Answer
Type: Parabola
Vertex $(0,-3)$
Focus: $\left(0,-2\frac{15}{16}\right)$
Work Step by Step
Rewrite the equation:
$4x^2=y+4$
$y+4=4x^2$
$\frac{1}{4}(y+4)=x^2$
$\frac{1}{4}(y-(-3))=(x-0)^2$
Recall: The equation $4p(y-b)=(x-a)^2$ describes a vertical parabola whose vertex is $(a,b)$ and the focus is $(a,p+b)$.
We have
$a=0$ and $b=-3$
$4p=\frac{1}{4}\to p=\frac{1}{16}$
$p+b=\frac{1}{16}+(-3)=-2\frac{15}{16}$
Then, the given equation describes a vertical parabola whose vertex is $(0,-3)$ and the focus is $\left(0,-2\frac{15}{16}\right)$.
