Question

Are the vertices for the hyperbola (y-2)^2/16-(x+1)^2/144=1,

A. (-1, 6) and (-1, -2)
or
B. (0, 6) and (0, -2) ?

Answer 1

\(\bf \cfrac{(y-2)^2}{16}-\cfrac{(x+1)^2}{144}=1\implies \cfrac{(y-2)^2}{4^2}-\cfrac{(x+1)^2}{12^2}=1\)

the positive fraction in this case is the one with the "y" variable, thus that means the hyperbola is moving over the y-axis, or vertically
vertex is at ( -1, 2 ) and from there it goes UP "a" units, or 4 units UP and 4 units DOWN

and the vertices are at those points, \(\bf (-1, 2\pm 4)\)

Answer 2

oh okay, so the correct answer is A?

Answer 3

yes

Answer 4

thank you :)

Answer 5

yw