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C3 graph transformations

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Reply 20
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raheem94
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Original post by Smiley Face :)
Basically you can never have a negative y value with modulus. So reflect the negative portion of the Lnx graph into the +/ve quadrant of


But why does the graph transformation method fails?

By transforming we can clearly see that we need to reflect in the y-axis.
Original post by raheem94
But why does the graph transformation method fails?

By transforming we can clearly see that we need to reflect in the y-axis.


Um, im not 100% but it may be to do with the fact that the modulus actually ^2 the equation, thus, u can only get a positive outcome
Original post by raheem94
But why does the graph transformation method fails?

By transforming we can clearly see that we need to reflect in the y-axis.


So yeah, it was a lot simpler than we tried to make it (sorry to above posters who were in fact right).

f(x) -> f(|x| + 1) is shifting the graph to the left by 1 and then reflecting all the positive values of x in the y-axis.
Reply 23
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raheem94
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Original post by hassi94
So yeah, it was a lot simpler than we tried to make it (sorry to above posters who were in fact right).

f(x) -> f(|x| + 1) is shifting the graph to the left by 1 and then reflecting all the positive values of x in the y-axis.


Thanks for taking your time.

I get it now.
Reply 24
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raheem94
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Original post by hassi94
Okay this isn't right but neither was I.

f(-(x-1)) is translating x one to the right and then reflecting in the line x = 1.

Or equivalently (and more simply) as someone else said - it's f(-x + 1) which means a shift to the left and then a reflection in the y axis.


This post is wrong, right?
Reply 25
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raheem94
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Original post by hassi94
Okay this isn't right but neither was I.

f(-(x-1)) is translating x one to the right and then reflecting in the line x = 1.

Or equivalently (and more simply) as someone else said - it's f(-x + 1) which means a shift to the left and then a reflection in the y axis.


My transformation technique was correct, right?
Original post by raheem94
My transformation technique was correct, right?



Original post by raheem94
This post is wrong, right?


Which one's sorry? My latest 2 posts are 2 correct methods.
Reply 27
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raheem94
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Original post by hassi94
Which one's sorry? My latest 2 posts are 2 correct methods.


I have doubts on post#20. I don't see why we reflect in the line x=1 x =1

As you said your initial method was wrong, but was the technique used in post#12 of drawing the graph of f(1−x)f(1-x) correct?

I am not talking about getting the right answer, just drawing the graph of f(1−x) f(1-x)
Original post by raheem94
I have doubts on post#20. I don't see why we reflect in the line x=1 x =1

As you said your initial method was wrong, but was the technique used in post#12 of drawing the graph of f(1−x)f(1-x) correct?

I am not talking about getting the right answer, just drawing the graph of f(1−x) f(1-x)


No I meant what I said. Think about y = x then transforming it to y = 1 - x

Sketch if it helps. We would sketch y = x - 1 and then reflect it in the line x = 1 (think about the expression x-1: x = 2 would give 1, and for y = 1 - x would give -1 and so on).

I'm pretty certain I'm right here (but I could be making a mistake). Just draw y = x and y = 1-x and you'll see what I mean.
Reply 29
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raheem94
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Original post by hassi94
No I meant what I said. Think about y = x then transforming it to y = 1 - x

Sketch if it helps. We would sketch y = x - 1 and then reflect it in the line x = 1 (think about the expression x-1: x = 2 would give 1, and for y = 1 - x would give -1 and so on).

I'm pretty certain I'm right here (but I could be making a mistake). Just draw y = x and y = 1-x and you'll see what I mean.


In that post you wrote:
f(-(x-1)) is translating x one to the right and then reflecting in the line x = 1.


The graph of f(x)=x(12−x2) f(x) = x(12-x^2) looks like this:




The graph of f(x−1) f(x-1) looks like this,



The graph of f(−(x−1))=f(1−x) f(-(x-1)) = f(1-x) looks like this,



At the end, i will have to agree with you here :smile:

Though i am slightly confused why my graph transforming technique fails.

And another thing i have noticed is. If you first reflect the graph of f(x) f(x) in the y-axis and then shift it by 1 units towards the right, this gives the correct graph.
Original post by raheem94

And another thing i have noticed is. If you first reflect the graph of f(x) f(x) in the y-axis and then shift it by 1 units towards the right, this gives the correct graph.


Yep because we have -x so reflect then add 1 (which is moving to the right when we have -x)
Reply 31
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raheem94
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Original post by hassi94
Yep because we have -x so reflect then add 1 (which is moving to the right when we have -x)


Thanks, i see where the problem was.

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