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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    \[f(x)= \frac{ x^{2}+x-2 }{ x ^{2-}3x-4 }\]

  2. karatechopper
    • one year ago
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    Lets factor!

  3. karatechopper
    • one year ago
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    Do you know how to factor?

  4. anonymous
    • one year ago
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    Yes \[\frac{ (x+1)(x-2) }{ (x+1)(x-4) }\]

  5. karatechopper
    • one year ago
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    Cool! Okay now, cross out the common factors and what are you left with?

  6. anonymous
    • one year ago
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    \[\frac{ (x-2) }{ (x-4) }\]

  7. karatechopper
    • one year ago
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    Right, okay. Now do you have the slightest idea of what we do now?

  8. anonymous
    • one year ago
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    Not really haha

  9. karatechopper
    • one year ago
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    Alright haha. Well the most simple way of going about this problem is to make an X,Y chart. Just some numbers for your x value, plug into the equation, pop out a Y value. - fill that into the chart. Once you get a few good points, plot. You are set!

  10. karatechopper
    • one year ago
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    Would you like me to help you make a X,Y chart?

  11. anonymous
    • one year ago
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    Yes please

  12. anonymous
    • one year ago
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    @karatechopper

  13. karatechopper
    • one year ago
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    Alright. Let's start.

  14. anonymous
    • one year ago
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    Thank you :D

  15. karatechopper
    • one year ago
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    |dw:1434953607858:dw| Pick 5 x-values for me. Keep them a good range. :)

  16. anonymous
    • one year ago
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    -2,-1,0,1,2

  17. anonymous
    • one year ago
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    Isn't that what people usually do

  18. karatechopper
    • one year ago
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    |dw:1434953723896:dw| Now, go through each X value and plug each of them into the above equation we simplified to: (x-2)/(x-4), to find each y-value. I will check them after you finish.

  19. karatechopper
    • one year ago
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    Also yes, those are common points, sometimes people pick different points because they know what the shape of the graph will look like, but there is not a requirement about which points you pick haha.

  20. anonymous
    • one year ago
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    \[\frac{ 2 }{ 3 }, -\frac{ 3 }{ 5 }, \frac{ 1 }{ 2 },-\frac{ 1 }{ 3 }, \frac{ 0 }{ -2 }\]

  21. anonymous
    • one year ago
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    I'm pretty sure some of those are wrong haha

  22. karatechopper
    • one year ago
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    Could you show me your work for x-values -1 and 1 please?

  23. campbell_st
    • one year ago
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    can I just say... ignore the table of values.... look at the asymptotes the vertical asymptote is when x - 4 = 0 so x = 4 is a vertical asymptote

  24. anonymous
    • one year ago
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    (-1-2)=-3 (-1-4)=-5

  25. anonymous
    • one year ago
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    Is that correct?

  26. karatechopper
    • one year ago
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    Negatives cancel, leaving you with positive 3/5. Same with x-value you 1, you get a positive 1/3. Also Campbell is right.

  27. campbell_st
    • one year ago
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    then the numerator and denominator are both degree 1 polynomials so the horizontal asymptote is at y = x/x or y = 1 not look at the numerator to see the where the cruve cuts the x-axis x - 2 = 0 so the x-intercept is x = 2 lastly the y- intercept let x = 0 and you get y = 1/2 so the graph|dw:1434954017902:dw|

  28. anonymous
    • one year ago
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    Its two separate lines?

  29. campbell_st
    • one year ago
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    a table of values is a waste of time... you need to recognise the curve is a hyperbola... find asymptotes and intercepts then sketch the curve

  30. campbell_st
    • one year ago
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    there are 2 parts to the curve

  31. campbell_st
    • one year ago
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    and there would be a point of discontinuity at x = -1... in my humble opinion

  32. UsukiDoll
    • one year ago
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    asymptotes is a calculus topic though... what if the original poster is in an elementary algebra course? Then a chart / table of values is needed.

  33. anonymous
    • one year ago
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    pre cal

  34. campbell_st
    • one year ago
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    but Uki will probably clean up my solution and meake it more readable

  35. UsukiDoll
    • one year ago
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    ah...so.. ihateschool18 is getting there to asymptotes..

  36. UsukiDoll
    • one year ago
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    ha ha I need to eat dinner... no seriously...can't do math without food

  37. karatechopper
    • one year ago
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    Asymptotes is an Algebra 2 Subject for me. But that's interesting to see how its in Pre-calc too. I remember it being recovered there too.

  38. campbell_st
    • one year ago
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    the great is complex... and would need a very large table to get an idea of the whole graph

  39. UsukiDoll
    • one year ago
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    @karatechopper wow that's early...I was first exposed to it in Calculus I O_O

  40. karatechopper
    • one year ago
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    I need sleep. Lol yeah Campbell I thought a table of value would get some of the point across as to how such points can't cross a certain line lol.

  41. campbell_st
    • one year ago
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    and I tend to use quick and dirty methods rather than the long and boring ones

  42. anonymous
    • one year ago
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    Thank you guys so much btw

  43. karatechopper
    • one year ago
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    Oye, old school is cool sometimes when getting the point across. xD

  44. karatechopper
    • one year ago
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    Welp, I tried haha. Have a good day guys.

  45. karatechopper
    • one year ago
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    :)

  46. anonymous
    • one year ago
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    Thanks and you too

  47. campbell_st
    • one year ago
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    the method is sound... just tedious... and in a question of this type.. I would expect the asker has some knowledge of asymptotes

  48. karatechopper
    • one year ago
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    Also @UsukiDoll I'm surprised as to how that's popping up in Calculus. Asymptotes appeared in Conics, which absolutely blew my mind to pieces. Conics was not much of a fun subject with all that memorization.

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