Question

Solve the system of equations. Verify each solution graphically.

y = 2x + 4

y = x2 − 4

Answer 1

To solve the system of equations:

1. **Set the equations equal to each other:**
\[
2x + 4 = x^2 - 4
\]

2. **Rearrange to form a standard quadratic equation:**
\[
x^2 - 2x - 8 = 0
\]

3. **Factor the quadratic equation:**
To factor \(x^2 - 2x - 8\), we look for two numbers that multiply to \(-8\) and add to \(-2\):
\[
(x - 4)(x + 2) = 0
\]

4. **Solve for \(x\):**
Setting each factor equal to zero gives us:
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4
\]
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\]

5. **Find the corresponding \(y\) values:**
For \(x = 4\):
\[
y = 2(4) + 4 = 8 + 4 = 12
\]
So, one solution is \((4, 12)\).

For \(x = -2\):
\[
y = 2(-2) + 4 = -4 + 4 = 0
\]
So, another solution is \((-2, 0)\).

The solutions to the system of equations are:
\[
(4, 12) \quad \text{and} \quad (-2, 0)
\]

### Verification Graphically

- **Graph the first equation \(y = 2x + 4\) (a straight line):**
- This line has a slope of 2 and a y-intercept of 4.

- **Graph the second equation \(y = x^2 - 4\) (a parabola):**
- This parabola opens upwards with a vertex at \((0, -4)\).

### Systems of Equations Graph

To visualize this, plot the following points:
- For \(y = 2x + 4\):
- When \(x = 0\), \(y = 4\) → point \((0, 4)\)
- When \(x = 2\), \(y = 8\) → point \((2, 8)\)
- When \(x = 4\), \(y = 12\) → point \((4, 12)\)

- For \(y = x^2 - 4\):
- When \(x = -2\), \(y = 0\) → point \((-2, 0)\)
- When \(x = 0\), \(y = -4\) → point \((0, -4)\)
- When \(x = 2\), \(y = 0\) → point \((2, 0)\)
- When \(x = 4\), \(y = 12\) → point \((4, 12)\)

Plot these points on the same graph, and you will see:
- The line \(y = 2x + 4\) crosses the parabola \(y = x^2 - 4\) at the points \((4, 12)\) and \((-2, 0)\), confirming that the solutions are correct.

The solutions \((4, 12)\) and \((-2, 0)\) have been verified.