since we know the function is linear, then we can just use any two points off the table, hmmm say let's use -2, -6 and 2, 14 then,
[tex]\bf \begin{array}{ccll}
x&y\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
-4&-16\\
\boxed{-2}&\boxed{-6}\\
0&4\\
\boxed{2}&\boxed{14}\\
4&24
\end{array}\qquad \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ -2 &,& -6~)
% (c,d)
&&(~ 2 &,& 14~)
\end{array}
\\\\\\
% slope = m
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{14-(-6)}{2-(-2)}\implies \cfrac{14+6}{2+2}\implies \cfrac{20}{4}\implies 5[/tex]
