To determine which graph represents the solution set of the equations \( y = x^2 - 4 \) and \( x + y + 2 = 0 \), let’s solve these equations step by step.
### 1. Expressing \( y \) from one of the equations:
From the second equation \( x + y + 2 = 0 \):
[tex]\[ y = -x - 2 \][/tex]
### 2. Substitute \( y = -x - 2 \) into the first equation \( y = x^2 - 4 \):
[tex]\[
-x - 2 = x^2 - 4
\][/tex]
### 3. Rearrange the equation:
[tex]\[
x^2 + x - 2 = 0
\][/tex]
### 4. Solve the quadratic equation \( x^2 + x - 2 = 0 \):
To solve \( x^2 + x - 2 = 0 \), we can factorize it as:
[tex]\[
(x + 2)(x - 1) = 0
\][/tex]
So, the solutions for \( x \) are:
[tex]\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\][/tex]
[tex]\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\][/tex]
### 5. Find the corresponding \( y \)-values using \( y = -x - 2 \):
For \( x = -2 \):
[tex]\[
y = -(-2) - 2 = 2 - 2 = 0
\][/tex]
For \( x = 1 \):
[tex]\[
y = -(1) - 2 = -1 - 2 = -3
\][/tex]
### 6. Summary of solutions:
The solutions to the system of equations are:
[tex]\[
(x, y) = (-2, 0) \quad \text{and} \quad (1, -3)
\][/tex]
### Interpretation:
On a graph, this means the points \((-2, 0)\) and \((1, -3)\) must satisfy both equations.
To decide between graph II and graph III, you should check these points on the graphs. The graph that shows these intersection points accurately represents the solution set of the given equations.
