# How do you find the critical numbers of an absolute value equation f(x) = |x + 3| - 1?

May 20, 2017

There is one critical point: $x = - 3 , y = - 1$
Critical points are minima, maxima, and points where $f ' \left(x\right) = 0$.
Since this is a v-shaped absolute value function, there is no point where f'(x) = 0. However, there is one extremum. It is a minimum located at $x = - 3$, and it is the point $\left(- 3 , - 1\right)$.
You can figure this out by looking at how $f \left(x\right) = | x + 3 | - 1$ is a modification of its parent function $f \left(x\right) = | x |$. It is shifted left 3 and down 1, so the local minimum from the parent function is also shifted left 3 and down 1.