Alright, let's solve the given system of equations algebraically.
We are given two equations:
1. [tex]\( y = 2x + 4 \)[/tex]
2. [tex]\( y = x^2 + 6x + 8 \)[/tex]
To find the points of intersection between these two equations, we set them equal to each other, as they both represent [tex]\( y \)[/tex]. Therefore:
[tex]\[ 2x + 4 = x^2 + 6x + 8 \][/tex]
Now, let's solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 + 6x + 8 = 2x + 4 \][/tex]
Subtract [tex]\( 2x + 4 \)[/tex] from both sides to set the equation to zero:
[tex]\[ x^2 + 6x + 8 - 2x - 4 = 0 \][/tex]
Combine like terms:
[tex]\[ x^2 + 4x + 4 = 0 \][/tex]
This is a quadratic equation. We can simplify it by factoring:
[tex]\[ (x + 2)^2 = 0 \][/tex]
To solve for [tex]\( x \)[/tex], we set the factor equal to zero:
[tex]\[ x + 2 = 0 \][/tex]
Thus:
[tex]\[ x = -2 \][/tex]
Now that we have the [tex]\( x \)[/tex]-coordinate of the intersection point(s), we substitute [tex]\( x = -2 \)[/tex] back into one of the original equations to find the corresponding [tex]\( y \)[/tex]-coordinate. Let's use the first equation [tex]\( y = 2x + 4 \)[/tex]:
[tex]\[ y = 2(-2) + 4 \][/tex]
[tex]\[ y = -4 + 4 \][/tex]
[tex]\[ y = 0 \][/tex]
So, the point of intersection is [tex]\((-2, 0)\)[/tex].
There is only one intersection point, so for the second point, we will type "none".
Therefore, the solution to the system of equations is:
[tex]\[ (-2, 0) \][/tex]
[tex]\[ \square \text{ none} \][/tex]
