algebra2sucks Group Title What are the points of discontinuity? Are they all removable? y = (x-3)/x^2 - 12x + 27 one year ago one year ago

1. myininaya Group Title

When is the bottom 0?

2. myininaya Group Title

For what values of x make the bottom 0?

3. algebra2sucks Group Title

not sure?

4. myininaya Group Title

So you should be able to solve an equation in this form ax^2+bx+c=0 by now. I'm guessing this since this is a type of problem that occurs after learning that.

5. myininaya Group Title

Do you know how to factor?

6. myininaya Group Title

7. myininaya Group Title

You can even do completing the square.

8. algebra2sucks Group Title

so x^2+12x+27=0?

9. algebra2sucks Group Title

how do i do that?

10. myininaya Group Title

Do you have -12x or +12x?

11. algebra2sucks Group Title

its postive 12x

12. myininaya Group Title

Hmmm...So the problem above is written/typed wrongly I assume. Okay You should know how to factor the polynomial x^2+12x+27

13. myininaya Group Title

Find two factors of c that add up to be b (you can do this when the coefficient of x^2 is 1 which in this problem it is)

14. myininaya Group Title

In other words find two factors of 27 that add up to be 12.

15. algebra2sucks Group Title

its like this |dw:1370027288868:dw|

16. myininaya Group Title

err so it is -12x?

17. algebra2sucks Group Title

noo x^2 minus 12x

18. myininaya Group Title

I'm really confused what problem we are doing because twice now you have written +12x

19. myininaya Group Title

You have also said twice that we have -12x

20. myininaya Group Title

Anyways, we take the bottom polynomial whatever it may be and set that equal to zero and solve for x. I will let you do it since I don't know what problem we are doing anymore.

21. algebra2sucks Group Title

sorry

22. algebra2sucks Group Title

this is the problem were doing

23. myininaya Group Title

Set bottom polynomial equal to zero. $x^2-12x+27=0$ You will notice this is in the form $ax^2+bx+c=0$ You have learned if you cannot factor this easily you could also use completing the square or quadratic formula. The easy way to solve this quadratic is by factoring in my opinion. Since the coefficient of x squared is 1, all we have to really do is find two factors of c that add up to be b; called them m and n. So The factored form would be (x-m)(x-n)=0 Then you set both factors equal to 0 like so x-m=0 or x-n=0 Solve both for x like so x=m or x=n

24. algebra2sucks Group Title

so which numbers do we use for x-m and x-n?

25. myininaya Group Title

Have you found factors of 27 that have product 27 and sum -12?

26. myininaya Group Title

Those are your m and n.

27. myininaya Group Title

Recall this part that I said: Since the coefficient of x squared is 1, all we have to really do is find two factors of c that add up to be b; called them m and n.

28. myininaya Group Title

c is 27 and b is -12

29. myininaya Group Title

|dw:1370027980807:dw|

30. myininaya Group Title

See how I factored 27?

31. myininaya Group Title

1(27)=27 3(9)=27 -1(-27)=27 -3(-9)=27

32. algebra2sucks Group Title

Yes I see on my answes I have x= -9, x =-3 is that right?

33. myininaya Group Title

Do any of those factors above that multiplied to be 27 also add up to be -12?

34. algebra2sucks Group Title

are these numbers removable?

35. algebra2sucks Group Title

and -3?

36. myininaya Group Title

(x-3)(x-9)=0 Set both equal to 0.

37. myininaya Group Title

I put (x-m)(x-n)=0 where I meant (x+m)(x+n)=0. So you would have x+m=0 or x+n=0 Which means you would have x=-m or x=-n. But anyways we have (x-3)(x-9)=0 And you are to set both factors equal to 0 and solve for x for both equations

38. myininaya Group Title

x-3=0 or x-9=0 I leaving solving these linear equations to you.

39. algebra2sucks Group Title

do we also use -1 or no?

40. myininaya Group Title

Where does that come from?

41. myininaya Group Title

Have you solve x-3=0 or x-9=0 yet?

42. algebra2sucks Group Title

no idk how

43. myininaya Group Title

x-3=0 You can't solve this for x? You don't know how to get x by itself?

44. myininaya Group Title

To undo subtraction by three you add three.

45. myininaya Group Title

Try adding three to both sides and tell me what the result is.

46. myininaya Group Title

x-3=0 add 3 to both sides +3 +3 --------- what is the result?

47. myininaya Group Title

x-3+3=0+3

48. myininaya Group Title

what is -3+3? what is 0+3?

49. algebra2sucks Group Title

0 and 3

50. myininaya Group Title

yes so x-3+3=0+3 gives us x+0=3 Or just x=3

51. myininaya Group Title

Now you solve x-9=0

52. myininaya Group Title

What are you going to add to both sides?

53. algebra2sucks Group Title

9?

54. myininaya Group Title

yes so x=?

55. algebra2sucks Group Title

9 and 0?

56. algebra2sucks Group Title

so x=9

57. myininaya Group Title

yes

58. myininaya Group Title

Discontinuous at x=3 or x=9

59. myininaya Group Title

Now go back to the problem where you had everything factored.

60. myininaya Group Title

$\frac{x-3}{(x-3)(x-9)}$

61. myininaya Group Title

This was the problem where everything is factored, correct?

62. myininaya Group Title

Does anything cancel? If so that is where you have your removable discontinuity.

63. myininaya Group Title

x-3 is just a number so is x-9 If we had 5/(5*9) you would say the 5's cancel right? Like this 1/9 Does anything cancel? Is there a factor on top that you see on bottom?

64. myininaya Group Title

$\frac{(x-3)}{(x-3)(x-9)}$ If you want you can put parenthesis around the x-3 on top, Do you see any factors that match on top and bottom?

65. myininaya Group Title

This is just like matching socks.

66. myininaya Group Title

I have to go to the grocery store. Good luck.

67. algebra2sucks Group Title

Thanks for all ur help