Question

Hi, how would I do this problem?
Graph f(x) = log2(x – 3) + 1.
Find its domain, its range, and the equation of its asymptote.

Answer 1

Well, in general \(log_2(x)\) has real values only for \(x > 0\) because there is no real power of 2 that will result in 0 or a negative number.
In our case we have \(log_2(x-3)\) and knowing log2 gives real values only for positive numbers we say \(x > 3\) is our domain.
The function is not bounded from either side it's range is from \(-\infty\) to \(\infty\), just like log2 itself. The +1 doesn't really make any difference.
\(log_2(x)\) has a vertical asymptote at \(x=0\) and $$ \lim_{x \to 0+} log_2(x) = -\infty
$$In our case we have \(log_2(x-3)\) so our asymptote is at \(x=3\):
$$ \lim_{x \to 3+} log_2(x - 3) = -\infty$$

Look at here for the graph:
http://www.wolframalpha.com/input/?i=+log2%28x+%E2%80%93+3%29+%2B+1.
Notice they like to show 'complex-valued plots' so change it to 'real-value plots' if you're not interested in the imaginary part.