anonymous
  • anonymous
What is the equation of the vertical asymptote of the graph of f(x) = 2 log3(x + 1) 3?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
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amistre64
  • amistre64
shuold be x = -1
amistre64
  • amistre64
is that an exponent of 3 on the end?
anonymous
  • anonymous
It is a sub unit of 3

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anonymous
  • anonymous
I'm trying to make quiz corrections in my math class and I had this one wrong :/
anonymous
  • anonymous
the answer IS -1 but I don't understand how
amistre64
  • amistre64
what were the choices?
amistre64
  • amistre64
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
anonymous
  • anonymous
Okay, so why would the correct answer choice be -1? I didn't understand it
anonymous
  • anonymous
When you say that the graph is shifted by (x + 1), how does the x component GET modified, could you explain that?
amistre64
  • amistre64
i can try :) all graphs are examined at the origin at (0,0)
amistre64
  • amistre64
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..
anonymous
  • anonymous
Right
amistre64
  • amistre64
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)
amistre64
  • amistre64
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?
anonymous
  • anonymous
we'd move it 1 place to the right, correct?
amistre64
  • amistre64
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?
anonymous
  • anonymous
Okay
amistre64
  • amistre64
suppose x=3 originally; what is our note look like then?
anonymous
  • anonymous
3 + 1 = 4
amistre64
  • amistre64
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.
anonymous
  • anonymous
so if x = 3, how far to the origin? that would be 3 units to the left, -3
amistre64
  • amistre64
correct :) so our note becomes (x-3) to remind ouselves where we got it from.
anonymous
  • anonymous
Okay
amistre64
  • amistre64
so your problem; the log is normally at x = 0 with an Vasymptote of x=0
amistre64
  • amistre64
but they got this graph from log(x+1); so where was it originally?
anonymous
  • anonymous
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?
amistre64
  • amistre64
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
anonymous
  • anonymous
OH! Okay!!!
anonymous
  • anonymous
That makes sense lol! Thank you!!
amistre64
  • amistre64
youre welcome :)

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