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March 02, 2026, 08:21:10 pm

Author Topic: absolute value graph II  (Read 1085 times)  Share 

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googoo

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absolute value graph II
« on: December 25, 2010, 11:36:40 pm »
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T or F    2-5|f(x)| = |2-5f(x)| ?

taiga

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Re: absolute value graph II
« Reply #1 on: December 25, 2010, 11:38:35 pm »
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F
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Andiio

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Re: absolute value graph II
« Reply #2 on: December 25, 2010, 11:51:41 pm »
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The first one is just the graph of |f(x)| dilated and translated accordingly, and the second one is a whole 'flipping' of the original f(x) function after it is dilated and translated, is that right dev?

Edit: My description is kinda obscure LOL, just realised, ceebs changing it though; I mean the 'flipping' as in the flipping 'over' the imaginary axis though!
« Last Edit: December 25, 2010, 11:54:58 pm by Andiio »
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googoo

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Re: absolute value graph II
« Reply #3 on: December 25, 2010, 11:55:55 pm »
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F

Didn't you say the opposite in absolute value graphs thread.
Let me quote 'All the transformations and dilations mentioned can be done before the "modding" as well'

taiga

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Re: absolute value graph II
« Reply #4 on: December 26, 2010, 12:20:21 am »
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ouch, good job man!

I can't be bothered going back and checking it, but congrats if you're correct :)

But for some specific graphs they can
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m@tty

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Re: absolute value graph II
« Reply #5 on: December 26, 2010, 12:53:36 am »
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T or F    2-5|f(x)| = |2-5f(x)| ?

Test f(x)=1

LHS=2-5=-3

RHS=|2-5|=3

Therefore false by example.
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googoo

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Re: absolute value graph II
« Reply #6 on: December 26, 2010, 08:47:08 am »
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T or F    2-5|f(x)| = |2-5f(x)| ?

Test f(x)=1

LHS=2-5=-3

RHS=|2-5|=3

Therefore false by example.

Thanks for clarifying that. But someone disagreed in the locked Absolute value graphs thread.

Also, I quote
'You're thinking of the word "reflect" in the way a 5 year old would, not how you are supposed to in methods.

Draw y=sin(x)

then y=|sin(x)|

that's not a reflection.

Anyway, I'm going to lock this on the basis of the OP's question being answered'

It is the neg sign and not |f(x)| that reflects f(x).


taiga

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Re: absolute value graph II
« Reply #7 on: December 26, 2010, 02:30:39 pm »
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T or F    2-5|f(x)| = |2-5f(x)| ?

Test f(x)=1

LHS=2-5=-3

RHS=|2-5|=3

Therefore false by example.

Thanks for clarifying that. But someone disagreed in the locked Absolute value graphs thread.

Also, I quote
'You're thinking of the word "reflect" in the way a 5 year old would, not how you are supposed to in methods.

Draw y=sin(x)

then y=|sin(x)|

that's not a reflection.

Anyway, I'm going to lock this on the basis of the OP's question being answered'

It is the neg sign and not |f(x)| that reflects f(x).



I have no idea what you're trying to prove.

I didn't say that was applicable for ALL cases.

I also find it amusing how you managed to agree with my statement about the reflection whilst phrasing it as a disagreement. You deserve two prizes/
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