Given the equation of a straight line
[tex]y=mx+c[/tex]m is the slope and c is the intercept on the y axis
From the table in the question, the equation of the line can be obtained using two points form the equation of a straight line
[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]Pick two points on the table and substitute for the x values and y values
[tex](-6,-24)\text{ and (3, -12)}[/tex][tex]\begin{gathered} x_1=-6,y_1=-24 \\ x_2=-3,y_2=-12 \end{gathered}[/tex][tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{-12-(-24)}{-3-(-6)}=\frac{-12+24}{-3+6}=\frac{12}{3}=4 \end{gathered}[/tex][tex]\begin{gathered} y=mx+c \\ -24=4(-6)+c \\ -24=-24+c \\ c=-24+24=0 \end{gathered}[/tex]Hence, the equation of the table is y = 4x