1. **State the problem:** We analyze a parabola that opens downward, symmetric about the vertical line $x=2$, with x-intercepts near 0 and 4, vertex at $(2,9)$, and y-intercept near 5. 2. **Axis of symmetry:** The axis of symmetry of a parabola is a vertical line through the vertex's x-coordinate. Here, it is $x=2$. 3. **Direction of opening:** Since the parabola opens downward, the coefficient of $x^2$ in its equation is negative. 4. **Find the equation of the parabola:** Using vertex form: $$y = a(x - h)^2 + k$$ where $(h,k)$ is the vertex. Here, $h=2$, $k=9$. 5. **Find $a$ using a known point:** Use the y-intercept at $(0,5)$: $$5 = a(0 - 2)^2 + 9$$ $$5 = 4a + 9$$ $$4a = 5 - 9 = -4$$ $$a = -1$$ 6. **Equation of the parabola:** $$y = -1(x - 2)^2 + 9 = -(x - 2)^2 + 9$$ 7. **Find x-intercepts:** Set $y=0$: $$0 = -(x - 2)^2 + 9$$ $$(x - 2)^2 = 9$$ $$x - 2 = \\pm 3$$ $$x = 2 \\pm 3$$ So, $x= -1$ or $x=5$ (approximate from graph was 0 and 4, exact is -1 and 5). 8. **Find y-intercept:** Set $x=0$: $$y = -(0 - 2)^2 + 9 = -4 + 9 = 5$$ 9. **Vertex coordinates:** Given as $(2,9)$. **Final answers:** (a) Parabola opens downward. (b) Axis of symmetry: $x=2$ (c) x-intercepts: $-1, 5$ y-intercept: $5$ (d) Vertex: $(2,9)$