Question

Please help me Pre cal!!

Answer 1

Wow. Those are some tough questions. Sorry, I need to check out Google for some clues.

Hang in there....

Answer 2

Okay thanks

Answer 3

Did you get anything?

Answer 4

Getting close and at the worse time my internet is my internet is very slow. Might have to reboot.

Answer 5

Sorry typing with low bandwidth is a challenge.

Answer 6

No no your fine haha

Answer 7

hi!!
the value under the y^2 term is b^2=16
the value under the x^2 term is a^2=25
so b=4
and a=5
asymptotes
y=plus or minus b/a x
y=+ or - 4/5 x

Answer 8

can you help me with the 2nd part?

Answer 9

to find the center we take the opposite sign of the squared terms so,
center = (2, -1)

Answer 10

ok

Answer 11

the foci.
c^2=a^2+b^2
c^2=81+16
c=+ or - sqrt 97

Answer 12

i dont get what the foci is ?

Answer 13

i this this explains it well... Foci of a Hyperbola. Two fixed points located inside each curve of a hyperbola that are used in the curve's formal definition. A hyperbola is defined as follows: For two given points, the foci, a hyperbola is the locus of points such that the difference between the distance to each focus is constant.

Answer 14

Thank LynFran, my system was really slow. Even slower than me trying to figure this out.

I have one question about these hyperbolas. How can you tell a horizontal one from a vertical one? Is it Y minis X and or Y - X order?

LegandaryNikki it looks like you are in good hands.

Answer 15

to tell if its horizontal or vertical hyperbola we look at the positive term ... like example
y^2/b^2- x^2/a^2=1
this is a vertical hyperbola meaning it opens up and down (y axis)...likewise

x^2/a^2 - y^2/b^2=1
this is a horizontal hyperbola meaning it opens left and right (x- axis)

Answer 16

Did we do B? the vertices ?

Answer 17

Thanks that's what I thought, except you use much better terminology.

Answer 18

??

Answer 19

for the vertices we have
A^2=81
A=9

and B^2=16
B=4
with center (2,-1)
our vertices are (2+9, -1) and ((2-9), -1)

Answer 20

so our vertices are (-7,-1) and (11,-1)

Answer 21

ohh okay

Answer 22

we just used the "A" and add that to the x term (x+ or -A,y) since the hyperbola opens to the left and right of the x axis ....if it opens up and down the y axis we add or subtract "B" from the y term (x, y + or -B)

Answer 23

the graph https://www.desmos.com/calculator/nn5od4u9fs

Answer 24

the asymptotes...

Answer 25

okay so the center is (2, -1), vertices is (-7,-1) and (11,-1) what are the foci?

Answer 26

@LynFran

Answer 27

to find the asymptotes we solve the function for x

Answer 28

Wait what about Foci???

Answer 29

sorry it was uncompleted
frm above...
c=+ or - sq.rt 97
with center (2,-1)
so foci (+ or - sq.rt 97 +2, -1)
so foci are (-7.85, -1) and (11.85, -1)

Answer 30

oh okay i was confused thanks so now onto the asymptotes

Answer 31

@LynFran

Answer 32

please fan me ,.ill send it through a message my pc keeps freaking out.. and im not getting to type out it...

Answer 33

or like here its on it https://www.desmos.com/calculator/ab0rbdhaha

Answer 34

the two lines

Answer 35

which one is asymptotes?

Answer 36

im confused

Answer 37

hello??

Answer 38

we solve for x that is
(x-2)^2/81 - (y+1)^2/16=1

(x-2)^2/81=(y+1)^1/16+1
(x-2)^2=81( y+1)^2/16+81
(x-2)=+ or - sq.rt (81(y+1)^2/16+81)
and as y approach infinity we have
x-2=+ or - sq.rt 81/16(y+1)^2
so,
x-2=+ or - 9/4(y+1)
x=+ or - 9/4(y+1)+ 2

Answer 39

so,
x=9y/4+9/4+2
x=9y/4+17/4

and
x=-9y/4-9/4+2
x=-9y/4 -1/4
and these are you asymptotes

Answer 40

am i supposed to graph the asymptotes? or can i leave those out

Answer 41

ur suppose to graph it and label it

Answer 42

how do i graph
x=9y/4+9/4+2
x=9y/4+17/4

and
x=-9y/4-9/4+2
x=-9y/4 -1/4

Answer 43

there just lines as shown here https://www.desmos.com/calculator/ab0rbdhaha

Answer 44

they never touches the hyperbola and always passes through the center

Answer 45

the center here being (2,-1)

Answer 46

any questions? :)

Answer 47

no but thank you so much

Answer 48

welcome