Question

Use the center, vertices, and asymptotes to graph the hyperbola.

(x - 1)2 - 9(y - 2)2 = 9

Answer 1

I am totally stuck if someone can help me find the center I can figure out the rest

Answer 2

is the center 0,0

Answer 3

http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx
is a good link to help you!

Answer 4

I am just really confused because there is nothing on the bottom for a denominator

Answer 5

I'm not sure myself, but it looks like it's (1,2) for the center.

Answer 6

First divide the whole equation by 9...
(x - 1)^2/(9) - (y - 2)^2 =1
now U know a=3 and b=1
So focus is (+-c,o) because it is on x-axis..... Can u do now ?

Answer 7

so the center is (1,3)

Answer 8

\[\frac{9}{1}=\frac{1}{\frac{1}{9}}=\frac{1}{(\frac{1}{3})^2}\]

Answer 9

so this would make it the last picture right

Answer 10

this one

Answer 11

If the center is (1,2) that is your only option!

Answer 12

Did you check out the link?
http://tutorial.math.lamar.edu/Classes/Alg/Hyperbolas.aspx

Answer 13

yes

Answer 14

centre would be (1,2) i think so

Answer 15

Well that's two of us. I say it's a good bet.

Answer 16

sometimes the pictures are just hard to go by

Answer 17

but we know that it is not (0,0) or (-1,2) so the last one is the best

Answer 18

When you look at formulas that have something like \[(x+h)\], you often want x to be modified only by addition or subtraction. If x is multiplied or divided by anything, get it out of the parenthesis! Anything multiplied or divided by \[(x+h)\]is then something that can really be expressed as just division if you want.
By doing so, the equation you have will start to match up with the general formula for the shape of the curve.

Answer 19

thank you

Answer 20

We are sure it is (1,2) when we compare it to the general formula.

The position of the center of any shape can be found when you see how all x's and all y's are modified with addition or subtraction. This seems to "shift" graphs.

Answer 21

You're welcome! :)

Back to the "shifting", if you have y = (x), then y = (x+5) is looks to be shifted 5 up. Same with y = 9(x) seeming to shift 5 up when you look at y = 9(x+5).

Answer 22

When you look at \[(x-h)^2\], \[(x-h)\]is how you are modifying x. You are finding the difference between them with subtraction.
Then that difference is squared, so only the difference between them matters, not at all whether x>h or x<h.

Answer 23

Lastly, for your future typed-up math discussions, it helps to express "to the power of 2" as "^2". It's a very common notation used on the internet.