Answer: We have to find the algebraic solution for the system of equations:
[tex]\begin{gathered} y=-2x-4\Rightarrow(1) \\ \\ y=x^2+6x+8\Rightarrow(2) \end{gathered}[/tex]
The solution is simply a point where (1) and (2) have the same coordinates, which basically means that we have to set two equations equal, therefore the solution is as follows:
[tex]\begin{gathered} (1)=(2) \\ \\ -2x-4=x^2+6x+8 \\ \\ x^2+8x+12=0\Rightarrow(3) \end{gathered}[/tex][tex]\begin{gathered} (1)=(2) \\ \\ -2x-4=x^2+6x+8 \\ \\ x^2+8x+12=0\Rightarrow(3) \end{gathered}[/tex]
The solution to (3) is determined through a quadratic formula, the steps are as follows:
[tex]\begin{gathered} \begin{equation*} x^2+8x+12=0 \end{equation*} \\ \\ \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ \\ a=1,b=8,c=12 \\ \\ x=\frac{-(8)\pm\sqrt{(8)_^2-4(1)(12)}}{2(1)}=\frac{-8\pm\sqrt{64-48}}{2} \\ \\ \text{ The solutions are:} \\ \\ x=\frac{-8+\sqrt{16}}{2}=\frac{-8+4}{2}=\frac{4}{2}=-2\Rightarrow(i) \\ \\ x=\frac{-8-\sqrt{64-48}}{2}=\frac{-8-4}{2}=-\frac{12}{2}=-6\Rightarrow(ii) \end{gathered}[/tex]
Therefore the values of the two functions at these two x values are as follows:
[tex]\begin{gathered} y(-2)=-2(-2)-4=0 \\ \\ y(-6)=-2(-6)-4=8 \\ \\ -------------------------- \\ y(-2)=(-2)^2+6(-2)+8=4-12+8=0 \\ \\ y(-6)=(-6)^2+6(-6)+8=36-36+8=8 \\ \end{gathered}[/tex]
The two solutions therefore are:
[tex]\begin{gathered} (x,y)=(-2,0)\Rightarrow\text{ Solution First} \\ \\ (x,y)=(-6,8)\Rightarrow\text{ Solution Second} \\ \end{gathered}[/tex]
The graph confirmation:
