# How do you graph f(x)=abs(x-2)+ 3?

Mar 30, 2015

Note that $x = 2$ is a critical point of the function $f \left(x\right)$ since for $x < 2$ the absolute value operation reverses the sign and for $x > 2$ the absolute value function has no effect.

Evaluate f(x) for $x = 0$ and for one value $< 2$ (say $x = 0$) and for one value $> 2$ (say $x = 4$)

$f \left(0\right) = 5$
$f \left(2\right) = 3$
$f \left(4\right) = 5$

draw a line from $\left(2 , 3\right)$ (the critical point) through $\left(0 , 5\right)$
and
a second line from $\left(2 , 3\right)$ (the critical point) through $\left(4 , 5\right)$
graph{|x-2|+3 [-5.03, 7.46, 0.4, 6.64]}