Understanding $y=|mx+n|$

Question

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

Answer 1

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

Answer 2

• 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 $y= | x - \frac{3}{2}|$.

Answer 3

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

Answer 4

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|