Answer
Foci: $(\pm 2\sqrt{5},0)$
Vertices: $(\pm 2,0)$
Work Step by Step
Recall: The hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ has the foci $(\pm c,0)$ where $c^2=a^2+b^2$ and the vertices $(\pm a,0)$.
We have $4x^2-y^2=16$.
Rewrite the equation above:
$\frac{4x^2-y^2}{16}=\frac{16}{16}$
$\frac{x^2}{4}-\frac{y^2}{16}=1$
Then,
$a^2=4\rightarrow a=2$
$b^2=16\rightarrow b=4$
$c=\sqrt{a^2+b^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$
So, the foci are $(\pm 2\sqrt{5},0)$ and the vertices are $(\pm 2,0)$.
