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anonymous

  • 5 years ago

how do i graph y=-1 over x + 1 on a graph??

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  1. anonymous
    • 5 years ago
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    you mean you want to sketch y= -1 on a graph right?

  2. anonymous
    • 5 years ago
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    im not sure, thats why im looking for help where ever i can get it. i have an assignment due wednesday and that is one of the questions. the teacher is very little help and hasnt told me what to do. can you help?

  3. anonymous
    • 5 years ago
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    what does the question really say?

  4. anonymous
    • 5 years ago
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    the question says: on graph D draw y=-1 over x+1

  5. anonymous
    • 5 years ago
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    graph D is x + 1 right?

  6. anonymous
    • 5 years ago
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    well i know it isnt equal to negative one.

  7. anonymous
    • 5 years ago
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    hmmm if so, then you have point y as -1 and you can find x : y = x + 1 -1 = x + 1 x = -2 so , your point on that graph must be at (-2,-1) I guess.

  8. anonymous
    • 5 years ago
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    since you want to graph y= -1 on x + 1, then they are equal at some point

  9. myininaya
    • 5 years ago
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    ? when you over do you mean it is a fraction like -1/(x+1)

  10. anonymous
    • 5 years ago
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    I think that's what he meant! then I have misread the question again lol, sorry

  11. anonymous
    • 5 years ago
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    yes

  12. anonymous
    • 5 years ago
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    then @matto forget my answer, I have misread the question, dearest apology

  13. myininaya
    • 5 years ago
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    Do you know what 1/x looks like?

  14. anonymous
    • 5 years ago
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    hyperbola

  15. anonymous
    • 5 years ago
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    to graph such question you have to find the limits to find the HAs and VAs

  16. myininaya
    • 5 years ago
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    just move the graph over 1 in the positive direction

  17. anonymous
    • 5 years ago
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    yes, it is bacically a an open c shape that only takes up one quater of a quadratic graph. im not sure how t explain it

  18. myininaya
    • 5 years ago
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    sorry over 1 in the negative direction

  19. anonymous
    • 5 years ago
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    he can't do that without finding the HAs and VAs

  20. anonymous
    • 5 years ago
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    you have to find the limit : \[\lim_{x \rightarrow \pm \infty} y \] to find the HAs and \[\lim_{x \rightarrow -1\pm}\] to find the VAs

  21. anonymous
    • 5 years ago
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    to see where the function's position will be in the end

  22. anonymous
    • 5 years ago
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    that's it :)

  23. anonymous
    • 5 years ago
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    im sorry, but in all seriousness, i have absolutly no idea what that means.

  24. anonymous
    • 5 years ago
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    lol, alright did you take limits?

  25. anonymous
    • 5 years ago
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    thank you! actually I have been strugling with this problem too... thanks now i am starting to get it... btw, I think you could be a GREAT lecturer

  26. anonymous
    • 5 years ago
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    thanks andy ^_^ I'm flattered

  27. anonymous
    • 5 years ago
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    maths is easily my weakest subject, im doing 4 3A/B subjects for year 12 and i was hoping this migh help.

  28. anonymous
    • 5 years ago
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    don't worry, it will help. I was weak in mathematics , but then I got on my feet and worked hard and look where have I reached :)

  29. anonymous
    • 5 years ago
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    @ matto, did you take limits?

  30. anonymous
    • 5 years ago
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    what is a limit?

  31. anonymous
    • 5 years ago
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    this :\[\lim_{a \rightarrow b} f(x) \] = L do you recognize this?

  32. anonymous
    • 5 years ago
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    im sorry, no. as i have said, my teacher has done very little teaching over the last 8 weeks.

  33. anonymous
    • 5 years ago
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    oh, btw, I have a book called "calculus" =D its like 1000 pages, anyone want it? If it feels too big, I also have Calculus Essentials, maybe you want this?

  34. anonymous
    • 5 years ago
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    andy, can you explain the problem for him please? :) brb

  35. anonymous
    • 5 years ago
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    the problem looks simple, but also looks uncomplete... I cant really get what the question means..

  36. anonymous
    • 5 years ago
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    it also asks to include all asymptotes if tha helps. i dont know what an asymptote is.

  37. anonymous
    • 5 years ago
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    In analytic geometry, an asymptote of a curve is a line such that the distance between the curve and the line approaches zero as they tend to infinity.

  38. anonymous
    • 5 years ago
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    http://upload.wikimedia.org/wikipedia/commons/d/d3/1-over-x-plus-x_abs.svg like this

  39. anonymous
    • 5 years ago
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    ok, well, its ok if you cant help. i appreciate the help. this was basically a last resort for me, i will seek help else where.

  40. anonymous
    • 5 years ago
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    you know, i think you should go to your friends ask them to help if they dont, go to your teacher if she doesnt, go to library

  41. anonymous
    • 5 years ago
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    if library doesnt, drop it, go do something fun...

  42. anonymous
    • 5 years ago
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    back, alright @matto

  43. anonymous
    • 5 years ago
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    i will ask my friends

  44. anonymous
    • 5 years ago
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    An asymptote is a line you draw on the graph that limits the graph from extending , contracting etc we have 2 types of asymptotes 1) Horizontal asymptotes are the lines that cut the y-axis horizontaly, you find them by taking the limit of f(x) as x goes to infinity like this: \[\lim_{x \rightarrow \pm \infty} f(x)\] and you take the limit from the left and right, if you end up with the limit = : if you end up with an answer that is a number, then that number is you Horizontal asymptote. 2) Vertical asymptotes are the lines that cut the x-axis, and you find them by taking the limit of x, as x goes to a number that makes your denominator 0, in your function which is y = -1/x+1 , x=-1 your number : \[\lim_{x \rightarrow 1\pm} f(x)\] \[\infty , -\infty\] then x = 1 is your Vertical Asymptote

  45. anonymous
    • 5 years ago
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    I hope he got the meaning and difference between HAs and VAs

  46. anonymous
    • 5 years ago
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    i hope so too

  47. anonymous
    • 5 years ago
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    it's not simple to explain it without showing him the graph by hand

  48. anonymous
    • 5 years ago
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    x = -1 * instead of x=1 sorry ^^"

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