#1

hi. i forgot to find the verticies of the hyperbolas. i really feel stupid

(Y-3)^2 /25 - (X-2)^2 /16

(Y-3)^2 /25 - (X-2)^2 /16

#2

lol wut?

lols i said lol

lols i said lol

**WAT***Last edited by dean.guitar.man at May 27, 2008,*

#3

well, the denominators give you the axes of the hyperbola so in this case the major axis would be parallel to the y-axis and have a length of ten, the minor parallel to x with a length of 8 (I'm pretty sure). Then the vertices would be at y=3, -3 along the major axis, I believe

#4

^ what he said

#5

Why 8?

im my book its givin me 2,8 and 2,-2

im my book its givin me 2,8 and 2,-2

#6

lol wut

i remember this stuff from from like 3-4 years ago in high school and haven't seen it since.

i remember this stuff from from like 3-4 years ago in high school and haven't seen it since.

#7

I don't know, I'm not much of a math person. Sorry if I confused you, I was just goin off the top of me head.

#8

i got this one, just did this today in math. so, what you first need to do is find the center of the hyperbolas. since there's a (Y-3) and an (X-2), the center is (2,3). ya dig? next, find the square roots of your denominators, in this case, 5 and 4. the square root of the first terms denominator is the distance from the center to the vertices, so therefore, your vertices are (2, 8) and (2, -2).

that's only for this problem though. if the X term had come first, it would have been (7,3) and (-3,3)

that's only for this problem though. if the X term had come first, it would have been (7,3) and (-3,3)

#9

Maths...

*shudders uncontrollably*

*shudders uncontrollably*

#10

square root of the denominators.

#11

Wait, what is the function? and is (Y-3) meaning F(x)=3, or that the function is y-3?

If you have the function, simply use -b/2a to find the x coordinate, and then using the function find the image...

If you don't have the function, find it....

If you have the function, simply use -b/2a to find the x coordinate, and then using the function find the image...

If you don't have the function, find it....

#12

I am out of school! Weeee!

#13

The square root of the denominators will help determine the asymptotes, not the vertices's.

When i did hyperbolas, the vertex was always (0,0). According to this, the vertex is (2,3), which confuses me, because i never did that.

Now what does make sense is that you (or the book) made a mistake and meant to put a plus between the two fractions, creating an ellipse, not a hyperbola.

Call me crazy, but i think i am right.

When i did hyperbolas, the vertex was always (0,0). According to this, the vertex is (2,3), which confuses me, because i never did that.

Now what does make sense is that you (or the book) made a mistake and meant to put a plus between the two fractions, creating an ellipse, not a hyperbola.

Call me crazy, but i think i am right.