Use the center, vertices, and asymptotes to graph the hyperbola.
(x - 1)2 - 9(y - 2)2 = 9

- anonymous

Use the center, vertices, and asymptotes to graph the hyperbola.
(x - 1)2 - 9(y - 2)2 = 9

- chestercat

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- anonymous

##### 3 Attachments

- anonymous

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

- anonymous

is the center 0,0

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## More answers

- theEric

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

- anonymous

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

- theEric

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

- waleed_imtiaz

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 ?

- anonymous

so the center is (1,3)

- theEric

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

- anonymous

so this would make it the last picture right

- anonymous

this one

##### 1 Attachment

- theEric

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

- theEric

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

- anonymous

yes

- waleed_imtiaz

centre would be (1,2) i think so

- theEric

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

- anonymous

sometimes the pictures are just hard to go by

- anonymous

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

- theEric

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.

- anonymous

thank you

- theEric

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.

- theEric

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).

- theEric

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

- theEric

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.

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