Observations:
- The parabola opens upwards (the arms go upward).
- The vertex of the parabola is at the origin $(0,0)$.
- The parabola is symmetric about the y-axis.
- When $x=1, y=1$ approximately, and when $x=-1, y=1$, which fits the form $y=x^2$.

Checking the options:
- (A) $\(y^2+x^2=5\)$ : This is a circle centered at the origin with radius $\(\sqrt{5}\)$, not a parabola.
- (B) $\(y=(x-5)^2\)$ : This parabola has vertex at $(5,0)$, which does not match the graph.
- (C) $\(y=x^2+4\)$ : This parabola opens upwards but with vertex at $(0,4)$, so it is shifted up, not touching the origin.
- (D) $\(y+4+x^2=0\)$ which means $\(y=-x^2-4\)$ : This parabola opens downwards and shifted down, not matching the graph.
- (E) $\(y+4+x=0\)$ or $\(y=-x-4\)$ : This is a straight line, not a parabola.

Conclusion:
The parabola shown in the image fits best to the basic form $y=x^2$, which is closest to Option (C) if we ignore the +4 (since $y=x^2+4$ is shifted up).

Since none of the given options exactly matches $\(y=x^2\)$, and (C) is shifted up, and (A), (B), (D), and (E) are clearly different shapes or positions, the best match is (C) $\(y=x^2+4\)$ assuming the scale might be different or the parabola is shifted slightly.

If the exact vertex is at the origin, the parabola is $\(y=x^2\)$, but since that option is not given, pick option:

$$
\(\text { (C) } y=x^2+4\)
$$