Question

How do you find the vertices of a hyperbola?

Answer 1

A hyperbola has two vertices.

Answer 2

Ah, my bad.

Answer 3

Your answer depends entirely upon the context.

Answer 4

You must know first what data you are given in order to properly answer this question.

Answer 5

Im gonna post a pic of the question

Answer 6

Good idea. I'll be right back.

Answer 7

help!!! please! :(

Answer 8

The center is \((3,2)\). It's just the two numbers in the top of each fraction. You are given that one vertex is at \((0, \text{ something})\). Let \(x=0\) and solve for \(y\) in the original equation.

Answer 9

Can you explain to me how to do this problem specifically? Please! I am confused..

Answer 10

Ok. I'll try. It will take a few minutes.

Answer 11

thank you

Answer 12

A hyperbola in the form
\[\frac{(x-x_0)^2}{a^2}-\frac{(y-y_0)^2}{b^2}=1\]
has a center at \((x_0,y_0)\).
Since we have
\[\frac{(x-3)^2}{9}-\frac{(y-2)^2}{16}=1\]
the center must be at \((3,2)\) because this equation is our original with \(x_0=3\) and \(y_0=2\). So that deals with one part.

The vertices of a hyperbola are located at \(([x_0\pm a],y_0)\). (\(\pm\) just means "plus or minus".)
In our original equation, we have that \(a^2=9\). So, \(a=\pm 3\). And, from before, we have that \(x_0=3\) and \(y_0=2\).

Therefore, our vertices are \(([3\pm \pm 3], 2)\) or \(([3\pm 3],2)\).
The expression \(3 \pm 3\) has two values: \(3+3=6\) and \(3-3=0\).
So, our two vertices are \((6,2)\) and \((0,2)\).

Answer 13

thank you so much!!!! :)

Answer 14

You're welcome.