#### Answer

The required parabola is as shown below.

#### Work Step by Step

So, for the provided quadratic equation $f\left( x \right)=3{{x}^{2}}-2x-4$ , when compared with the standard form of the quadratic equation $f\left( x \right)=a{{x}^{2}}+bx+c$ , one gets:
$\begin{align}
& a=3, \\
& b=-2, \\
& c=-4
\end{align}$
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 downward 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:
$\begin{align}
& x=-\frac{b}{2a} \\
& =-\frac{-2}{2\times 3} \\
& =\frac{1}{3}
\end{align}$
And, the y-coordinate can be calculated as:
$\begin{align}
& y=3{{\left( \frac{1}{3} \right)}^{2}}-2\left( \frac{1}{3} \right)-4 \\
& =\frac{1}{3}-\frac{2}{3}-4 \\
& =-\frac{13}{3}
\end{align}$
So, the vertex is $\left( \frac{1}{3},-\frac{13}{3} \right)$.
Step 3:
Above steps lead to the parabola that opens upwards and has a vertex at $\left( \frac{1}{3},-\frac{13}{3} \right)$ and it intersects x-axis at $\left( 1+\frac{\sqrt{13}}{3} \right)$ and $\left( 1-\frac{\sqrt{13}}{3} \right)$ , and y-axis at $-4$.
Thus, the required parabola is as shown above.