Question

use the values of a or b to locate the coordinates of the vertices for (y+3)^2/9-(x-2)^2/16=1

Answer 1

Oh you posted another one, my bad o.o

Answer 2

I see xD yeah, thought you might have posted the other problem in the same forum we were in, but oh well. You're doing the same hw as someone else in here, so I can copy and paste some of what I put in that one xD

Answer 3

ow. my bad can u go back to the other problem?

Answer 4

Well, the center would be (x-h) and (y-k). So your y-coordinate for the center would be (y-(-3)) and the x-coordinate would be (x-2). Now the negative sign is supposed to be there, so that means that the 2 there is actually positive.

when y^2 comes first (the one that is positive), your verticies are up and down a units from where your center is. So it looks like a^2 is going to be 9, meaning a is 3. As for the foci, those are just:

\[c = \sqrt{(a)^{2}+ (b)^{2}}\]
This time it is plus because the foci are going to be further out than the verticies. They go inside of the little bowls that the hyperbola makes.

As for the verticies, when y^2 is the first term (the positive one), the verticies are up and down a units from the center. the a^2 denominator is always with the positive fraction, meaning a^2 is 9 making a = 3.

For asymptotes, when y^2 is the positive term, they are:
\[y = \pm\frac{ ax }{ b }\]

this was just a quick run down of the things, but we can go in more detail of course : )

Answer 5

@Rosa34

we can go back to the closed problem and talk in there, but what I just posted here is relevant, too, lol.

Answer 6

how long are u online for?

Answer 7

Until I decide I need to sleep xD

Answer 8

and when do u usually start getting sleepy?

Answer 9

What time is it where you are?

Answer 10

its 10:33 here what about where your at? Where do you live?

Answer 11

Same, lol. I just wanted to know so I could say it in your time. I live in vegas.

Answer 12

ok in my time. Did u graduate from collage or university yet.?

Answer 13

Yep, so I live somewhere near ya ish I guess. And I have an associates. Im still trying to get a second associates in math. I'll be going to a 4 year university to work on my bachelors/minor for the first time on the 26th. So I'm working on it xD

Answer 14

That's good. So for the problem center is (2,-3) and the vertices are (2,3), (2,6), (9,-3) and (-3,-3)

Answer 15

Well from -3 you need to go up 3 and down 3. Up 3 would be 0 and down 3 would be -6. there are no real verticies using the b-values, but those WOULD be (6,-3) and (-2,-3) o.o

Answer 16

how would the graph look?

Answer 17

the dashed lines are the asymptotes.