anonymous
  • anonymous
Graph a hyperbola with foci at (0, ±10) and vertices at (0, ±8) using graphing technology.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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hba
  • hba
@brookester6 can help you on this :P
anonymous
  • anonymous
umm yeah JAS!
anonymous
  • anonymous
wow a lot of people on one question lol party

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More answers

anonymous
  • anonymous
well, I did tag a lot of people. I really need help.
anonymous
  • anonymous
:( everyone gone.
Loser66
  • Loser66
I witnessed many people stopped by to help you, why did they go all?
anonymous
  • anonymous
they, left, It's too hard of a question
anonymous
  • anonymous
@brookester6 said just a second, and then left.
Loser66
  • Loser66
not that girl!! I think because you called too many people. so, this one thinks that the other's duty. hahahaa
anonymous
  • anonymous
:( aw man, i just need an equation
anonymous
  • anonymous
Let's start from scratch again
Loser66
  • Loser66
ok, my question is " what does graphing technology?" I always graph by hand, don't know how to use "technology"
anonymous
  • anonymous
lol using a website, but I just need an equation, I already have a website to use
Loser66
  • Loser66
so, you can solve it by yourself? no need to stay ? right? actually, I am a super lazy guy. If you don't need help anymore, I glad to hear that. hehehe
anonymous
  • anonymous
no, i need to have the numbers put into an equation, I don't know how to do that
Loser66
  • Loser66
shoot.what is your equation? for hyperbola?
anonymous
  • anonymous
foci at (0, ±10) and vertices at (0, ±8) Make this into an equation
Loser66
  • Loser66
remember that foci (0,+-c) and vertices (0,+-a) and c^2 = a^2 + b^2( use this form to find out b) and put everything into the standard form of hyperbola \[\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\]
Loser66
  • Loser66
and draw the asymptotes by \[y= (+-)\frac{a}{b}*x\]
Loser66
  • Loser66
that's it
anonymous
  • anonymous
10^2=8^2+b^2
Loser66
  • Loser66
remember that if your foci is (0,+- c) your hyperbola has the shape like|dw:1369325453179:dw|
Loser66
  • Loser66
so b^2 = 100 - 64 = 36 so ?
anonymous
  • anonymous
100=64+b^2 36=b^2 6=b
Loser66
  • Loser66
\[\frac{y^2}{64}-\frac{x^2}{36}=1\] your equation form is a^2 and b^2 , you need a, b when you have to figure out the asymptotes only. to get the form of hyperbola, no need to get that far
Loser66
  • Loser66
got what i mean?
anonymous
  • anonymous
ok, but how do i find x or y?
Loser66
  • Loser66
why you have to know x, y?
anonymous
  • anonymous
oh wait, that's it
Loser66
  • Loser66
yeap!!!
Loser66
  • Loser66
if you have to sketch the graph, you have to sketch the asymptotes, too. but it's quite easy when you have a, b, just draw the lines out.
anonymous
  • anonymous
I got the graph
anonymous
  • anonymous
http://www3.wolframalpha.com/Calculate/MSP/MSP52421f29bdchhib18c60000436031h3b4eaa8g7?MSPStoreType=image/gif&s=37&w=200.&h=204.&cdf=RangeControl
anonymous
  • anonymous
But, I have online class
anonymous
  • anonymous
girl?
anonymous
  • anonymous
no you say girl at the end of sentence, I CONFUSE
Loser66
  • Loser66
i think you are a girl

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