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Transformation of graphs - C3

Hello everyone,

I'm getting confused with the order to perform graph transformations when there are two or more to perform. Is there an easy set of rules to follow to get them right all the time?

Thanks in advance
Original post by Electrogeek
Hello everyone,

I'm getting confused with the order to perform graph transformations when there are two or more to perform. Is there an easy set of rules to follow to get them right all the time?

Thanks in advance


Set of rules? No, this isn't Decision maths.

Just put a bracket around x and that can help transform it in a more accessible way for you.
Both inside the brackets? Shift first.
Both outside the brackets? Stretch first.
One in, one out? Order doesn't matter.
Reply 3
Maybe if you provide a specific example?
Not my thread but oh well. These... :smile:

y= |x^2 + x - 3| and y= |x|^2 + |x| - 3
Reply 5
Original post by Ze Witcher
Not my thread but oh well. These... :smile:

y= |x^2 + x - 3| and y= |x|^2 + |x| - 3


What about them?
Original post by Ze Witcher
Not my thread but oh well. These... :smile:

y= |x^2 + x - 3| and y= |x|^2 + |x| - 3


First one: the graph is 'reflected' (so to speak) in y=0 because y is always positive.
Second one: The graph is reflected in x=0 because x is always positive.

Graph them both to see what I mean.
Original post by B_9710
What about them?


Although they're the same equation, why does the |x| in the second one change the shape of the function?
Original post by RDKGames
First one: the graph is 'reflected' (so to speak) in y=0.
Second one: The graph is reflected in x=0.

Graph them both to see what I mean.


ooh so they're reflections. Thanks i'll try it:smile:
Reply 9
Original post by Ze Witcher
Although they're the same equation, why does the |x| in the second one change the shape of the function?


Because the |x| term is always positive but if it was just x on its own then it can be positive or negative depending if x>0 or x<0.
For the curve x2+x3 |x|^2 + |x| -3 you have to consider two separate cases, when x<0 x<0 and when x0 x\geq 0 .
(edited 7 years ago)
Original post by Ze Witcher
ooh so they're reflections. Thanks i'll try it:smile:
I wouldn't use that word lightly. Just consider what the modulus function does, and what you are applying it to.

Pics below are for f(x)=x22x1f(x)=x^2-2x-1 then f(x)=x22x1\lvert{f(x)}\lvert=\lvert{x^2-2x-1}\lvert and finally f(x)=x22x1f(\lvert x \lvert)=\lvert x\lvert^2-2\lvert x\lvert-1
(edited 7 years ago)
Thanks for the help! I get it a lot better now. I shall have a go at a few questions and see how they go. :smile:
Original post by Ze Witcher
Although they're the same equation, why does the |x| in the second one change the shape of the function?


They aren't the same equation though. You have given two *different* composite functions, namely f(g(x))f(g(x)) and g(f(x))g(f(x)) where g(x)=x,f(x)=x2+x3g(x)=|x|, f(x)=x^2+x-3

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