Answer
The vertex for the parabola is at $\left( 3,2 \right)$.
Work Step by Step
Consider the formula:
$y={{\left( x-3 \right)}^{2}}+2$
Now, simplify the above formula:
$\begin{align}
& y={{\left( x-3 \right)}^{2}}+2 \\
& ={{x}^{2}}-6x+{{3}^{2}}+2 \\
& ={{x}^{2}}-6x+11
\end{align}$
Now, consider the x-coordinate of the vertex for the quadratic formula:
$x=\frac{-b}{2a}$
Substitute the values in the above equation:
$\begin{align}
& x=\frac{-\left( -6 \right)}{2\left( 1 \right)} \\
& =3
\end{align}$
This shows that the vertex is located when x is equal to 3.
Substitute x as 3 in the given equation:
$\begin{align}
& y={{\left( 3 \right)}^{2}}-6\left( 3 \right)+11 \\
& =9-18+11 \\
& =2
\end{align}$
This shows that the vertex occurs at the coordinate $\left( 3,2 \right)$.
Thus, the vertex of the equation is $\left( 3,2 \right)$.
