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

- anonymous

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

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- amistre64

shuold be x = -1

- amistre64

is that an exponent of 3 on the end?

- anonymous

It is a sub unit of 3

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## More answers

- anonymous

I'm trying to make quiz corrections in my math class and I had this one wrong :/

- anonymous

the answer IS -1 but I don't understand how

- amistre64

what were the choices?

- 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

Okay, so why would the correct answer choice be -1? I didn't understand it

- anonymous

When you say that the graph is shifted by (x + 1), how does the x component GET modified, could you explain that?

- amistre64

i can try :)
all graphs are examined at the origin at (0,0)

- 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

Right

- 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

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

we'd move it 1 place to the right, correct?

- 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

Okay

- amistre64

suppose x=3 originally; what is our note look like then?

- anonymous

3 + 1 = 4

- 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

so if x = 3, how far to the origin? that would be 3 units to the left, -3

- amistre64

correct :) so our note becomes (x-3) to remind ouselves where we got it from.

- anonymous

Okay

- amistre64

so your problem; the log is normally at x = 0 with an Vasymptote of x=0

- amistre64

but they got this graph from log(x+1); so where was it originally?

- 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

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

OH! Okay!!!

- anonymous

That makes sense lol! Thank you!!

- amistre64

youre welcome :)

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