Question

Find the domain, range, and vertical asymptote of y=log3 (x+2).

Answer 1

For the domain,
\[x+2 > 0\]
For the range;
Is there any value that can't take???
and a vertical asymptote, well, are you familiarized with limits?

Answer 2

okay so the range would be all real numbers..
not really

Answer 3

Well, anyway, by appllying limits you can know where is the vertical asymptote of any function
in the case of logaritms, the vertical asymptote will exist in the "x" for which the argument (y) of this example is 0
\[\log_z y =f(x)\]
so when y=0 in our case
x+2=0
that means that in x=-2 we have our vertical asymptote

Answer 4

So we have the logarithmic function \[y = \log3(x+2)\]

We can say that \[y \in \mathbb{N} \]

And

\[x \in \mathbb{N} / x \neq -2\]

So yea...

Answer 5

I think you should graph it to check. Its easy just plug in values remember that when you put it in your calculator it looks like \[\frac{ \log(x+2) }{ \log3 }\]

Answer 6

i think

Answer 7

here's what i think the graph looks like

Answer 8

alright, makes sense! thanks so much!!