Answer
\[\left( 3,\ -1 \right)\]
Work Step by Step
We know that parabolas always have a lowest point. This point, where the parabola changes direction, is called the `vertex'. If the quadratic is written in the form $y=a{{\left( x-h \right)}^{2}}+k,$ then the vertex is the point $\left( h,\ k \right)$.
For a given $y=a{{x}^{2}}+bx+c\,,$ the vertex $\left( h,\ k \right)$ is found by computing $h=\frac{-b}{2a},$ and then equation $y$ at $h$ to find $k.$ If you have already learned the quadratic formula, you may find it for $k,$ since it is related to both the formula for $h$ and the discriminant in the quadratic formula: $k=\frac{\left( 4ac-{{b}^{2}} \right)}{4a}.$
By equating the coefficients $a,$ $b,$ and $c,$ that is
$a=1\,\,\,\,\,\,\,\,b=-6\,\,\,\,\,\,\,\,c=8$
The formula for the vertex gives
$h=\frac{-b}{2a}=\frac{-\left( -6 \right)}{2\left( 1 \right)}=3$
Then we can find $k$ by equating $y$ at $h=3$
$\begin{align}
& k=\frac{\left( 4ac-{{b}^{2}} \right)}{4a} \\
& =\frac{\left( 4\times 1\times 8 \right)-{{\left( -6 \right)}^{2}}}{4\times 1} \\
& =-1 \\
\end{align}$
Hence, the vertex of the parabola is $\left( 3,-1 \right).$
