# How do you find the vertex of the parabola whose equation is y = x^2 - 4x + 3?

Vertex is at $\left(2 , - 1\right)$

#### Explanation:

the given
$y = {x}^{2} - 4 x + 3$
completing the square method

add and subtract 4 to form a "perfect square trinomial"

$y = {x}^{2} - 4 x + 4 - 4 + 3$

$y = \left({x}^{2} - 4 x + 4\right) - 4 + 3$ group the trinomial

$y = {\left(x - 2\right)}^{2} - 4 + 3$ write the equivalent

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

$y + 1 = {\left(x - 2\right)}^{2}$ transpose the -1 to the left side

$y - - 1 = {\left(x - 2\right)}^{2}$ rearrange the signs

following the form $4 p \left(y - k\right) = {\left(x - h\right)}^{2}$

it is clearly seen that vertex $\left(h , k\right) = \left(2 , - 1\right)$