Question 5894
i wouldn't do anything like that myself..too easy to get confused.


What i would do is know what y = ln(x) looks like and then use this as a basis to work out the question.


So, i assume you know what y=ln(x) looks like? no -ve x-values allowed. The curve starts very close to zero for large -ve y-values and comes through (1,0) before curving away as x increases.


So, y=ln(1-x) is going to be the same shape more or less.


if x=0, then y=ln(1) --> 0, so the curve passes through the origin.
if x=1, then y=ln(0) --> asymptote at x=1.


So now you can draw the curve: basically a mirror image of ln(x) in shape.

If you want to put a large -ve x-value in, just to check your shape:

eg if x=-10, then y=ln(11) --> large +ve value: CORRECT.
eg if x=+10, then y=ln(-9) --> ERROR: CORRECT.


Can you see how simpler it is to know the basic curve and then just mess with intercepts and asymptotes to get the "more complicated" version?


jon.