1. **Problem Statement:** We analyze a graph of a downward-opening parabola starting near $x=0$, peaking near $x=2$, and falling by $x=4$. 2. **Determine if the graph represents a function:** A graph represents a function if every $x$-value corresponds to exactly one $y$-value. 3. **Check the vertical line test:** Since the graph is a parabola opening downward, each vertical line intersects the curve at most once. Therefore, it passes the vertical line test and is a function. 4. **Domain of the function:** The parabola starts near $x=0$ and ends near $x=4$, so the domain is approximately $$0 \leq x \leq 4$$ 5. **Range of the function:** The parabola opens downward with a peak near $x=2$. Let the peak $y$-value be $y_{max}$. The range is all $y$-values from the minimum at the ends up to the maximum at the peak. Assuming the minimum $y$-value is about $-6$ (from the problem context), the range is $$-6 \leq y \leq y_{max}$$ 6. **Find $y$-value(s) when $x=2$:** At the peak $x=2$, the parabola has a single $y$-value, the maximum $y_{max}$. 7. **Find $x$-value(s) when $y=-6$:** Since the parabola opens downward and $y=-6$ is near the minimum, there are two $x$-values where $y=-6$. These correspond to the start and end points of the parabola, approximately $x=0$ and $x=4$. **Final answers:** - The graph represents a function. - Domain: $$0 \leq x \leq 4$$ - Range: $$-6 \leq y \leq y_{max}$$ - $y$ at $x=2$ is $y_{max}$ (the peak). - $x$ at $y=-6$ are approximately $0$ and $4$.