Answer
The standard form is \[f\left( x \right)=3{{\left( x-11 \right)}^{2}}\]
Work Step by Step
We know that a quadratic function can be expressed in the standard form as $f\left( x \right)=a{{\left( x-h \right)}^{2}}+k$ corresponding to which the graph is a parabola whose vertex is the point $\left( h,k \right)$. Since, the parabola attains its minimum value of 0 so it must be opening upwards. We have $a>0$ , therefore, $f\left( h \right)=k$ is the minimum. Therefore, $k=0$.
Since the parabola attains its maximum at 11, thus $h=11$
Because, the parabola is upwards so the shape of the parabola is $f\left( x \right)=3{{x}^{2}}$.
Now, put the obtained results in the standard form to get:
$\begin{align}
& f\left( x \right)=3{{\left( x-h \right)}^{2}}+k \\
& =3{{\left( x-11 \right)}^{2}}+0 \\
& =3{{\left( x-11 \right)}^{2}}
\end{align}$
Hence, the parabola is expressed in standard form as $f\left( x \right)=3{{\left( x-11 \right)}^{2}}$.
