[Geometry]
Problem: The parabola with equation $y=-2 x^{2}+4 x+c$ has vertex $V(1,18)$. The parabola intersects the $y$-axis at $D$ and the $x$-axis at $E$ and $F$. Determine the area of $\triangle D E F$.

Solution: Since $V(1,18)$ is on the parabola, then $18=-2\left(1^{2}\right)+4(1)+c$ and so $c=18+2-4=16$.

Thus, the equation of the parabola is $y=-2 x^{2}+4 x+16$.

The $y$-intercept occurs when $x=0$, and so $y=16$. Thus, $D$ has coordinates $(0,16)$.

The $x$-intercepts occur when $y=0$. Here,

$$
\begin{array}{r}
-2 x^{2}+4 x+16=0 \\
-2\left(x^{2}-2 x-8\right)=0 \\
-2(x-4)(x+2)=0
\end{array}
$$



and so $x=4$ and $x=-2$.

This means that $E$ and $F$, in some order, have coordinates $(4,0)$ and $(-2,0)$.

Therefore, $\triangle D E F$ has base $E F$ of length $4-(-2)=6$ and height 16 (vertical distance from the $x$-axis to the point $D$ ).

Finally, the area of $\triangle D E F$ is $\frac{1}{2} \cdot 6 \cdot 16=48$.

Final answer: 48