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

Hint:
The above given problem is a very simple problem of coordinate geometry. The given question is related to the concept of parabola and for solving these types of questions we first need to understand the different forms of parabolas. There are four types of parabola and for each of them, the corresponding vertex and foci are as follows:



Complete step by step solution:
Given is $x = {y^2} - 4y + 3$ and we have to find the vertex of this parabola.
As we can clearly see that the equation has a ${y^2}$ term, which means that the given equation indicates a horizontal opening parabola with equation
${\left( {y - k} \right)^2} = 4a\left( {x - h} \right)$, where $\left( {h,k} \right)$ are the coordinates of the vertex and $a$ is a multiplier.
In order to obtain the vertex form for this given equation, we use the method of completing the square.
$
   \Rightarrow x = {\left( {y - 2} \right)^2} - 4 + 3 \\
   \Rightarrow {\left( {y - 2} \right)^2} = x + 1 \\
 $
Now, after comparing the above obtained equation with ${\left( {y - k} \right)^2} = 4a\left( {x - h} \right)$, where $\left( {h,k} \right)$ are the coordinates of the vertex, we get $h = - 1,k = 2$. So, the vertex is $\left( { - 1,2} \right)$.

Hence, the vertex of the parabola is $\left( { - 1,2} \right)$.

Note:
We can also find the vertex using the method of shifting of origin or origin transformation, but it is a very complex as well as a lengthy process. The solution shown above is the simplest one. In origin transformation, we shift the origin of the actual coordinate system, to some other arbitrary system, so as to simplify the equation and make it similar to that of the general form. However, to find the required point, we need to revert back to the original coordinate system.