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The diagram shows the graph of $y=|mx+n|$

(i tried my best to do the same thing as my exercise book, actually 1 is propotional to 1 and 3 is propotional to 3, but 2 is not propotional to 2)

Find the value of $m$ and $n$.

Anyone can help me solve this problem? Thanks

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    Rather than say "1 is propotional to 1 and 3 is propotional to 3, but 2 is not propotional to 2" it would be more accurate to say (1,1) and (3,3) are on the graph, but (2,2) is not. Being on the graph is quite different from being proportional2011-02-27

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Hints: What value inside the absolute value sign can make $y=0$? If you set $x=0$, what is $y$?

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enter image description here

  • Hint. This is the graph of $y=|x|$. The figure which you have has shifted the bend by $\frac{3}{2}$ units on the right hand side. So i think your graph should $y= | x - \frac{3}{2}|$.
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    @Theo: May be the OP is giving an approx idea of how the graph looks like. From that, i could only figure this out2011-02-26
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If it was without modulus graph will go through Oy at (0;-3) so n=-3, I assume we know that it touch Ox on (1.5;0) so

y=mx+n 0=1.5m-3 m=2 

y=|2x-3|

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HINT $\ $ Inspecting the diagram reveals the slope $\rm\:m\:$ and $\rm\:y$-intercept $\rm\:n\:$ of the line segment(s).