instantaneously not moving in the vertical direction. I would keep it open for the inflection points, too.
f''(x)=12x^2+12x, and I agree with the inflection points.
y-intercept without a calculator.
x^4+2x^3+1=0
(x+1)(x^3+x^2-x+1)=0
By synthetic division, x= -1 is a root, so (x+1) is a factor.  The other function does not have an integer root.  Minus 2 is almost a root but isn't.
The function goes to +oo for large minus x and large +x.
The y-intercept is (0,1)
There are two real roots, -1 and the other needs a calculator, but it is between -1.5 and -2 (it is -1.83)
The inflection point does occur where the graph crosses the x-axis.  
You have a function that goes to positive infinity rapidly for positive and negative x.
The y-intercept and roots are known.  
The only change I would make is that the inflection point is where the graph starts rising rapidly, rather than rising more slowly.  You know that, because the first derivative is large negative or large positive, so you have almost a vertical line when x is not very large.
The minimum is known, (-3/2,-11/16)

While you can't use a calculator, for learning purposes, it's worth seeing the graph anyway.


{{{graph(300,300,-5,5,-10,10,x^4+2x^3+1)}}}