#### Answer

The required parabola is as shown below.

#### Work Step by Step

For the provided quadratic equation $f\left( x \right)={{x}^{2}}+6x+3$ , when compared with the standard form of the quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$ , we get:
$\begin{align}
& a=1, \\
& b=6, \\
& c=3
\end{align}$
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 upwards and if $a<0$ then the parabola opens downwards. 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:
$\begin{align}
& x=-\frac{b}{2a} \\
& =-\frac{6}{2\times 1} \\
& =-3
\end{align}$
Or, the y-coordinate can be calculated as:
$\begin{align}
& y={{\left( -3 \right)}^{2}}+6\left( -3 \right)+3 \\
& =9-18+3 \\
& =-6
\end{align}$
Therefore, the vertex is $\left( -3,-6 \right)$
Step 3:
So, the above steps lead to the parabola that opens upwards and has a vertex at $\left( -3,-6 \right)$ and it intersects x-axis at $\left( -3+\sqrt{6} \right)$ and $\left( -3-\sqrt{6} \right)$ , and y-axis at 3.
Therefore, the required parabola is as shown above.