Question 5894: I'm trying to figure out how to plot the graph y=ln(1-x) by shifting the basic graph for y=ln(x). including the equation of the asymptote, and the coordinates of the y-intercept. Can I think of it as y=ln(-x+1)? Then I could reflect the graph across the y axis and shift it up 1? Would that work?
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! 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.
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