First, note that the domain of ln(x) is x>0. Then, the domain of ln(x-k) is x-k>0, or x>k. We can neglect the +2 in determining this because it has no bearing on domain--it constitutes a vertical shift.
Furthermore, note that ln(x) is asymptotic at x=0. Symbolically, lim_(xrarr0^+)ln(x)=-oo. Here, this becomes:
- lim_(xrarrk^+)ln(x-k)=-oo
One other valuable thing to note is that ln(x)=0 at x=1. Then, ln(x-k)=0 at x-k=1, or at x=k+1.
- ln(x-k) has a zero at x=k+1.
Assuming we know the general shape of the logarithmic curve (which for ln(x) is a reflection of the exponential curve e^x across the line y=x), we can draw the function ln(x-k).
Arbitrarily setting k=5 for the sake of graphing the function in the Socratic graphing system, we know that ln(x-5) will have the domain x>5, have an asymptote to -oo at x=5 and have a zero at x=5+1=6. This looks like:
ln(x-5)
graph{ln(x-5) [1.01, 21.01, -6, 4]}
The missing piece here is that the function we want is ln(x-k)+2. So, all we do is we take each point on the graph of ln(x-k) and shift it up 2. Again using our test function with k=5:
ln(x-5)+2
graph{ln(x-5)+2 [2.76, 22.76, -4.28, 5.72]}
If we're unsatisfied with this process of shifting vertically, we can also find the formal point at which ln(x-k)+2 has a zero. This is when ln(x-k)=-2, so x=k+e^-2.
When k=5, this translates to a zero at x~~5.1353.