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

Apr 20, 2015

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

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
$y = a \cdot {\left(x + b\right)}^{2} + c$.

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
$y = a \cdot {\left(x + b\right)}^{2}$.

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
$y = a \cdot {x}^{2}$.

Since the parabola $y = a \cdot {x}^{2}$ has a vertex at $\left(0 , 0\right)$, the parabola $y = a \cdot {\left(x + b\right)}^{2}$ has a vertex at $\left(- b , 0\right)$.

Then the parabola $y = a \cdot {\left(x + b\right)}^{2} + c$ has a vertex at $\left(- b , c\right)$.

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