1. **State the problem:** We are given the function $f(y) = 3 \cdot (y + 4)^2$ defined for $0 \leq y \leq 4$. We want to understand its behavior and graph shape. 2. **Formula and rules:** The function is a quadratic function in terms of $y$, specifically a vertical stretch of the basic parabola $y^2$ shifted left by 4 units and scaled by 3. 3. **Intermediate work:** - The expression inside the square is $(y + 4)$, so the vertex of the parabola is at $y = -4$, but since the domain is restricted to $0 \leq y \leq 4$, we only consider this interval. - Calculate values at the endpoints: - At $y=0$: $f(0) = 3 \cdot (0 + 4)^2 = 3 \cdot 16 = 48$ - At $y=4$: $f(4) = 3 \cdot (4 + 4)^2 = 3 \cdot 64 = 192$ 4. **Explanation:** The function is increasing on the interval $[0,4]$ because as $y$ increases, $(y+4)^2$ increases. The graph is a portion of a parabola opening upwards, starting at $48$ when $y=0$ and rising to $192$ when $y=4$. **Final answer:** The function $f(y) = 3 \cdot (y + 4)^2$ on $0 \leq y \leq 4$ is an increasing parabola segment starting at $48$ and ending at $192$.