1. The problem is to analyze the function $f(x) = -(x+1)^2 + 16$ and understand its graph. 2. This is a quadratic function in vertex form: $f(x) = a(x-h)^2 + k$, where $(h,k)$ is the vertex. 3. Here, $a = -1$, $h = -1$, and $k = 16$. Since $a$ is negative, the parabola opens downward. 4. The vertex is at $(-1,16)$, which is the highest point on the graph. 5. The axis of symmetry is the vertical line $x = -1$. 6. To find the y-intercept, set $x=0$: $$f(0) = -(0+1)^2 + 16 = -1 + 16 = 15$$ 7. To find the x-intercepts, set $f(x) = 0$: $$0 = -(x+1)^2 + 16$$ $$ (x+1)^2 = 16$$ $$x+1 = \pm 4$$ $$x = -1 \pm 4$$ So, $x = 3$ or $x = -5$. 8. Summary: The parabola opens downward, vertex at $(-1,16)$, y-intercept at $(0,15)$, and x-intercepts at $(3,0)$ and $(-5,0)$. Final answer: The graph is a downward-opening parabola with vertex $(-1,16)$, intercepts at $(0,15)$, $(3,0)$, and $(-5,0)$.