# How do you write the vertex form equation of the parabola y=x^2-2x-5?

$y = {\left(x - 1\right)}^{2} - 6$
$y = {x}^{2} - 2 x - 5$
$y = {\left(x - 1\right)}^{2} - {\left(- 1\right)}^{2} - 5$
$y = {\left(x - 1\right)}^{2} - 6$
since a coeficient of ${\left(x - 1\right)}^{2}$ is positive value, it has minimum at $- 6$ and it axis of simetry at $x = 1$.