Answer
$y=(x-2)^2+2$
Work Step by Step
Recall:
(1) The vertex form of $y=ax^2+bx+c$ is $y=a(x-h)^2+k$ where $(h, k)$ is the vertex.
(2) The vertex of the graph of $y=ax^2+bx+c$ is at $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$.
The given function has $a=1$ and $b=-4$.
Thus, the $x$-coordinate of the vertex is:
\begin{align*}
x&=-\frac{b}{2a}\\\\
x&=-\frac{-4}{2(1)}\\\\
x&=\frac{4}{2}\\\\
x&=2
\end{align*}
Find the $y$-coordinate of the vertex by substituting $2$ to $x$ to obtain:
\begin{align*}
y&=x^2-4x+6\\
y&=2^2-4(2)+6\\
y&=4-8+6\\
y&=2
\end{align*}
The vertex is at $(2, 2)$ so $h=2$ and $k=2$.
Substitute the values of $h$ and $k$ into the vertex form in the recall part (1) above to obtain:
$$y=(x-2)^2+2$$
