Opps similiar to asymptote problem but different, vertex....

[maggie0160]

Opps, I just posted one similiar to this now this one is asking for:

A vertex for the hyperbola x^2/4 - y^2/16 = 1 is

The numbers are kinda the same and it is the same situation my book gives a few examples, when I do them I get: (0,2) and the other time I got (4,0). Now I may be switching numbers around or something. Not sure can someone tell me what I may be doing wrong so I know which formula to use from now on

 

[tkhunny]

A vertex must actually be ON the graph.

This is an hyperbola that intersects the x-axis twice and does not intersect the y-axis. (0,2) can't be a vertex. The vertices are on the x-axis.

Try y = 0 and see where that leads you.