Help on sketching on log functions with absolute

[duhherro]

Lets try y= |ln(x-2)|
y= ln (x-2) <0
and

ln(x-2) >= 0     ,     where >= means greater or equal to
 
Sketching the ln(x-2) graph gives us an x-int of 3

so we can make new equations:

y= -loge (x-2), 2 < x < 3
and

loge (x-2) >= 3

The above working out makes me think there is an easier way of  working it out.


Another example was y=|-loge (1-x) -2| , where I did the same above working out , but for y=|log10 (x+3)| +1 , i cannot find the hybrid equations with the method I do , so I was hoping you could provide me an easier way of doing these kinds of questions (like how you would do the above questions with your method)

[nubs]

If all you need to do is sketch it, then just pretend there are no modulus signs, sketch that graph, and for all the parts of the graph that are negative, reflect them in the x-axis to make them positive, and that will give you the graph you wanted initially

For example,
y=|-loge (1-x) -2|

First, just ignore the modulus signs, so you get y=-loge (1-x) -2
Then sketch that graph
Now look at the graph, and focus on the parts of the graph that are negative, or under the x-axis
redraw these parts by reflecting them in the x-axis so they are now on top of the x-axis

That will give you the graph of y=|-loge (1-x) -2|

EDIT: Sorry, I didn't read all of what you said

For the graph y=|log10 (x+3)| +1

Ignore the plus 1 on the end, and just sketch the graph of y=|log10 (x+3)| they way you normally would by making up a hybrid function
Once you have that graph, just shift everything up by 1, because that's all the +1 on the end is describing

[duhherro]

If all you need to do is sketch it, then just pretend there are no modulus signs, sketch that graph, and for all the parts of the graph that are negative, reflect them in the x-axis to make them positive, and that will give you the graph you wanted initially

For example,
y=|-loge (1-x) -2|

First, just ignore the modulus signs, so you get y=-loge (1-x) -2
Then sketch that graph
Now look at the graph, and focus on the parts of the graph that are negative, or under the x-axis
redraw these parts by reflecting them in the x-axis so they are now on top of the x-axis

That will give you the graph of y=|-loge (1-x) -2|

EDIT: Sorry, I didn't read all of what you said

For the graph y=|log10 (x+3)| +1

Ignore the plus 1 on the end, and just sketch the graph of y=|log10 (x+3)| they way you normally would by making up a hybrid function
Once you have that graph, just shift everything up by 1, because that's all the +1 on the end is describing

Wow thank you for your fast replies! I see, so doesn't really matter to the fact that I have to go through all these working out when I can just do a rough sketch etc etc?

Btw, wouold you know how I could practise for my upcoming Sac on Logs + graphs, Exp + graphs, and then inverse functions. If you are familiar with Math quests, they are Chp 3, 4, and 5. !

[nubs]

Well if all it is asking you to do is sketch the graph, then you won't need to show all the working out, like you won't have to divide it into a hybrid function if you don't find it necessary, you could just simply sketch it

But if it asked you to show working out, then for sketching something like y=|log10 (x+3)| +1, I would just let the examiner know what I'm doing

i.e, I would show the working out for the graph y=|log10 (x+3)|, and do a rough sketch of that somewhere else on the page (which might be unnecessary), and then assert that the graph that I am required to sketch is just the graph of y=|log10 (x+3)| shifted upwards by 1 unit, and then sketch that.

Depends on how much working you need to show, doubt you would need to do something like that, however, most of the time it's just asking you to sketch the graph and show asymptotes and axis intercepts

and no, sorry, I've never even looked at Maths Quest before
all the best dude

[duhherro]

Well if all it is asking you to do is sketch the graph, then you won't need to show all the working out, like you won't have to divide it into a hybrid function if you don't find it necessary, you could just simply sketch it

But if it asked you to show working out, then for sketching something like y=|log10 (x+3)| +1, I would just let the examiner know what I'm doing

i.e, I would show the working out for the graph y=|log10 (x+3)|, and do a rough sketch of that somewhere else on the page (which might be unnecessary), and then assert that the graph that I am required to sketch is just the graph of y=|log10 (x+3)| shifted upwards by 1 unit, and then sketch that.

Depends on how much working you need to show, doubt you would need to do something like that, however, most of the time it's just asking you to sketch the graph and show asymptotes and axis intercepts

and no, sorry, I've never even looked at Maths Quest before
all the best dude

Thanks again for your replies!!!
I'm just learning abs on the power of the exponentials:

like y= e^|x-1| + 4, how the hell would  you do them haha

[Keynesian]

I'm just learning abs on the power of the exponentials:

like y= e^|x-1| + 4, how the hell would  you do them haha
[/quote]

Hi,
Yeah im learning the same stuff and have a SAC coming up on those exact chapters, and using the math quest book.

y=e^|x-1| + 4 is sketched the same as the log graph. Except e is a real number and stands for a value which is 2.71828

basically draw y=e^(x-1) then reflect what ever is under the x-axis which is the negative part of the graph, then after that move the graph up 4 units.

[yawho]

I'm just learning abs on the power of the exponentials:

like y= e^|x-1| + 4, how the hell would  you do them haha

Hi,
Yeah im learning the same stuff and have a SAC coming up on those exact chapters, and using the math quest book.

y=e^|x-1| + 4 is sketched the same as the log graph. Except e is a real number and stands for a value which is 2.71828

basically draw y=e^(x-1) then reflect what ever is under the x-axis which is the negative part of the graph, then after that move the graph up 4 units.

[/quote]
but there is nothing under the x-axis

[b^3]

When you have something of the form then you filp what is below the x-axis in the x-axis, as it is making anything that is negative, positive.

When you have something of the form , then you take what is on the right hand side of the y-axis, copy it and flip it in the y-axis, that is because you are taking all the input values of x that are negative, and replacing the corresponding y value with the y value at the positive value of that x.

So for , it is the second case above. So we are going to apply the transformations.
Firstly we'll start with , which looks like

Now we will transform it to by copying, then flipping everything on the right side of the y-axis in the y-axis, which results in

Next we will transform it to , so that is a shift (or 'translation') of 1 unit in the positive direction of the x-axis. So we end up with

Finally we will transform it to by a translation of 4 units in the positive direction of the y-axis, which finally results in


Really the best way to do it is first to work out whether you have a or a , then do the appropriate transformations.

You don't have to draw all these graphs out above, I just did them all to show each step. Hope that helps

[duhherro]

When you have something of the form then you fill what is below the x-axis in the x-axis, as it is making anything that is negative, positive.

When you have something of the form , then you take what is on the right hand side of the y-axis, copy it and flip it in the y-axis, that is because you are taking all the input values of x that are negative, and replacing the corresponding y value with the y value at the positive value of that x.

So for , it is the second case above. So we are going to apply the transformations.
Firstly we'll start with , which looks like
(Image removed from quote.)
Now we will transform it to by copying, then flipping everything on the right side of the y-axis in the y-axis, which results in
(Image removed from quote.)
Next we will transform it to , so that is a shift (or 'translation') of 1 unit in the positive direction of the x-axis. So we end up with
(Image removed from quote.)
Finally we will transform it to by a translation of 4 units in the positive direction of the y-axis, which finally results in
(Image removed from quote.)

Really the best way to do it is first to work out whether you have a or a , then do the appropriate transformations.

You don't have to draw all these graphs out above, I just did them all to show each step. Hope that helps

Wow, thanks a lot! Really helped about the part where you flip the left side of the Y-axis part! Also how come I sometimes can't see your equations, its like blocked out.

Also what happens if its applied towards logs?

so loge abs       and       abs loge   abs

[b^3]

Wow, thanks a lot! Really helped about the part where you flip the left side of the Y-axis part! Also how come I sometimes can't see your equations, its like blocked out.

Not sure, it might just be the LaTeX rendering playing up again.

Also what happens if its applied towards logs?

so loge abs       and       abs loge   abs

Ok so for is of the form , so that is flipping in the y-axis.
So we originally start off with

So now we take what is on the right side of the y-axis and flipped it in the y-axis, leaving the original function on the right side of the axis.

Now if you look at it, it makes sense in that you now have a modulus making all the negative values of x that are 'put in', positive, and so you can take the log of them, and that the value of y that comes out will corresspond to the same value of y for the positive value of x.

Now for , now this has both forms in it, so apply the first. That will result in the graph above.
Now that we have that we apply the form, so remember that the will effectively flip everything that is below the x-axis, in the x-axis. So that is everything between -1 and 1. So we get the graph below. (RED is the original function, BLUE is the new function after its been flipped, purple is the overlap where there isn't a change).


Hope that helps

[pi]

b^3 is the god of absolute functions!

[duhherro]

Haha thanks B^3, GREAT HELP!!! but also, what happens if the x coefficient is a negative for all of the 4 above scenarios.

like i asked my tutor about this...

eg:

y = log10 |3-x| + 1
y = log 10 |-(x-3)| + 1

and he then said it will be the same as y = log 10 |x-3| +1, you can simply get rid of the negative in front of the bracket.

[TrueTears]

|-(x-3)| = |-1|*|x-3| = |x-3|

[duhherro]

|-(x-3)| = |-1|*|x-3| = |x-3|

ahh i see , so that |-1| gets modulus'd and becomes 1.

what if its like... |-(3-x)|, so like 2 negatives?

[TrueTears]

just keep using |a*b| = |a|*|b|