SOLUTION
Equation of a parabola is given by
[tex]\begin{gathered} (x-h)^2=4p(y-k) \\ where\text{ the focus is \lparen h, k + p\rparen} \\ directrix\text{ is y = k - p } \end{gathered}[/tex]
From the question, we have been given
[tex]focus\text{ at \lparen0, -4\rparen and a directrix of y = 4}[/tex]
This means that the vertex is halfway at (0, 0) and the focal length p = 4
The equation becomes
[tex]\begin{gathered} (x-h)^2=4p(y-k) \\ (x-h)^2=4py-4k \\ (x-h)^2+4k=4py \\ (x-h)^2\times\frac{1}{4}+k=py \end{gathered}[/tex]
Now, note that k and h = 0, so we have
[tex]\begin{gathered} \frac{1}{4}(x-0)^2+0=py \\ \frac{1}{4}(x^2)=py \\ y=\frac{1}{4p}(x^2) \end{gathered}[/tex]
So because this graph will be an "n" shape, so the focus will be negative, hence, we have
[tex]\begin{gathered} y=-\frac{1}{4p}(x^2) \\ y=-\frac{1}{4\times4}x^2 \\ y=-\frac{1}{16}x^2 \\ Hence\text{ } \\ f(x)=-\frac{1}{16}x^2 \end{gathered}[/tex]
Hence the answer is the third option
