# How do you graph x= - (y-2)^2 + 3?

Vertex $\left(3 , 2\right)$ and it opens to the LEFT
with intercepts at $\left(- 1 , 0\right) , \left(0 , 2 + \sqrt{3}\right) , \left(0 , 2 - \sqrt{3}\right)$

#### Explanation:

The graph is a parabola which opens to the left
$x = - {\left(y - 2\right)}^{2} + 3$. When we transform the equation to its vertex form
$\pm 4 p \left(x - h\right) = {\left(y - k\right)}^{2}$, it is noticable that there is a negative sign

$- \left(x - 3\right) = {\left(y - 2\right)}^{2}$ that is why this parabola opens to the left.
graph{x=-(y-2)^2+3[-20,20,-10,10]}

God bless...I hope the explanation is useful.