Rationale
(2, 4)
The function f is even, as indicated by the property f(x) = f(-x) for all real numbers x. This symmetry means that if the point (-2, 4) is on the graph of y = f(x), then the corresponding point (2, 4) must also be present due to the nature of even functions.
A) (-2, -4)
This point does not satisfy the property of the function. Since f(-2) = f(2) and we know f(-2) = 4, it follows that f(2) must also equal 4, not -4. Therefore, (-2, -4) cannot be on the graph.
B) (0, 0)
While the point (0, 0) could lie on the graph of an even function, there is no given information to confirm its presence. The function could take on any value at x = 0, and without specific information about f(0), we cannot assert that (0, 0) is on the graph.
C) (0, 4)
Similar to (0, 0), the point (0, 4) may or may not be on the graph. The evenness of the function does not imply that f(0) equals 4 unless explicitly stated; hence we cannot conclude its presence on the graph.
D) (2, -4)
This point contradicts the even function property. Since f(-2) = 4, it follows that f(2) = 4 as well. Therefore, (2, -4) cannot be a valid point on the graph of y = f(x).
E) (2, 4)
This point must lie on the graph, as established by the even function property. Since the function is symmetric and f(-2) = 4, it follows that f(2) must also equal 4, confirming that (2, 4) is indeed on the graph.
Conclusion
The even nature of the function f ensures that for any point (a, b) on the graph, the point (-a, b) is also present. Given that (-2, 4) is on the graph, the corresponding point (2, 4) must also be included, while the other options either contradict the function's properties or lack sufficient information to confirm their inclusion.
