how to find asymtote of log2(x-4)+1

Question

the 2 in "log2" is smaller ps not exponent 

Answer 1

For some functions, there are certain values that "break" the function. For example, if you have the function f(x) = 1/x, f(0) "breaks" the function, since division by zero is undefined. Since values of x slightly bigger or smaller than x=0 work just fine, it's a good sign that this function has a vertical asymptote at x=0.

 

Similar things happen for similar functions: g(x) = 1/(x-2) has a vertical asymptote at x=2, because the function g(x) is just the function f(x) shifted 2 units to the right.

So something to think about: what are the values for x that cause the function log₂(x) to "break", when slightly bigger or slightly smaller values work just fine? One way to answer this question is to think about the domain of the function log2(x). Once you know where the vertical asymptote of log2(x) is, you'll be most of the way to answering your question.