To solve the given system of equations:
1. [tex]\( y = x + 4 \)[/tex]
2. [tex]\( y = x^2 + 2 \)[/tex]
We need to find the points of intersection, which means we set the two equations equal to each other.
[tex]\[ x + 4 = x^2 + 2 \][/tex]
Re-arrange the equation to standard quadratic form:
[tex]\[ x^2 - x - 2 = 0 \][/tex]
We solve this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -2 \)[/tex].
Calculate the discriminant ([tex]\( \Delta \)[/tex]):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = (-1)^2 - 4(1)(-2) \][/tex]
[tex]\[ \Delta = 1 + 8 \][/tex]
[tex]\[ \Delta = 9 \][/tex]
Now, find the roots using the quadratic formula:
[tex]\[ x = \frac{-(-1) \pm \sqrt{9}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{1 \pm 3}{2} \][/tex]
This gives us two solutions for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{1 + 3}{2} = 2 \][/tex]
[tex]\[ x = \frac{1 - 3}{2} = -1 \][/tex]
Now, we'll find the corresponding [tex]\( y \)[/tex] values using the first equation [tex]\( y = x + 4 \)[/tex]:
For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = 2 + 4 = 6 \][/tex]
For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = -1 + 4 = 3 \][/tex]
Therefore, the points of intersection are:
[tex]\[ (-1, 3) \][/tex]
[tex]\[ (2, 6) \][/tex]
Looking at the given answer options:
A. [tex]\((-1, 3)\)[/tex] and [tex]\((2, 2)\)[/tex]
B. [tex]\((1, 3)\)[/tex] and [tex]\((-2, 2)\)[/tex]
C. [tex]\((-1, 3)\)[/tex] and [tex]\((2, 6)\)[/tex]
The correct answer that matches our solution is:
C. [tex]\((-1, 3)\)[/tex] and [tex]\((2, 6)\)[/tex]
