Why is there $(2,3)$ corner coordinates on the graph, using this function: $f(x)=2x-1+|x-2|?$ Currently I am studying about absolute values and I had to consider this function and draw it's graph:
$$f(x)=2x-1+|x-2|$$
I helped myself with symbolab, here is the link: https://www.symbolab.com/solver/step-by-step/f%5Cleft(x%5Cright)%3D2x-1%2B%5Cleft%7Cx-2%5Cright%7C/?origin=button
I understood everything about getting the $x$ coordinate $(-1, 0)$ and $y$ coordinate $(0,1)$ but I don't see where did $(2,3)$ corner coordinates come from or what does it mean - could anyone please tell that to me$?$
I am also including the picture: graph and it's coordinates
 A: $(2,3)$ is clearly a point on the graph because it satisfies the equation:
$$3=2\cdot2-1+|2-2|$$
The reason why the graph has an angle at that point is because it is the minimum of the second part of the expression, $|x-2|.$ Compare this to the minimum of the function $|x|$ itself at $x=0.$
A: Since 
$$|x - 2| = \begin{cases}
            x - 2 & \text{if $x \geq 2$}\\
            -(x - 2) & \text{if $x < 2$}
            \end{cases}
$$
the function $f(x) = 2x - 1 + |x - 2|$ can be defined piecewise as 
\begin{align*}
f(x) & = \begin{cases}
        2x - 1 + x - 2 & \text{if $x \geq 2$}\\
        2x - 1 - (x - 2) & \text{if $x < 2$}
        \end{cases}\\
     & = \begin{cases}
         3x - 3 & \text{if $x \geq 2$}\\
         x + 1 & \text{if $x < 2$}
         \end{cases}
\end{align*}
Thus, if $x > 2$, the graph has slope $3$, while the graph has slope $1$ if $x < 2$.  The graph has a corner at $x = 3$ since the slope to the right of $x = 2$ is different from the slope to the left of $x = 2$.
