## Babynini one year ago Vertex.

1. Babynini

Write the equation in standard form then find and list the vertex, focus, and directrix of the parabola. sketch its graph showing the focus and the directrix.(note: Choose the locations of the coordinate axes and draw them on the grid and choose scales so that your graph is, if possible, about three by three inches in size.) x^2+12x+4y+44=0

2. Babynini

@jim_thompson5910 If you have time :)

3. jim_thompson5910

are you able to complete the square for the x terms?

4. Babynini

I'm not sure how.

5. jim_thompson5910

x^2+12x+_____ what goes in the blank to make that a perfect square?

6. Babynini

36

7. jim_thompson5910

yep

8. jim_thompson5910

so add and subtract 36 on the same side or add 36 to both sides

9. jim_thompson5910

then group (x^2-12x+36) and factor that to get (x-6)^2

10. Babynini

kks, I chose to add 36 to both sides (x-6)^2+4y+44=36

11. Babynini

(x-6)^2+4y=-8 (subtracted 44 from both sides)

12. jim_thompson5910

now isolate y

13. Babynini

y=-(x-6)^2/4 - 8/4

14. jim_thompson5910

yes

15. jim_thompson5910

you can simplify that further

16. Babynini

17. Babynini

hrm the original and that one don't graph the same o.0

18. Babynini
19. Babynini
20. Babynini

Would it work to do: 4y=-x^2-12x-44 y= -x^2/4 -3x -11 ?

21. Babynini

Or does the equation we previously made look better?

22. Babynini
23. jim_thompson5910

let me check

24. Babynini

kks

25. jim_thompson5910

oh I made a mistake, it's not -12x it's +12x

26. jim_thompson5910

so it should factor to (x+6)^2

27. jim_thompson5910

you should get $\Large y = -\frac{1}{4}(x+6)^2 - 2$

28. Babynini

Yep that's what I have :)

29. Babynini

that is still a different graph than the original

30. jim_thompson5910

I have it matching up

31. Babynini

oh yeah, sorry. It is. Wolfram was doing a close up of the graph and I hadn't noticed.

32. jim_thompson5910

you forgot the negative in http://www.wolframalpha.com/input/?i=Parabola+y%3D%28%28x-6%29%5E2%29%2F4%29+-2

33. Babynini

ah, sorry.

34. Babynini

so next is finding foci, vertex, and directrix.

35. jim_thompson5910

have a look at this http://www.mathwords.com/f/focus_parabola.htm

36. Babynini

k, so mines a vertical parabola.

37. jim_thompson5910

yes

38. Babynini

um and I use x^2=4py to find everything?

39. jim_thompson5910

Convert $\Large y = -\frac{1}{4}(x+6)^2 - 2$ to 4p(y-k) = (x-h)^2 form

40. Babynini

would that be (y+2)/4=(x+6)^2 ??

41. jim_thompson5910

more like -4(y+2) = (x+6)^2

42. jim_thompson5910

-4(y+2) = (x+6)^2 is in the form 4p(y-k) = (x-h)^2 p = -1 h = -6 k = -2

43. Babynini

ahh ok

44. jim_thompson5910

|dw:1433470391325:dw|

45. jim_thompson5910

|dw:1433470408839:dw|

46. jim_thompson5910

|dw:1433470431794:dw|

47. Babynini

f: = (-6,-3)

48. jim_thompson5910

yes

49. Babynini

d: y = - 3

50. jim_thompson5910

no

51. Babynini

k-p ?

52. Babynini

oh sorry, -1

53. jim_thompson5910

yeah directrix is y = -1

54. anonymous

hi can i borrow jim_thompson5910

55. anonymous

?

56. Babynini

umm so the graph looks like |dw:1433470684289:dw|

57. Babynini

|dw:1433470710284:dw|

58. Babynini

|dw:1433470757831:dw| ish lol

59. Babynini

the prof loves complete accuracy. so if I wanted to find the exact points that the parabola passes though do I just substitute numbers for x into the equation and find out y?

60. jim_thompson5910

yeah you can do that to plot points or use desmos to get an accurate graph desmos.com/calculator

61. jim_thompson5910
62. Babynini

well, i've got to show work. but i'll use desmos to make sure :) thanks!

63. Babynini

er it worked until I got to -7 and -8 values for x :o