What is the x-coordinate of the vertex #y=x^2+2x+1#?

It is always helpful to know how the graph of a function #y=F(x)# is transformed if we switch to a function #y=a*F(x+b)+c#. This transformation of the graph of #y=F(x)# can be represented in three steps:
(a) stretching along Y-axis by a factor of #a# getting #y=a*F(x)#;
(b) shifting to the left by #b# getting #y=a*F(x+b)#;
(c) shifting upwards by #c# getting #y=a*F(x+b)+c#.

To find a vertex of a parabola using this methodology, it is sufficient to transform the equation into a full square form that looks like
#y=a*(x+b)^2+c#.

Then we can say that this parabola is the result of a shift upwards by #c# (if #c<0#, it's actually downward by #|c|#) of a parabola with an equation
#y=a*(x+b)^2#.

That last one is a result of shifting to the left by #b# (if #b<0#, it's actually to the right by #|b|#) of a parabola with an equation
#y=a*x^2#.

Since the parabola #y=a*x^2# has a vertex at #(0,0)#, the parabola #y=a*(x+b)^2# has a vertex at #(-b,0)#.

Then the parabola #y=a*(x+b)^2+c# has a vertex at #(-b,c)#.

Let's apply it to our case:
#y=x^2+2x+1=(x+1)^2+0#
Hence, the vertex if this parabola is at #(-1,0)# and the graph looks like this:
graph{x^2+2x+1 [-10, 10, -5, 5]}