# How do you write y=x^2-2x+1 in vertex form and identify the vertex, y intercept and x intercept?

Jun 7, 2015

$y = {x}^{2} - 2 x + 1 = {\left(x - 1\right)}^{2} + 0$

This has vertex at $x = 1$, $y = 0$, that is $\left(1 , 0\right)$.

The intercept with the $y$ axis is where $x = 0$, so substituting $x = 0$ into the formula for $y$ we get:

$y = {\left(- 1\right)}^{2} + 0 = 1$

So the $y$ intercept is at $\left(0 , 1\right)$

The intercept with the $x$ axis is at the vertex $\left(1 , 0\right)$ where the parabola touches it.

graph{x^2-2x+1 [-10, 10, -5, 5]}