Answer
center $(0,0)$, transverse axis $x$-axis, vertices $(-2,0),(2,0)$, foci $(-2\sqrt {5},0),(2\sqrt {5},0)$, and asymptotes $y=\pm2x$.
See graph.
Work Step by Step
1. Given $4x^2-y^2=16\Longrightarrow \frac{x^2}{4}-\frac{y^2}{16}=1$, we can find the center $(0,0)$, transverse axis $x$-axis, $a=2,b=4$, vertices $(-2,0),(2,0)$, $c=\sqrt {a^2+b^2}=\sqrt {4+16}=2\sqrt {5}$, foci $(-2\sqrt {5},0),(2\sqrt {5},0)$, and asymptotes $y=\pm\frac{b}{a}x=\pm2x$.
2. See graph.
