sloppycanada
  • sloppycanada
This little guy is annoying. http://gyazo.com/ae8382ea043cf4d43c5dfc637f165dbf (x+11.5)^2/30.25 - (y-1.5)^2/4 = 1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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jim_thompson5910
  • jim_thompson5910
you're trying to find the equation based on the graph?
sloppycanada
  • sloppycanada
I am @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
do you see where the left edge of the box is? what x coordinate is this?

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sloppycanada
  • sloppycanada
-11.5
jim_thompson5910
  • jim_thompson5910
I'm thinking -11
jim_thompson5910
  • jim_thompson5910
since it's halfway between -12 and -10
jim_thompson5910
  • jim_thompson5910
how about the right edge?
sloppycanada
  • sloppycanada
Oh yeah... okay so -11 and 1
jim_thompson5910
  • jim_thompson5910
add them up and divide by 2
jim_thompson5910
  • jim_thompson5910
doing that gives you what?
sloppycanada
  • sloppycanada
(x+11)^2/9 - (y-1)^2/4 = 1
jim_thompson5910
  • jim_thompson5910
\[\Large \frac{-11+1}{2} = ??\]
jim_thompson5910
  • jim_thompson5910
doing that calculation gives the x coordinate of the center
sloppycanada
  • sloppycanada
-5
jim_thompson5910
  • jim_thompson5910
good
jim_thompson5910
  • jim_thompson5910
now do the same for the top and bottom edges
sloppycanada
  • sloppycanada
9 - -7/2 8?
jim_thompson5910
  • jim_thompson5910
Add not subtract \[\Large \frac{9 + (-7)}{2} = \frac{2}{2} = 1\]
jim_thompson5910
  • jim_thompson5910
So the center is (-5,1)
jim_thompson5910
  • jim_thompson5910
\[\Large \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\] (h,k) is the center a = half of the horizontal length of the box b = half of the vertical length of the box hyperbola looks like this |dw:1435720969381:dw|
sloppycanada
  • sloppycanada
(x+5)^2/9 - (y-1)^2/4 = 1
sloppycanada
  • sloppycanada
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
I don't agree with the 9 and 4 down below
jim_thompson5910
  • jim_thompson5910
|dw:1435721888993:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1435721928414:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1435721972999:dw|
sloppycanada
  • sloppycanada
But don't I have to square the bottom?
jim_thompson5910
  • jim_thompson5910
what do you mean?
sloppycanada
  • sloppycanada
a^2 and b^2 36 and 64
jim_thompson5910
  • jim_thompson5910
yeah a = 6 and b = 8 so a^2 = 36 and b^2 = 64
sloppycanada
  • sloppycanada
Final equation- (x+5)^2/36 - (y-1)^2/64 = 1
jim_thompson5910
  • jim_thompson5910
yes

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