What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x^2 – 4y^2 = 64

need it explained step by step
thank you in advance!

16x^2-4y^2=64=>
(x/2)^2-(y/4)^2=1
y=0=>
x/2=+/-1=>x=+/-2
x=0=>
y/4=+/-i=>
y=+/-4i
=>
the vertices are
(2,0) & (-2, 0)
Let (c,0) be the focus, then
c=+/-sqrt(2^2+4^2)=>
c=+/-sqrt(20)=+/-2sqrt(5)=>
the foci are
(2sqrt(5), 0) & (-2sqrt(5),0).

Let (x/2)^2-(y/4)^2=0, then
The equations of the asymptotes are
y=+/-(4/2)x=>
y=+/-2x