1. **State the problem:** We are given the quadratic function $$f(x) = 4 - (x - 3)^2$$ and asked to sketch its graph using the vertex and intercepts, find the equation of the axis of symmetry, and identify the domain and range. 2. **Identify the vertex:** The function is in vertex form $$f(x) = a(x - h)^2 + k$$ where the vertex is at $$(h, k)$$. Here, $$a = -1$$ (since the squared term is subtracted), $$h = 3$$, and $$k = 4$$. So, the vertex is at $$(3, 4)$$. 3. **Axis of symmetry:** The axis of symmetry is the vertical line passing through the vertex, given by $$x = h$$. Thus, the axis of symmetry is $$x = 3$$. 4. **Find intercepts:** - **y-intercept:** Set $$x = 0$$: $$f(0) = 4 - (0 - 3)^2 = 4 - 9 = -5$$ So, the y-intercept is $$(0, -5)$$. - **x-intercepts:** Set $$f(x) = 0$$: $$0 = 4 - (x - 3)^2$$ Rearranged: $$(x - 3)^2 = 4$$ Take square root: $$x - 3 = \pm 2$$ So, $$x = 3 + 2 = 5$$ or $$x = 3 - 2 = 1$$ The x-intercepts are $$(1, 0)$$ and $$(5, 0)$$. 5. **Domain:** Quadratic functions are defined for all real numbers, so the domain is: $$\text{Domain} = (-\infty, \infty)$$ 6. **Range:** Since $$a = -1 < 0$$, the parabola opens downward, and the vertex is the maximum point. The maximum value of $$f(x)$$ is $$4$$ at $$x = 3$$. Thus, the range is: $$\text{Range} = (-\infty, 4]$$ --- **Final answers:** - Vertex: $$(3, 4)$$ - Axis of symmetry: $$x = 3$$ - x-intercepts: $$(1, 0)$$ and $$(5, 0)$$ - y-intercept: $$(0, -5)$$ - Domain: $$(-\infty, \infty)$$ - Range: $$(-\infty, 4]$$