## sloppycanada one year ago This little guy is annoying. http://gyazo.com/ae8382ea043cf4d43c5dfc637f165dbf (x+11.5)^2/30.25 - (y-1.5)^2/4 = 1

1. jim_thompson5910

you're trying to find the equation based on the graph?

I am @jim_thompson5910

3. jim_thompson5910

do you see where the left edge of the box is? what x coordinate is this?

-11.5

5. jim_thompson5910

I'm thinking -11

6. jim_thompson5910

since it's halfway between -12 and -10

7. jim_thompson5910

Oh yeah... okay so -11 and 1

9. jim_thompson5910

add them up and divide by 2

10. jim_thompson5910

doing that gives you what?

(x+11)^2/9 - (y-1)^2/4 = 1

12. jim_thompson5910

$\Large \frac{-11+1}{2} = ??$

13. jim_thompson5910

doing that calculation gives the x coordinate of the center

-5

15. jim_thompson5910

good

16. jim_thompson5910

now do the same for the top and bottom edges

9 - -7/2 8?

18. jim_thompson5910

Add not subtract $\Large \frac{9 + (-7)}{2} = \frac{2}{2} = 1$

19. jim_thompson5910

So the center is (-5,1)

20. 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|

(x+5)^2/9 - (y-1)^2/4 = 1

@jim_thompson5910

23. jim_thompson5910

I don't agree with the 9 and 4 down below

24. jim_thompson5910

|dw:1435721888993:dw|

25. jim_thompson5910

|dw:1435721928414:dw|

26. jim_thompson5910

|dw:1435721972999:dw|

But don't I have to square the bottom?

28. jim_thompson5910

what do you mean?

a^2 and b^2 36 and 64

30. jim_thompson5910

yeah a = 6 and b = 8 so a^2 = 36 and b^2 = 64

Final equation- (x+5)^2/36 - (y-1)^2/64 = 1

32. jim_thompson5910

yes