First, we can take the points (0,10), (-4,8) and (2,11) to find the slope, and then find the linear equation using the third point.
[tex]\begin{gathered} (x_1,y_1)=(0,10) \\ (x_2,y_2)=(-4,8) \\ m=\frac{8-10}{-4-0}=\frac{-2}{-4}=\frac{1}{2} \\ m=\frac{1}{2} \end{gathered}[/tex]
now that we have that the first slope is m = 1/2, we can use the point (2,11) to find the linear equation:
[tex]\begin{gathered} y-11=\frac{1}{2}(x-2)=\frac{1}{2}x-1 \\ \Rightarrow y=\frac{1}{2}x-1+11=\frac{1}{2}x+10 \\ y=\frac{1}{2}x+10 \end{gathered}[/tex]
we can see that the three points fit the equation:
[tex]\begin{gathered} (0,10) \\ 10=\frac{1}{2}(0)+10\Rightarrow10=10 \\ (-4,8) \\ 8=\frac{1}{2}(-4)+10=-2+10=8 \\ (2,11) \\ 11=\frac{1}{2}(2)+10=1+10=11 \end{gathered}[/tex]
if we do the same with the points (-3,-3), (-1,1) and (2,7), we get:
[tex]\begin{gathered} m=\frac{1-(-3)}{-1-(-3)}=\frac{1+3}{-1+3}=\frac{4}{2}=2 \\ \Rightarrow y-7=2(x-2)=2x-4 \\ \Rightarrow y=2x-4+7=2x+3 \\ y=2x+3 \end{gathered}[/tex]
therefore, the two linear functions found are y = 1/2 x +10 and y=2x+3
