#### Answer

The equation of the required parabola is\[f\left( x \right)=\frac{1}{2}{{\left( x+3 \right)}^{2}}-4\].

#### Work Step by Step

Putting in the value of the vertex, that is, $\left( -3,-4 \right)$ in the general form of parabola, we get:
$\begin{align}
& f\left( x \right)=a{{\left( x-\left( -3 \right) \right)}^{2}}+\left( -4 \right) \\
& f\left( x \right)=a{{\left( x+3 \right)}^{2}}-4 \\
\end{align}$
Since the graph of $f\left( x \right)$ passes through the point $\left( 1,4 \right)$ , therefore the point $\left( 1,4 \right)$ satisfies the equation of $f\left( x \right)$.
Thus,
$\begin{align}
& 4=a{{\left( 1+3 \right)}^{2}}-4 \\
& =a{{\left( 4 \right)}^{2}}-4 \\
& =16a-4
\end{align}$
Simplifying,
$\begin{align}
& 16a=8 \\
& a=\frac{1}{2}
\end{align}$
Putting in the value of a in $f\left( x \right)=a{{\left( x+3 \right)}^{2}}-4$.
Therefore, the equation of the parabola is $f\left( x \right)=\frac{1}{2}{{\left( x+3 \right)}^{2}}-4$.