Question

?

Answer 1

vertices are on the y-axis so I'm thinking it's oriented along the y-axis for part I) [not entirely sure but it will become clear as we finish the calculations]

Answer 2

center is the midpoint of the vertices

Answer 3

so maybe i should just leave part i blank for now until we finish

Answer 4

sure

for II) center is midpoint of the vertices

Answer 5

(0,0)

Answer 6

awesome

for part III) we use the right column (since our hyperbola is y-oriented)

Answer 7

our asymptotes have slope 3/4 and -3/4 so a = 3 and b = 4

Answer 8

now for part IV) just need to plug in a and b into the hyperbola equation on the right hand column, given that our h and k values are 0 and 0 (since it's centered at origin)

Answer 9

THIS ONE?

Answer 10

yes

Answer 11

\[\frac{ (y-0)^2 }{ 3^2 }-\frac{ (x-0)^2 }{ 4^2}=1\]

Answer 12

good, just make sure to write things in simplest terms (so don't leave those 0's there, and simplify those exponents

Answer 13

would this be the simplest form

Answer 14

don't leave those 0's there (x-0 and y-0 are just x and y

Answer 15

so the equation as it isz but the numerator is just x and y

Answer 16

x^2 and y^2

Answer 17

okay so how it is x^2 and y^2

Answer 18

how it is with just x^2 and y^2

Answer 19

the square exponents don't just disappear

Answer 20

good

Answer 21

for part I I'm not sure how they want you to phrase it, you could say it "opens up and down along the y-axis" something like that