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anonymous

  • 5 years ago

how do i find the vertical asymptote of the function f(x)=4/(x+5)

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  1. amistre64
    • 5 years ago
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    you first cross out like terms from top to bottom, then what ever value makes the bottom a zero is your VA

  2. amistre64
    • 5 years ago
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    in this case....x =-5

  3. anonymous
    • 5 years ago
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    so you just make x+5=0

  4. anonymous
    • 5 years ago
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    then solve for x?

  5. anonymous
    • 5 years ago
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    Try finding the lim as x goes to 0

  6. amistre64
    • 5 years ago
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    that is correct, you want to know what vale of x can never be used

  7. amistre64
    • 5 years ago
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    x=-5 is a nono in this equation, so never use it, draw a lint to mark its position and stay away from it, that your VA

  8. amistre64
    • 5 years ago
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    line not lint, but if it is in macaroni art, use zitis

  9. anonymous
    • 5 years ago
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    so to find the equation of the vertical asymptote of the function f(x)=4/x+5 what is the answer?

  10. anonymous
    • 5 years ago
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    i am so confused?

  11. amistre64
    • 5 years ago
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    lie down on my couch right here and lets talk about this confusion :) tell me, can the bottom of a fraction ever be a zero?

  12. anonymous
    • 5 years ago
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    no

  13. amistre64
    • 5 years ago
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    then the line on a graph that marks the spot where the bottom would be zero is indictaed by the vertical asymptote. at the value of x that is bad.

  14. amistre64
    • 5 years ago
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    what value of x makes the bottom of your fraction there go bad?

  15. anonymous
    • 5 years ago
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    anything negative?

  16. amistre64
    • 5 years ago
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    negative numbers are not bad numbers they are just misunderstood :) there is on value in particular that is going to be offlimits: x+5 = 0 when x =?

  17. anonymous
    • 5 years ago
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    -5

  18. amistre64
    • 5 years ago
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    correct! so that is your VA

  19. amistre64
    • 5 years ago
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    x = -5 is the VA

  20. anonymous
    • 5 years ago
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    ok i get it now,

  21. anonymous
    • 5 years ago
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  22. anonymous
    • 5 years ago
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    there i graphed in it matlab shows you what is happening in case the visual aspect helps

  23. amistre64
    • 5 years ago
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    you got that in crayon? :)

  24. anonymous
    • 5 years ago
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    then how do i get horizontal asym to 5x^2-4/(x+1)

  25. amistre64
    • 5 years ago
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    we divide the top by the bottom like noremal long division to get an exact answer, but its not a horizontal asymptote.... itll be a line that is slanted

  26. anonymous
    • 5 years ago
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  27. anonymous
    • 5 years ago
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    look at the graph that should help you visualize it.

  28. amistre64
    • 5 years ago
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    the right side there seems off, but that might be the zoom limit

  29. anonymous
    • 5 years ago
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    ill try to fix it

  30. amistre64
    • 5 years ago
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    if we simply do a shortcut and divide all the numbers by x, we get the top to be 5x

  31. anonymous
    • 5 years ago
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  32. anonymous
    • 5 years ago
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    there you go now its a bit better

  33. amistre64
    • 5 years ago
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    thats better, but the your asymptotes gone parabolic instead of linear :)

  34. amistre64
    • 5 years ago
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    5x^2-4/(x+1) the top and bottom differ by 1 degree, so the asymptote is linear

  35. anonymous
    • 5 years ago
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  36. anonymous
    • 5 years ago
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    sorry i had the equation not copy and pasted wrong lol here is the correct result

  37. amistre64
    • 5 years ago
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    yay!! i knew you could do it :)

  38. anonymous
    • 5 years ago
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    man im just trying to help him out?

  39. amistre64
    • 5 years ago
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    youre doing great....

  40. anonymous
    • 5 years ago
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    im pretty sure when you learn these things it helps to look at a graph

  41. amistre64
    • 5 years ago
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    it does, i am visual alot, the analysis part still gets my brain tied in knots

  42. anonymous
    • 5 years ago
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    sorry i thought you were being sarcastic there

  43. amistre64
    • 5 years ago
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    ;) wel it was in humour, but nothing cruel or hateful. we all have errors to keep us humble i beleive, i know i have my share

  44. anonymous
    • 5 years ago
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    well to be honest i did have the right equation i just got excited using matlab and didn bracket off the exponents properly

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