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Graph Transformation - Natural Logs?

Now I can sketch these graphs roughly by working out y and x intercepts etc, but I can't describe how the graph transforms from the original. I'll give you an example to show you what I mean. Here is a question:

Describe the transformations required to sketch the graph of y=e^(6x-2)-4 starting from the graph of y=e^(x).

Now I can sketch the graph by finding out what y is when x=0 and what x is when y=0 and what happens to the original y int and x int and whatever the case may be, but I can't describe how the graph transforms because I have the transformation of f(6x-2)-4 here, and I don't know where to go from there.

If you understand what I seem to be failing to articulate here, please help me out.

I'll appreciate any answers.

Thanks a lot!
Reply 1
Avatar for TheGoat
TheGoat
12
Isn't this successive transformations? like.......

y=e^(x) translated 2 units in the positive x direction becomes y=e^(x-2), etc.. Or something like that anyway, haven't revised that stuff yet :P but I think that's what it is?
Reply 2
Avatar for Hasufel
Hasufel
16
describe it one step at a time, i.e. ,first you know that e^6x is a stretch of e^x by a factor 6, ( -although it stretches, it actually looks like it gets smaller!)
then e^6x-2 is a translation of the previous curve by a factor of 2 in the +ve direction of the x-axis... -4 means previous graph translated 4 units in the negative y-axis directio...

the first graph is to the right, the second to the left of THAT ONE, the 3rd to the RIGHT of THAT one
(edited 12 years ago)
Reply 3
Avatar for fGDu
fGDu
14
I think that you know all of the rules to answer this question, but you are having trouble putting them together.

Try thinking in terms of :

f(x) = e^x
f(?) - 4 = (e^6x-2) - 4

Then you need to think about what this function of 6x-2 will do to the graph, I think you know that f(Ax+B) will stretch the graph parallel to the x axis with a scale factor of 1/A , then move it to the left by B units.
Reply 4
Avatar for tahseen1995
tahseen1995
OP
Thanks a lot, figured it out now, just thought it'd be a lot more complicated but I guess not :smile:

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