# Hyperbola

• Aug 17th 2008, 12:45 PM
zooanimal98
Hyperbola
(y+1)^2/4 - (x-2)^2/16 = 1

so i know how to graph it....

i know the center is (-1,2)

Am I right the to say the a-value is 2 or is it +and- 2
and the b-value is 4 or is it +and- 4

and finally how do you find the domain and range?

is the range 0 less than or equal to y greater than or equal 4
and the domain all real?
• Aug 17th 2008, 12:57 PM
mr fantastic
Quote:

Originally Posted by zooanimal98
(y+1)^2/4 - (x-2)^2/16 = 1

so i know how to graph it....

i know the center is (-1,2)

Am I right the to say the a-value is 2 or is it +and- 2
and the b-value is 4 or is it +and- 4

and finally how do you find the domain and range?

is the range 0 less than or equal to y greater than or equal 4
and the domain all real?

a = 4, b = 2. If you draw the graph (which you say you can do) the domain and range will be obvious.
• Aug 17th 2008, 01:08 PM
zooanimal98
Quote:

Originally Posted by mr fantastic
a = 4, b = 2. If you draw the graph (which you say you can do) the domain and range will be obvious.

obviously it is not obvious for me. i am staring at the graph in front of me... i looks like to wide u's, one facing up, one facing down...what I don't see is the domain and range...

do you know how to find it?
• Aug 17th 2008, 04:02 PM
mr fantastic
Quote:

Originally Posted by zooanimal98
obviously it is not obvious for me. i am staring at the graph in front of me... i looks like to wide u's, one facing up, one facing down...what I don't see is the domain and range...

do you know how to find it?

The domain is all real numbers.

The range is $\displaystyle \{ y : ~ - \infty < y \leq y_1 \} \cup \{ y : ~ y_2 \leq y < \infty \}$ where $\displaystyle y_1$ and $\displaystyle y_2$ are the y-coordinates of the maximum and minimum turning points respectively.

Do you know how to find the coordinates of the turning points (which are key features that should be labelled on your graph)?
• Aug 18th 2008, 07:03 AM
zooanimal98
Range
are the coordinates (-1,4) and (-1,0)?

so then range is (- infty < y less than or equal to 4)
( 0 < y less than or equal to infty)

ok is that wrong...

Quote:

Originally Posted by mr fantastic
The domain is all real numbers.

The range is $\displaystyle \{ y : ~ - \infty < y \leq y_1 \} \cup \{ y : ~ y_2 \leq y < \infty \}$ where $\displaystyle y_1$ and $\displaystyle y_2$ are the y-coordinates of the maximum and minimum turning points respectively.

Do you know how to find the coordinates of the turning points (which are key features that should be labelled on your graph)?

• Aug 18th 2008, 01:10 PM
mr fantastic
Quote:

Originally Posted by zooanimal98
are the coordinates (-1,4) and (-1,0)?

so then range is (- infty < y less than or equal to 4)
( 0 < y less than or equal to infty)

ok is that wrong...

No.

(2, 1) and (2, -3).

Draw $\displaystyle \frac{y^2}{4} - \frac{x^2}{16} = 1$ and apply the necessary translations.

And by the way, the centre is at (2, -1)

Quote:

so i know how to graph it....
Not until you can correctly find the vertices and centre you do ....
• Aug 20th 2008, 06:43 AM
zooanimal98
Quote:

Originally Posted by mr fantastic
No.

(2, 1) and (2, -3).

Draw $\displaystyle \frac{y^2}{4} - \frac{x^2}{16} = 1$ and apply the necessary translations.

And by the way, the centre is at (2, -1)

Not until you can correctly find the vertices and centre you do ....

so sorry that I accidentaly switched y for x when I was graphing. Thanks for the help, btw you are not fantastic.