Answer
The required value is\[\left( 2,\ -11 \right)\]
Work Step by Step
The quadratic function of the form, $f\left( x \right)=a{{x}^{2}}+bx+c$ , can be converted into its standard form as
$f\left( x \right)=a{{\left( x+\frac{b}{2a} \right)}^{2}}+c-\frac{{{b}^{2}}}{4a}$.
The quadratic function in its standard form can be written as
$f\left( x \right)=a{{\left( x-h \right)}^{2}}+k,\,\,\,\,\,\,\,\,a\ne 0$.
Here, $a,\ h,\ k$ are constants and $x$ is a variable.
The graph of $f\left( x \right)$ is a parabola which is symmetric about the line $x=h$.
Now, the coordinates of the vertex of the parabola are $\left( h,k \right)$.
Compare this with standard form to get $h=-\frac{b}{2a}$ and $k=c-\frac{{{b}^{2}}}{4a}$.
The coordinates of the vertex of the parabola can also be written as $\left( -\frac{b}{2a},f\left( -\frac{b}{2a} \right) \right)$.
Let us consider the given function $f\left( x \right)=3{{x}^{2}}-12x+1$.
Compare with the quadratic function $f\left( x \right)=a{{x}^{2}}+bx+c$ to get $a=3,\ b=-12$.
The $x$ -coordinates of vertex are
$\begin{align}
& h=-\frac{b}{2a} \\
& =-\left( \frac{-12}{6} \right) \\
& =2.
\end{align}$
Calculate $y$ -coordinate by determining $f\left( 2 \right)$:
$\begin{align}
& f\left( 2 \right)=3{{\left( 2 \right)}^{2}}-12\left( 2 \right)+1 \\
& =12-24+1 \\
& =-11
\end{align}$
Hence, the coordinates of the vertex are $\left( 2,-11 \right)$
