Posted by kikii on Wednesday, February 13, 2013 at 2:23am.
If your log is of (2x+3)/(x-5), then you must have
(2x+3)/(x-5) > 0
So, either both are positive or both are negative, so the domain is
x < -3/2 or x > 5
As x -> ∞, y->log2
For x < -3/2, y < log2
For x > 5, y > log2
The range is all reals except y=log2
That is the horizontal asymptote
The vertical asymptotes are at x = -3/2, 5
For the y-intercept, what is y when x=0?
Well, log(-3/5) is not defined, so there is no y-intercept
When is y=0? When (2x+3)/(x-5) = 1
That is, x = -8
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