## ChrisV 3 years ago graph the fuction f(x)= 1/x+8

1. mattt9

when x goes to zero y approaches 8

2. ChrisV

yea i am trying to determine what the graph would look like

3. MasterChief

But I am thy Master Chief? http://www.wolframalpha.com/input/?i=y%3D.5x%2B8

4. mattt9

it's 1/x not (1/2)x there chief

5. ChrisV

well it 1/(x+8)

6. MasterChief

Preposterous! How could I have made such a delinquent mistake? http://www.wolframalpha.com/input/?i=y%3D%281%2Fx%29%2B8

7. ChrisV

but (x-8) is different from x+8 when graphing

8. mattt9

you know anyone can type numbers in to wolfram

9. ChrisV

(x+8) oops

10. mattt9

i'm talking to chief

11. ChrisV

yea i want to know how to graph the thing

12. MasterChief

Well of course, although I could rule the Earth if it was off limits to everyone but me...

13. ChrisV

i can go to a graphing calculator and not come here but it doesnt teach me how to graph it

14. mattt9

i disagree chief. conversation over.

15. ChrisV

So i assume that I would substitute random numbers for X to graph it?

16. MasterChief

Make a table and input numbers for x like 0, and then see what comes out as y. Keep getting numbers until you can make a decent graph.

17. ChrisV

so if x= 0 y would = 1/8

18. MasterChief

Disagree with what? Obviously, no s**t anyone can use it! I linked a page to the graph, to make things easier even though I misunderstood the question.

19. MasterChief

Umm, Well, you put 0 in for x. This cancels out the numerator of 1 above it. The answer would be 8.

20. MasterChief

Because 1 divided 0 is nothing, 0.

21. ChrisV

1/(x+8)

22. ChrisV

i cant figure out how to put the one over both in here

23. ChrisV

in the original problem there is no parenthesis

24. MasterChief

Oh, that's a different story. Yes, it would com eout as \[1/8\]

25. ChrisV

because x+8 and (x+8) would be different shifts when graphing wouldnt it?

26. MasterChief

Yes, the solution graph would differ.

27. ChrisV

one would shift the graph up and one would shift the graph left

28. ChrisV

so i am assuming there would be a hole in my graph at -8

29. ChrisV

because -8+8 = 0

30. MasterChief

Sorry, it appears so. I have not taken this but look: http://www.wolframalpha.com/input/?i=y%3D1%2F%28x%2B8%29%2C+y%3D%281%2Fx%29%2B8

31. MasterChief

I entered the two equations.

32. ChrisV

the first one is correct. Thats where I was confused

33. ChrisV

because at x=-9

34. MasterChief

Yes, it does move the graph to the right if you seclude the +8

35. ChrisV

the equation would result in a negative

36. ChrisV

hence a verticle asymtope at -8

37. ChrisV

x can never be -8 )

38. MasterChief

I would love to answer your question here but I really don't know. I'm in eighth grade and enrolled in Algebra 2. Sorry!

39. ChrisV

you did great, and you helped me answer a college calculus question

40. ChrisV

just remember that the denominator can never = 0 so if it does you have a hole

41. ChrisV

hence for 1/(-8+8)

42. ChrisV

=1/0

43. MasterChief

Oh wow, Thank you! And yes, because you can't have, for instance, 4 slices of nothing, right?

44. ChrisV

the value doesnt exist

45. MasterChief

Like, 4/0

46. ChrisV

so in your graph that spot would be left open

47. ChrisV

it doesnt exist

48. MasterChief

Yes, I know what you mean. It's like trying to have one piece of nothing. That doesn't exist!

49. MasterChief

I understand now.

50. ChrisV

so in this case we have an asymptope which means we can get closer and closer to 8

51. MasterChief

Yup, I get it now thanks!

52. ChrisV

such as 7.999

53. ChrisV

so the graph goes up and down at 8 infinitely

54. MasterChief

I think so.