Please help

Question

http://photobucket.com/gallery/user/danidoll100/media/cGF0aDovU2NyZWVuIFNob3QgMjAxOC0wMy0wNyBhdCA1LjUzLjQwIFBNX3pwc2tiemZxaGlrLnBuZw==/?ref=1

 

the link includes the problem! thank you

Answer 1

Given f(x) = log_2(x-3)+1

The domain of the function is set of all positive values for x .

The range is all possible values for y .

Here observe the function  here s logarithm is involved.

(x-3) > 0 then x >3

Thus the domain is (3,∞).

Exchanging the position of x & y,

x= log_2(y-3)+1

log_2(y-3)=x-1

y-3=2^(x-1)

y=3+2^(x-1)

Thus the range is R.

The basic graph of f(x) = log_2(x-3)+1 has been translated

up 1 units and 3 unit to the right, so x=3 is the vertical asymptote.