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anonymous

  • 5 years ago

What is the equation of the vertical asymptote of the graph of f(x) = 2 log3(x + 1) 3?

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  1. amistre64
    • 5 years ago
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    shuold be x = -1

  2. amistre64
    • 5 years ago
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    is that an exponent of 3 on the end?

  3. anonymous
    • 5 years ago
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    It is a sub unit of 3

  4. anonymous
    • 5 years ago
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    I'm trying to make quiz corrections in my math class and I had this one wrong :/

  5. anonymous
    • 5 years ago
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    the answer IS -1 but I don't understand how

  6. amistre64
    • 5 years ago
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    what were the choices?

  7. amistre64
    • 5 years ago
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    all graphs of logs.... at least normal logs.... never reach zero or negative numbers; any modification of them tends to keep that same trait. when they shift the graph by (x+1) they modify the x component so that it reads x = -1

  8. anonymous
    • 5 years ago
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    Okay, so why would the correct answer choice be -1? I didn't understand it

  9. anonymous
    • 5 years ago
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    When you say that the graph is shifted by (x + 1), how does the x component GET modified, could you explain that?

  10. amistre64
    • 5 years ago
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    i can try :) all graphs are examined at the origin at (0,0)

  11. amistre64
    • 5 years ago
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    that is the easiest place to examine them up close and personal like; when they are NOT at the origin we have to move them..

  12. anonymous
    • 5 years ago
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    Right

  13. amistre64
    • 5 years ago
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    if the graph is at (x=-1)... we write ourselves a note to tell us where we got it from; or rather, how far we had to move it and in what direction to get it to (0,0)

  14. amistre64
    • 5 years ago
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    that way we can put it back when we are done looking at it. if the graph is at x=-1; how far and in what direction did we have to move it to get it to the origin?

  15. anonymous
    • 5 years ago
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    we'd move it 1 place to the right, correct?

  16. amistre64
    • 5 years ago
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    that correct; we move the "x" by +1. So we write ourselves a note (x+1); so we can put it back when we are done right?

  17. anonymous
    • 5 years ago
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    Okay

  18. amistre64
    • 5 years ago
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    suppose x=3 originally; what is our note look like then?

  19. anonymous
    • 5 years ago
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    3 + 1 = 4

  20. amistre64
    • 5 years ago
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    think about that again; how far and in what direction do we need to move the graph from x=3 to get it to the origin (0)? if we add (+1) to it it gets us to 4... not to zero.

  21. anonymous
    • 5 years ago
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    so if x = 3, how far to the origin? that would be 3 units to the left, -3

  22. amistre64
    • 5 years ago
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    correct :) so our note becomes (x-3) to remind ouselves where we got it from.

  23. anonymous
    • 5 years ago
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    Okay

  24. amistre64
    • 5 years ago
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    so your problem; the log is normally at x = 0 with an Vasymptote of x=0

  25. amistre64
    • 5 years ago
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    but they got this graph from log(x+1); so where was it originally?

  26. anonymous
    • 5 years ago
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    Ummm i don't understand it. with the log... would it be x = -1, so that it could move to the right to get to 0?

  27. amistre64
    • 5 years ago
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    thats absolutley correct :) they moved it over (+1) so it was originally at: x=-1 .... any answers we get from looking at it from x=0 have to be "adjusted" now...by -1. since the original Vasymp is at 0; the Vasymp has to be adjusted by (-1)..... Vasymp of log(x+1) is: x=(0-1) or simply x=-1

  28. anonymous
    • 5 years ago
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    OH! Okay!!!

  29. anonymous
    • 5 years ago
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    That makes sense lol! Thank you!!

  30. amistre64
    • 5 years ago
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    youre welcome :)

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