Graph of |y|=F(x)

This is the image of $|y|=F(x)$ , here the dotted part was the part of the original F(x) is the part that was cut off.

$$ y = \begin{cases} F(x), & \text{if $F(x) \ge 0$} \\ -F(x), & \text{if $F(x) < 0$} \end{cases} $$

if we use this then I afraid we get the wrong graph :?

Our graph in that case would like this : Mod F(x)

I am sorry if it is a silly doubt (Maybe am making a silly mistake :/) !


1 Answer 1


The graphs in the pictures are correct.

You have to interpret $|y|=f(x)$ and $y=|f(x)|$ as relations between real numbers. These relations can be graphed in the plane. They are different relations, as the pictures show.

When $f(x)<0$, there are no points $(x,y)$ such that $|y|=f(x)$.

When $f(x)\ge 0$, there are two points $(x,y)$ such that $|y|=f(x)$. They are $(x,\pm f(x))$.

  • $\begingroup$ when $F(x)<0$ it will become $y = -F(x)$ in that case , $(x,-y)$ would exist $\endgroup$
    – brainst
    Apr 1, 2016 at 11:50
  • $\begingroup$ Is it ... $F(x) = |y|$ therefore as the RHS is always positive the LHS should also be always positive ..... then I'm definitely wrong :( $\endgroup$
    – brainst
    Apr 1, 2016 at 11:52

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