Graphs of functions with absolute value

Text

Sketch the graph:

• (a) $y=|x-2|(x-2)-4$
• (b) $y=x+|x-2|+1$

Solution

Remove the absolute value case by case, with breakpoint $x=2$.

(a) For $x\ge2$: $y=(x-2)^2-4$, an upward parabolic arc with vertex $(2;-4)$. For $x<2$: $y=-(x-2)^2-4$, a downward arc, again through $(2;-4)$. The graph is continuous at $(2;-4)$ with an “S” shape (a shifted odd cubic). For $x\ge2$ it meets the $x$-axis where $(x-2)^2=4\Rightarrow x=4$.

(b) For $x\ge2$: $y=2x-1$ (slope $2$). For $x<2$: $y=3$ (horizontal). The two pieces join at $(2;3)$: the graph is constant $y=3$ up to $x=2$, then rises along $y=2x-1$.

$\boxed{\text{(a) } y=\begin{cases}(x-2)^2-4 & x\ge2\ -(x-2)^2-4 & x<2\end{cases}\quad \text{(b) } y=\begin{cases}2x-1 & x\ge2\ 3 & x<2\end{cases}}$