Graphs with absolute value and the term x times modulus of x

Text

Draw the graph of the following functions and find their intersections with the $x$- and $y$-axes:

• (a) $y=-|2x-1|+3x-4$
• (b) $y=-\dfrac{x\,|x|}{9}+2x+9$
• (c) $y=|x+2|+|x-2|$

Solution

(a) Breakpoint $x=\dfrac12$. $\ x\ge\tfrac12:\ y=x-3$; $\ x<\tfrac12:\ y=5x-5$. $\cap y:(0;-5)$; $\cap x:\ x=3$ (valid), so $(3;0)$ (the left branch’s $x=1$ is discarded, not $<\tfrac12$). (b) Using $x|x|$: $\ x\ge0:\ y=-\dfrac{x^2}{9}+2x+9$ (downward parabola); $\ x<0:\ y=\dfrac{x^2}{9}+2x+9$ (upward parabola). $\cap y:(0;9)$; $\cap x:\ x\ge0:\ x=9+9\sqrt2\approx21{,}73$; $\ x<0:\ (x+9)^2=0\Rightarrow x=-9$. Points $(9+9\sqrt2;0)$ and $(-9;0)$. (c) $x\ge2:\ y=2x$; $\ -2\le x<2:\ y=4$; $\ x<-2:\ y=-2x$. ”Tub” shape: horizontal segment at height $4$ between $-2$ and $2$, and branches $y=2|x|$ outside. $\cap y:(0;4)$; no $x$-intercept (minimum $4$). $\boxed{\text{Remove the absolute values piecewise; intersections: (a) }(3;0),(0;-5);\ (b)\ (-9;0),(9+9\sqrt2;0),(0;9);\ (c)\ (0;4).}$