# How do you write y = 3(x-2)^2 + 1 in standard form?

$y = 3 {x}^{2} - 12 x + 13$
${\left(x - 2\right)}^{2} = \left(x - 2\right) \left(x - 2\right) = {x}^{2} - 2 x - 2 x + 2 = {x}^{2} - 4 x + 4$
$3 {\left(x - 2\right)}^{2} = 3 \left({x}^{2} - 4 x + 4\right) = 3 {x}^{2} - 12 x + 12$
So $3 {\left(x - 2\right)}^{2} + 1 = 3 {x}^{2} - 12 x + 12 + 1 = 3 {x}^{2} - 12 x + 13$