Answer
Yes.
$$y= \begin{cases} \sqrt{25-x^2} & \text{ if } -5 \le x \le 0 \\ - \sqrt{25-x^2} & \text{ if } 0 < x \le 5 \end{cases}$$
Work Step by Step
Yes, it is a function.
This graph consists of two sections of the circle $x^2+y^2=25$:
First, the left half of the upper semicircle $y= \sqrt{25-x^2}$, including the points $(0,5)$ and $(-5,0)$, that is,$$y=\sqrt{25-x^2} \qquad -5 \le x \le 0.$$Second, the right half of the lower semicircle $x^2+y^2=25$, including the point $(5,0)$ but not the point $(0,-5)$, that is,$$y=- \sqrt{25-x^2} \qquad 0 \le x \le 5.$$
Thus, the function representing this graph is$$y= \begin{cases} \sqrt{25-x^2} & \text{ if } -5 \le x \le 0 \\ - \sqrt{25-x^2} & \text{ if } 0 < x \le 5 \end{cases}.$$
