Answer
$f(x) =\frac{1}{2}\left(x+3\right)^2-4$
Work Step by Step
Given
\begin{equation}
f(x)=a(x-h)^2+k
\end{equation}
We know that the vertex is $(h,k) = (-3,-4)$. This means that
$f(x) =a(x+3)^2-4 $.
Since the point $(1,4)$ lies on the graph, it follows that:
\begin{equation}
\begin{aligned}
a(1+3)^2-4 & = 4\\
16a&= 4+4\\
16a& = 8 \\
a & = \frac{8}{16}\\
a & = \frac{1}{2}\\
\end{aligned}
\end{equation} Hence
$$
f(x) =\frac{1}{2}\left(x+3\right)^2-4
$$
