Question

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

Answer 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]}