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

Jul 7, 2018

$\text{vertex } = \left(- 1 , 2\right)$

#### Explanation:

$\text{since the equation has a "y^2" term this indicates a}$
$\text{horizontal opening parabola with equation}$

${\left(y - k\right)}^{2} = 4 a \left(x - h\right)$

$\text{where "(h,k)" are the coordinates of the vertex}$

$\text{to obtain this form "color(blue)"complete the square}$

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

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

$\text{with "h=-1" and } k = 2$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(- 1 , 2\right)$
graph{x=y^2-4y+3 [-10, 10, -5, 5]}