#### Answer

$(2,-1)$

#### Work Step by Step

Completing the square of the right side of the given equation, $
y=x^2-4x+3
,$ results to
\begin{array}{l}\require{cancel}
y=(x^2-4x)+3
\\\\
y=\left( x^2-4x+\left( \dfrac{-4}{2} \right)^2 \right)+3-\left( \dfrac{-4}{2} \right)^2
\\\\
y=\left( x^2-4x+\left( -2 \right)^2 \right)+3-\left( -2 \right)^2
\\\\
y=\left( x^2-4x+4 \right)+3-4
\\\\
y=\left( x-2 \right)^2-1
.\end{array}
The vertex of the parabola whose quadratic equation is in the form $y=a(x-h)^2+k$ is $(h,k)$. Hence, the vertex of the equation above is
\begin{array}{l}\require{cancel}
(2,-1)
.\end{array}