1. **State the problem:** We need to graph the quadratic function $$y = -\frac{1}{2}(x - 3)^2 + 5$$ and understand its shape and key features. 2. **Formula and important rules:** The function is in vertex form $$y = a(x - h)^2 + k$$ where $a = -\frac{1}{2}$, $h = 3$, and $k = 5$. - The vertex is at $(h, k) = (3, 5)$. - Since $a < 0$, the parabola opens downward. - The axis of symmetry is the vertical line $x = h = 3$. 3. **Intermediate work:** - The vertex is $(3, 5)$. - The parabola opens downward because $a = -\frac{1}{2} < 0$. - The parabola is wider than the standard parabola $y = x^2$ because $|a| = \frac{1}{2} < 1$. 4. **Plot key points:** - Vertex: $(3, 5)$ - Find points on either side of the vertex by plugging in values for $x$: - For $x = 2$, $$y = -\frac{1}{2}(2 - 3)^2 + 5 = -\frac{1}{2}(1)^2 + 5 = -\frac{1}{2} + 5 = 4.5$$ - For $x = 4$, $$y = -\frac{1}{2}(4 - 3)^2 + 5 = -\frac{1}{2}(1)^2 + 5 = 4.5$$ - For $x = 1$, $$y = -\frac{1}{2}(1 - 3)^2 + 5 = -\frac{1}{2}(2)^2 + 5 = -\frac{1}{2} \times 4 + 5 = -2 + 5 = 3$$ - For $x = 5$, $$y = -\frac{1}{2}(5 - 3)^2 + 5 = -\frac{1}{2}(2)^2 + 5 = 3$$ 5. **Summary:** The graph is a downward-opening parabola with vertex at $(3, 5)$, axis of symmetry $x = 3$, and points symmetric about the axis at $(2, 4.5)$, $(4, 4.5)$, $(1, 3)$, and $(5, 3)$.