#### Answer

The required parabola is as shown below.

#### Work Step by Step

So, for the provided quadratic equation $f\left( x \right)=6-4x+{{x}^{2}}$ , when compared with the standard form of the quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$ , one gets:
$\begin{align}
& a=1, \\
& b=-4, \\
& c=6
\end{align}$
And use the steps shown below to determine the graph of the quadratic equation.
Step 1: Determine how the parabola opens:
Note that a, the coefficient of ${{x}^{2}}$ , is -1. If $a>0$ , the parabola opens in the upward direction. Also, if $\left| a \right|$ is small, the parabola opens more flatly than if $\left| a \right|$ is large.
Now, from the provided equation of the function, it is observed that the graph opens upwards as $a>0$.
Step 2: Calculate the vertex:
The x-coordinate can be calculated as:
$\begin{align}
& x=-\frac{b}{2a} \\
& =-\frac{-4}{2\times 1} \\
& =2
\end{align}$
And the y-coordinate can be calculated as:
$\begin{align}
& y\left( 2 \right)=6-4\left( 2 \right)+{{\left( 2 \right)}^{2}} \\
& =6-8+4 \\
& =2
\end{align}$
So, the vertex is $\left( 2,2 \right)$.
Step 3:
Above steps lead to the parabola that opens downwards and has a vertex at $\left( 2,2 \right)$ and y-axis at $6$.
So, the required parabola is as shown above.