Answer:
[tex]y(x)=-\frac{1}{2}x-5[/tex]Explanation: We have to find the standard equation of a line that passes through the two points, (-4,-3) and (8,-9):
The stand line equation:
[tex]\begin{gathered} y(x)=mx+b\Rightarrow(1) \\ m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1} \end{gathered}[/tex]The slope and the y-intercept of the equation are determined as follows:
[tex]\begin{gathered} (x_1,y_1)=(-4,-3) \\ (x_2,y_2)=(8,-9) \end{gathered}[/tex]The slope:
[tex]\begin{gathered} m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-9-(-3)}{8-(-4)} \\ m=\frac{-9-(-3)}{8-(-4)}=\frac{-6}{12} \\ m=-\frac{1}{2} \end{gathered}[/tex]y-intercept:
[tex]\begin{gathered} y(x)=-\frac{1}{2}x+b \\ (x,y)=(-4,-3) \\ -3=-\frac{1}{2}(-4)+b \\ -3=2+b \\ b=-5 \end{gathered}[/tex]The equation finally is as follows:
[tex]y(x)=-\frac{1}{2}x-5\Rightarrow(2)[/tex]The plot:
Follwoing plot confirms that the equation passes through the points (-4,-3) and (8,-9):
