1. **State the problem:** We need to graph the parabola given by the equation $$y = -x^2 + 4x - 2$$ and plot five points: the vertex, two points to the left, and two points to the right of the vertex. 2. **Formula and rules:** The general form of a quadratic function is $$y = ax^2 + bx + c$$ where $a$, $b$, and $c$ are constants. The vertex of the parabola is found using the formula $$x = -\frac{b}{2a}$$. 3. **Find the vertex:** Here, $a = -1$, $b = 4$, and $c = -2$. Calculate the $x$-coordinate of the vertex: $$x = -\frac{4}{2 \times (-1)} = -\frac{4}{-2} = 2$$ Calculate the $y$-coordinate by substituting $x=2$ into the equation: $$y = -(2)^2 + 4(2) - 2 = -4 + 8 - 2 = 2$$ So, the vertex is at $(2, 2)$. 4. **Find points to the left and right of the vertex:** - For $x=1$: $$y = -(1)^2 + 4(1) - 2 = -1 + 4 - 2 = 1$$ - For $x=3$: $$y = -(3)^2 + 4(3) - 2 = -9 + 12 - 2 = 1$$ - For $x=0$: $$y = -(0)^2 + 4(0) - 2 = 0 + 0 - 2 = -2$$ - For $x=4$: $$y = -(4)^2 + 4(4) - 2 = -16 + 16 - 2 = -2$$ 5. **Summary of points:** - Vertex: $(2, 2)$ - Left points: $(1, 1)$ and $(0, -2)$ - Right points: $(3, 1)$ and $(4, -2)$ 6. **Graph shape:** Since $a = -1 < 0$, the parabola opens downward. Final answer: The parabola $y = -x^2 + 4x - 2$ has vertex at $(2, 2)$ and passes through points $(1, 1)$, $(0, -2)$, $(3, 1)$, and $(4, -2)$, forming a downward-opening parabola.