# Homework Help: Drawing modulus graphs

1. Dec 23, 2011

### RCB

1. The problem statement, all variables and given/known data
how would I draw the graph
f(x) = ln|3x-6|

2. Relevant equations
NONE

3. The attempt at a solution
So I know that the graph ln(3x-6) would be a transformation of ln(x) 6 units to the right and the stretch by scale factor (1/3) along the x-axis (i.e: divide all x co-ordiantes on ln(x-6) by 3)

However I then have to modulus the function 3x-6
I am NOT saying |f(x)| therefore I don't reflect in the x-axis
I am NOT saying f(|x|) therefore I don't reflect in the y-axis

Therefore how do I continue this transformation.

Apparently I reflect in the asymptote (which moved to x=2) BUT WHY?

2. Dec 23, 2011

### DivisionByZro

ln(x) and even ln|x| have asymptotes at x=0. In your new function, ln|3x-6|, where is it undefined? That is, when does 3x-6=0?

3. Dec 23, 2011

### RCB

This happens when x = 2

4. Dec 23, 2011

### RCB

are you saying for any modulus function i just equate the modulus bit to 0 and reflect it in that?

5. Dec 23, 2011

### DivisionByZro

No. If by any modulus function you really mean any function composed with an absolute value function, then no. |x| has no asymptotes, so you don't need to do that. My advice to you: find where your function intersects with the x and y axes (when y=0 and x=0, respectively), and find the asymptote (Which you've already done), determine if there are horizontal asymptotes (there could be), determine the concavity, then try to graph it.

6. Dec 23, 2011

### RCB

ok but why do I reflect the graph above in the asymptotes ?

7. Dec 23, 2011

### Staff: Mentor

No it wouldn't. ln(3x - 6) = ln(3(x - 2)).

The translation (shift) to the right is by 2 units relative to the graph of y = ln(3x).

8. Dec 23, 2011

### Staff: Mentor

ln(3x - 6) is defined only where 3x - 6 > 0
ln|3x - 6| is defined everywhere except where 3x - 6 = 0.

It might be helpful to sketch the graph of y = |3x - 6| and then use it to get the graph of y = ln|3x - 6|.

9. Dec 24, 2011

### jsmith613

Ok here's why I am confused
|ln(3x-6)| is a reflection of f(x) > 0 (for f(x) < 0) in the x-axis

ln(3|x| - 6) is a reflection of x > 0 (for x < 0) in the y-axis

but lets say we took the graph
|3x-6|
(3x-6) is defined EVERYWHERE (for ALL values of y)

so why would it be different here
sorry I seem to be slow :(

10. Dec 24, 2011

### jsmith613

oh hold on
because |3x-6| is always positive, ln(u) is always positive, right?

11. Dec 24, 2011

### SammyS

Staff Emeritus
|3x-6| is always non-negative, not quite always positive.

ln(u) is not defined if u ≤ 0. |3x-6| = 0 if x = 2 .

12. Dec 24, 2011

### Staff: Mentor

I think I get what you're trying to say, but you're not saying it very well. Taking the absolute value reflects the part of the graph of y = ln(3x - 6) that is below the x-axis, across the x-axis.
It's not useful to consider this function in your problem.
y = 3x - 6 = 3(x - 2) is a straight line with a slope of 3 and a y-intercept of -6. The line has an x-intercept at (2, 0).

The graph of y = |3x - 6| = 3|x - 2| is identical to the graph of y = 3(x - 2) for x >= 2. For x < 2, the graph of y = 3(x - 2) is reflected across the x-axis.

The graph of f(x) = ln|3(x - 2)| is defined everywhere except at x = 2.

13. Dec 24, 2011

### SammyS

Staff Emeritus
I quickly skimmed through the posts to this thread and did not see the following way of looking at this problem.

ln(|3x-6|) = ln(|3| |x-2|) = ln(3) + ln(|x-2|)

|x-2| is symmetric w.r.t. the vertical line, x=2.

Take it from there.

14. Dec 24, 2011

### RCB

OK.
just to check I understand this for all any future problems:

|x| is the absolute value of x
if I took
|x|2 + 3|x| + 5
I then take the positive value of all y values.
this leads to a reflection in the y-axis because the f(x) x≥0 will be the same as x<0

for
|f(x)| the f(x) value for any -f(x) value is then mapped onto +f(x)
thus it is a reflection in the x axis

Now for a composite function:
f(x) = 3x+2
g(x) = ln(x)

Thus
gf(x) = ln(3x+2)
so, g(f(|x|)) = ln(|3x+2|)

Now the graph of
|3x+2| is symmetrical w.r.t the vertical line x = -2/3

now because ln|3x+2| is a composite function, we take the vertical line x = -2/3 to be the reflection point

15. Dec 24, 2011

### RCB

so for simple any graph (even composite graph) EXCLUDING THOSE for ln(x) and ex:
e.g:
f(x) = 3x-2
g(x) = 3x2

fg(x) = 9x2 -2
so
|9x2 -2| would just involve reflecting negative fg(x) values in the x-axis
etc...

HOWEVER for ln(x) graphs of ex graphs, when I have a composite function and I mod. ONLY the linear function NOT the composite / exponential or logarithmic function I look for where the composite function IS NOT DEFINED and reflect the graph in that vertical line

THESE ARE RULES I WILL JUST LEARN
are they correct?

thanks

Last edited: Dec 24, 2011
16. Dec 24, 2011

### Staff: Mentor

Instead of trying to memorize a bunch of rules, I think it would be better to have a basic understanding of what's going on.

For your example in the previous post, let's get rid of all the composite function stuff and just look at y = 9x2 - 2 and y = |9x2|. Bringing in composite functions needlessly complicates things.

The first graph can be obtained by sketching y = 9x2 (a parabola), and shifting it down by 2 units. The central part of this graph, between -√(2)/3 and +√(2)/3, lies below the x-axis. If x < -√(2)/3 or x > √(2)/3, the graph is above the x-axis.

To get the graph of y = |9x2| flip the central part of the graph, the part between x = -√(2)/3 and x = √(2)/3, across the x-axis. Voila, you're done!

This same process can be followed for an arbitrary function. To get the graph of y = |f(x)|, do this:

1. Sketch the graph of y = f(x)
2. If any parts of the graph are below the x-axis, reflect them across the x-axis to get the graph of y = |f(x)|. Any parts of the original graph that were above the x-axis don't get reflected.