You need to find the functions for each part of the graph.
All 3 parts are linear functions, given by the general form
[tex]y=mx+b\begin{cases}m=\text{slope} \\ b=y-\text{intercept}\end{cases}[/tex]
You can find the slope as follows:
[tex]m=\frac{y2-y1}{x2-x1}[/tex]
Let's evaluate a pair of points in each part of the graph to find the correspondent slope:
Part 1. Point 1 = (-2, 0) and Point 2= (0, 2)
[tex]m1=\frac{2-0}{0-(-2)}=\frac{2}{2}=1[/tex]
The y-intercept is the point where it crosses the y-axis, for the first part of the graph you can see it at y=2, then b=2
The function of this part is then:
[tex]f(x)=y=1\cdot x+2\text{ for x}\leq0[/tex]
Part 2. Point 1 = (0, 1) and Point 2= (1, 1.4)
[tex]m2=\frac{1.4-1}{1-0}=\frac{0.4}{1}=0.4[/tex]
The y-intercept is at b=1, then the function for the second part is:
[tex]f(x)=y=0.4x+1\text{ for 0
Part 3. Point 1 = (3, -6) and Point 2= (4, -7)[tex]m3=\frac{-7-(-6)}{4-3}=\frac{-7+6}{1}=-1[/tex]
In this case, we can't see the point where the function crosses the y-axis, thus we need to evaluate the general form of the linear equation, replacing the slope and one of the known points
[tex]\begin{gathered} y=mx+b\text{ replace the values} \\ -6=-1\cdot3+b\text{ add +3 to both sides} \\ -6+3=-3+3+b \\ -3=b\text{ reorder the terms} \\ b=-3\text{ this is the y-intercept for the third part of the function} \end{gathered}[/tex]
The function for the third part is:
[tex]f(x)=y=-1x-3\text{ for x>2}[/tex]
