1. The problem is to analyze the function $f(y) = 3 \cdot (y + 4)^2$ for the domain $0 \leq y \leq 4$. 2. The function is a quadratic function in vertex form: $f(y) = a(y - h)^2 + k$, where $a=3$, $h=-4$, and $k=0$. 3. Since $a=3 > 0$, the parabola opens upwards, meaning the function has a minimum point at $y = -4$. 4. However, the domain is restricted to $0 \leq y \leq 4$, so we only consider values of $y$ in this interval. 5. To find the range on this domain, evaluate $f(y)$ at the endpoints: - At $y=0$: $f(0) = 3 \cdot (0 + 4)^2 = 3 \cdot 16 = 48$ - At $y=4$: $f(4) = 3 \cdot (4 + 4)^2 = 3 \cdot 64 = 192$ 6. Since the parabola opens upward and the vertex $y=-4$ is outside the domain, the minimum value on $[0,4]$ is at $y=0$ with $f(0)=48$, and the maximum is at $y=4$ with $f(4)=192$. 7. Therefore, the range of $f$ on $0 \leq y \leq 4$ is $[48, 192]$. 8. The graph of $f(y)$ is a parabola opening upwards starting at $(0,48)$ and ending at $(4,192)$ within the domain. Final answer: The function $f(y) = 3(y+4)^2$ on $0 \leq y \leq 4$ has range $[48, 192]$.