To solve this system of equations, we need to identify where the two curves intersect.

The two equations are:

1. $y = x^2$ (a parabola)
2. $y = 4$ (a horizontal line)

The solution(s) will be the points where the graphs of these two equations intersect. From the graph:

• The parabola $y = x^2$ opens upwards and passes through the origin.
• The line $y = 4$ is a horizontal line.

To find the intersections, we set the two equations equal to each other:

$x^2 = 4$

Solving for $x$:

$x = \pm 2$

Thus, the points of intersection are $(-2, 4)$ and $(2, 4)$.

Therefore, the correct answer is B. (-2, 4) and (2, 4).

Would you like any more details or have questions?

Here are five related questions you might find useful:

1. What are the steps to solve a system of non-linear equations algebraically?
2. How do you determine the number of solutions for a quadratic and linear system?
3. What other methods can be used to solve a system of equations graphically?
4. Can the quadratic equation have fewer or more than two solutions when intersecting a line?
5. How does the vertex of a parabola help in identifying possible solutions?

Tip: Always check both algebraic and graphical solutions to confirm the accuracy of intersections.