Graph y=log 3(x+2)-1. Then, find the domain/range

Question

anonymous · Answer

what does "solve" mean in this context?

anonymous · Answer

Whoops! Not solve. Just graph and find the domain/range

jim_thompson5910 · Answer

I would use a graphing calculator like desmos https://www.desmos.com/calculator to graph this out. You will have to use the change of base formula to make sure the expression is typed in properly you will have to type in log(x+2)/log(3) - 1

anonymous · Answer

Ok thanks :) Would the domain be all real numbers?

jim_thompson5910 · Answer

why all real numbers?

jim_thompson5910 · Answer

you should have this graph https://www.desmos.com/calculator/fs1pnhr9hx

jim_thompson5910 · Answer

the left side of the graph seems to hit a "wall" of some sort

anonymous · Answer

Ok, now I do. And it does seem like that

jim_thompson5910 · Answer

it turns out that you cannot plug in 0 or negative numbers into a log function

so x+2 has to be positive

x+2 > 0 leads to x > ???

anonymous · Answer

x=-2 ? x+2=0

jim_thompson5910 · Answer

yeah x = -2 is your vertical asymptote. The graph does NOT touch the vertical asymptote. It only gets closer and closer and closer.

jim_thompson5910 · Answer

solving x+2 > 0 for x gives x > -2, which is the domain

anonymous · Answer

Alright, I see :) Now what about the range?

jim_thompson5910 · Answer

the left side shoots off to negative infinity (very quickly)

the right side slowly approaches positive infinity. it just keeps growing forever

anonymous · Answer

so the range would be all real #'s or infinity

jim_thompson5910 · Answer

all real numbers

I think you're thinking of this notation $\Large (-\infty,\infty)$

anonymous · Answer

Yes :)

anonymous · Answer

Thanks!

jim_thompson5910 · Answer

you're welcome