Where should I draw the asymptotes of $(x+2)^2/4-(y+1)^2/16=1$ ?

Question

2887 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-11T21:43:52

Since the positive term includes the variable x, this hyperbola opens up along the x-axis. The equation for the asymptotes is $y=+-b/a(x-h)+k$

$a=sqrt4 = 2$
$b=sqrt16=4$
$h=-2$ , $k=-1$

$y=+-b/a(x-h)+k$

$y=+-4/2(x+2)-1$

enter image source here

hope that helped

Where should I draw the asymptotes of (x+2)^2/4-(y+1)^2/16=1(x+2)^2/4-(y+1)^2/16=1?