1. **State the problem:** We are given a parabola with vertex at $(3,1)$ and need to find the x-intercepts $p$ and $q$ of the parabola. 2. **Identify the vertex form of the parabola:** The vertex form is $$y = a(x - h)^2 + k$$ where $(h,k)$ is the vertex. Here, $h=3$ and $k=1$, so $$y = a(x - 3)^2 + 1.$$ 3. **Find the value of $a$ using a point on the parabola:** From the graph, the parabola passes through $(0,8)$ (since at $x=0$, $y=8$). Substitute: $$8 = a(0 - 3)^2 + 1 = a(9) + 1,$$ which gives $$8 - 1 = 9a \\ 7 = 9a \\ a = \frac{7}{9}.$$ 4. **Write the full equation:** $$y = \frac{7}{9}(x - 3)^2 + 1.$$ 5. **Find the x-intercepts $p$ and $q$ by setting $y=0$:** $$0 = \frac{7}{9}(x - 3)^2 + 1 \\ \frac{7}{9}(x - 3)^2 = -1.$$ Since the right side is negative and the left side is always non-negative (square term), this equation has no real solutions. This contradicts the problem statement that $p$ and $q$ are x-intercepts. 6. **Re-examine the vertex y-value:** The vertex is at $(3,1)$, so the parabola opens upwards and the minimum y-value is 1, meaning it never crosses the x-axis. Therefore, there are no real x-intercepts $p$ and $q$. 7. **Conclusion:** The parabola does not intersect the x-axis, so $p$ and $q$ do not exist as real numbers. **Final answer:** There are no real x-intercepts $p$ and $q$ for the given parabola.