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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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    I don't understand the question...2011-02-26
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    edited the question2011-02-26
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    Still don't understand the question. What problem you want to solve? What exercise book you refer to? What does the title mean? Do you want to find values m and n such that $|m x + n|$ fulfills what?2011-02-26
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    You should explain what it is that you don't understand more precisely. Also, the title does not make a lot of sense. Edit: Your rephrasing of the question is helpful. You should consider the meaning of the coefficients $m$ and $n$. Maybe you could ignore the absolute value for a moment and just focus on the line $y = mx + n$ itself. How does the line change if you fix one constant and change the other?2011-02-26
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    i've edited the questionn2011-02-26
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    @Stijn: that should have been a comment.2011-02-26
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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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    There are several things that are wrong here. First, I find it confusing that a square in the picture does not have two equal sides, second, your solution gives for $x = 0$ the value $3/2$ while the OPs graph has $3$ for $x = 0$.2011-02-26
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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).