modulus question help - a level maths

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Bertybassett
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Hi, in regards to question 10.) ii.) on the following paper, I'm confused as to how to work it out. what I usually do is find the equation of the line in the reflection of the line in which the line if "reflected" (if you know what I mean) and then solve it as simultaneous equations. However, when the equation of the graph is in the form that it is, how do you find the equation of the second "part of the line" (would the equation of part of the line with positive gradient be y=x+1? Basically I'm just confused on how to solve it. https://pmt.physicsandmathstutor.com...dulus%20QP.pdf
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RDKGames
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(Original post by Bertybassett)
Hi, in regards to question 10.) ii.) on the following paper, I'm confused as to how to work it out. what I usually do is find the equation of the line in the reflection of the line in which the line if "reflected" (if you know what I mean) and then solve it as simultaneous equations. However, when the equation of the graph is in the form that it is, how do you find the equation of the second "part of the line" (would the equation of part of the line with positive gradient be y=x+1? Basically I'm just confused on how to solve it. https://pmt.physicsandmathstutor.com...dulus%20QP.pdf
I'm sure you know that a modulus function like y=|x| can actually be split into two functions for two joining regions like so: \displaystyle \begin{cases} x, & x \geq 0 \\ -x, & x <0.

You can do the same with y = |x-2|+1, this just means \displaystyle \begin{cases} (x-2)+1, & x \geq 2 \\ -(x-2)+1, & x <2

So then, what are the parts of both modulus curves that intersect?? It would be the one with the +ve gradient from y=|x| and the one from -ve gradient from y=|x-2| +1.

Can you see what your simultaneous equations are?
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