Help sketching this

[nacho]

how would one go about sketching:


Also, what does the hybrid function look like?
I ended up with:

My initial guess was to:
sketch : reflect this in y-axis
then sketch :
then take the modulus of that, which is where i figured i was wrong, since there is no change...

What have I done wrong, and generally what is the approach to these graphs?
Also, for a graph such as
Is there really any point in applying the transformations, or is it sufficient in finding a y-intercept and maybe another point on the graph? I would generally just draw: and list the new points on the transformed graph - anything wrong with this approach?


Another question:
you have:

Is the hybrid this? :


TO find the y-int for the graph, do we just:
? What happens of the modulus signs when we are looking for the algebraic solution? Perhaps this question is not the best example...

Okay, for like:

what would the algebraic solution be like? I am genuinely confused when solving for x and y-intercepts when there is a modulus function present...

Thanks

Edit: LaTex troubles, one sec..

[dooodyo]

For part 1 the easiest method I use is to simply sketch the graph without the modulus and then reflect the part where x>0 in the y-axis and for part 2 just sketch the graph y= mod( log(x-2,e) ) which is just the original graph with the negatiers reflected in the x-axis then translate 1 unit up.

For intercepts :

y-int let x=0

y= mod( log(-2,e) )
y= log(2,e)

X-int. let y=0

e^0= x-2
x= 3

PS. I'm a noob I can't use latex sorry  

[nacho]

For intercepts :

y-int let x=0

y= mod( log(-2,e) )
y= log(2,e)
X-int. let y=0
e^0= x-2
x= 3

Thanks for your reply, just making sure but there wouldn't actually be a y-intercept - given that we cannot take the log of a negative number, correct?
The modulus is outside the number which we are taking the log of.

Still a little bit unsure of the first graph.  And could someone identify mistakes in my workings and show me how to do them correctly? Thanks

[luffy]

how would one go about sketching:


Also, what does the hybrid function look like?
I ended up with:

My initial guess was to:
sketch : reflect this in y-axis
then sketch :
then take the modulus of that, which is where i figured i was wrong, since there is no change...

What have I done wrong, and generally what is the approach to these graphs?
Also, for a graph such as
Is there really any point in applying the transformations, or is it sufficient in finding a y-intercept and maybe another point on the graph? I would generally just draw: and list the new points on the transformed graph - anything wrong with this approach?


Another question:
you have:

Is the hybrid this? :


TO find the y-int for the graph, do we just:
? What happens of the modulus signs when we are looking for the algebraic solution? Perhaps this question is not the best example...

Okay, for like:

what would the algebraic solution be like? I am genuinely confused when solving for x and y-intercepts when there is a modulus function present...

Thanks

Edit: LaTex troubles, one sec..

1.
expressed as a hybrid would be:



The best way to "visualise" this question (for me):

Firstly, draw the graph of when -x-1 >0 (i.e. x<1).
-i.e. Transformations - reflection in the y - axis, translated 1 unit left and 2 units up.

Then imagine there is a vertical line at x = 1 and reflect the entire graph from that point. This will give you the graph of when x > 1.

Don't forget to show a distinct sharp point at x = 1.

2. When graphing, it is always best to look at all the transformations. However, as you are not drawing the exponential graph to scale, the horizontal translation will not be seen easily, nor will the dilation.. However, the vertical translation and reflections will be obvious and must be graphed accordingly (obviously).

3. Yes, your hybrid for the logarithm function is correct.
The y-int is just subbing in x=0 for the part of the hybrid that will exist at x=0.

i.e.



Therefore, y-intercept at
y-intercept (sub x = 0):


This value is undefined. Therefore, there is no y-intercept. If you graph this on your calculator or by hand, you will see why this is the case.

x-intercept (sub y = 0):









x-intercept at (3,0) [/tex]

This was a simple example, but often you might have something like:



If you were finding the x-intercept in this case, it is slightly more tricky:
So, x -int, sub y = 0.




Note that if the value here was -2, there would be no x-intercepts as it would be undefined.


It is plus or minus two because the value inside the modulus could be both positive or negative and both would yield the result of 2.  

    OR   

    OR   

Therefore, x intercepts at and
To see why you have two intercepts, draw a sketch of both parts of the hybrid functions.

Hope I helped. Tell me if I didn't explain anything clearly.

[nacho]

thanks so much man, everything is much clearer!
someone make him moderator of methods cas subforum asap, before i recite to its' hundreth decimal

Edit:
Just to clarify, i wrote the question wrong, it is actually:

Is the approach to sketching modulus functions different at times?
I thought it was that we always drew the inside of the modulus, then applied the refection, then any other transformations (in order of DRT)
With the change in question, would I:
reflect in y, reflect in x, translate 1 left, 2 up? What follows then?

And also, just curious as to how you figured the hybrid function?

I put:

and solved from there to obtain

[dooodyo]

Oh whoops my mistake haha sorry.

[luffy]

thanks so much man, everything is much clearer!
someone make him moderator of methods cas subforum asap, before i recite to its' hundreth decimal

Edit:
Just to clarify, i wrote the question wrong, it would be:


And also, just curious as to how you figured the hybrid function?

I put:

and solved from there to obtain

Yep, and for the other part of the hybrid,

-x - 1 < 0 and the equation would be:



, where

Hope I helped.

[nacho]

thanks so much man, everything is much clearer!
someone make him moderator of methods cas subforum asap, before i recite to its' hundreth decimal

Edit:
Just to clarify, i wrote the question wrong, it would be:


And also, just curious as to how you figured the hybrid function?

I put:

and solved from there to obtain

Yep, and for the other part of the hybrid,

-x - 1 < 0 and the equation would be:



, where

Hope I helped.

Do you mean
And for:

Is the approach to sketching modulus functions different at times?
I thought it was that we always drew the inside of the modulus, then applied the reflection in the x-axis, then any other transformations (in order of DRT).
With the change in question, would I:
reflect in y, reflect in x, translate 1 left, 2 up? What follows then? The final reflection in x because of the mod?

[luffy]

thanks so much man, everything is much clearer!
someone make him moderator of methods cas subforum asap, before i recite to its' hundreth decimal

Edit:
Just to clarify, i wrote the question wrong, it would be:


And also, just curious as to how you figured the hybrid function?

I put:

and solved from there to obtain

Yep, and for the other part of the hybrid,

-x - 1 < 0 and the equation would be:



, where

Hope I helped.

Do you mean

Haha yea. I was editing my post as you typed that.

Is the approach to sketching modulus functions different at times?
I thought it was that we always drew the inside of the modulus, then applied the refection, then any other transformations (in order of DRT)
With the change in question, would I:
reflect in y, reflect in x, translate 1 left, 2 up? What follows then?

Yes, it really just depends on the user. I use different methods depending on the question. Usually, if I am sketching the graph, I will simply draw the values where the modulus makes no difference to the equation (i.e the value inside the modulus is greater than 0) and then reflect it.

[nacho]

Yes, it really just depends on the user. I use different methods depending on the question. Usually, if I am sketching the graph, I will simply draw the values where the modulus makes no difference to the equation (i.e the value inside the modulus is greater than 0) and then reflect it.

hm? That would yield an incorrect answer wouldn't it?

Edit: It all makes sense now, thanks x100!

[luffy]

No... Draw the part where the modulus makes no difference. This will be one part of the hybrid function.

Then imagine a vertical line at the "sharp point". Then reflect the graph you drew in that line (this will give you the second part of the hybrid).

Its hard to explain in words.