# Modulus Graph

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why is the graph for y=2x+1 on the negative side of the quadrant

does a modulus mean that the graph has a v shape. what are the rules for that?

Question: question 4http://pmt.physicsandmathstutor.com/download/Maths/A-level/C3/Papers-OCR-MEI/June%202016%20QP%20-%20C3%20OCR%20MEI.pdf

Answer:

http://pmt.physicsandmathstutor.com/...0OCR%20MEI.pdf

does a modulus mean that the graph has a v shape. what are the rules for that?

Question: question 4http://pmt.physicsandmathstutor.com/download/Maths/A-level/C3/Papers-OCR-MEI/June%202016%20QP%20-%20C3%20OCR%20MEI.pdf

Answer:

http://pmt.physicsandmathstutor.com/...0OCR%20MEI.pdf

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(Original post by

why is the graph for y=2x+1 on the negative side of the quadrant

does a modulus mean that the graph has a v shape. what are the rules for that?

Question: question 4http://pmt.physicsandmathstutor.com/download/Maths/A-level/C3/Papers-OCR-MEI/June%202016%20QP%20-%20C3%20OCR%20MEI.pdf

Answer:

http://pmt.physicsandmathstutor.com/...0OCR%20MEI.pdf

**esmeralda123**)why is the graph for y=2x+1 on the negative side of the quadrant

does a modulus mean that the graph has a v shape. what are the rules for that?

Question: question 4http://pmt.physicsandmathstutor.com/download/Maths/A-level/C3/Papers-OCR-MEI/June%202016%20QP%20-%20C3%20OCR%20MEI.pdf

Answer:

http://pmt.physicsandmathstutor.com/...0OCR%20MEI.pdf

First draw f(x)=2x+1 which is a straight line then |f(x)| = |2x+1|

So this is a f(x) -> |f(x)| transformation which reflects anything below the x-axis in the x-axis.

The unreflected part of the graph has equation y=2x+1 and the reflected part has equation y=-(2x+1).

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#3

Is the modulus around the entire function, like |2x+1|? Then that would mean, as I think about it, "Only accept positive values of the output and force every negative value to be positive". Because |this| means the absolute value (so the absolute value of -6 = 6).

So, if you imagine the graph of f(x) = 2x+1, if you only took the absolute value of your output, |2x+1|, you would reflect the negative part of your output in the x-axis in order to actually get the absolute value once you get to the point where 2x+1 would be giving negative outputs. In my head I go from right to left.

So how do you solve equations with it, since it's not like a normal function that you can just rearrange?

Well, you can, for one of the roots. For the other you have to treat it differently.

Once the graph has "bounced" off the x-axis, it's the equivalent of the function -f(x). So you would look for the intercept of -f(x) and -x as well as the intercept of f(x) and -x.

So:

2x+1=-x

and

-(2x+1)=-x

Now, an equation like this can have no roots, such as |2x| = 0.5x-6. This is why you need to sketch when doing these problems.

So, if you imagine the graph of f(x) = 2x+1, if you only took the absolute value of your output, |2x+1|, you would reflect the negative part of your output in the x-axis in order to actually get the absolute value once you get to the point where 2x+1 would be giving negative outputs. In my head I go from right to left.

So how do you solve equations with it, since it's not like a normal function that you can just rearrange?

Well, you can, for one of the roots. For the other you have to treat it differently.

Once the graph has "bounced" off the x-axis, it's the equivalent of the function -f(x). So you would look for the intercept of -f(x) and -x as well as the intercept of f(x) and -x.

So:

2x+1=-x

and

-(2x+1)=-x

Now, an equation like this can have no roots, such as |2x| = 0.5x-6. This is why you need to sketch when doing these problems.

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