1. **State the problem:** We need to sketch the graph of the quadratic function $$y = -x^2 + 4x - 4$$. 2. **Recall the standard form and vertex formula:** A quadratic function can be written as $$y = ax^2 + bx + c$$. Here, $$a = -1$$, $$b = 4$$, and $$c = -4$$. The vertex of the parabola is given by $$x = -\frac{b}{2a}$$. 3. **Calculate the vertex:** $$x = -\frac{4}{2 \times (-1)} = -\frac{4}{-2} = 2$$. Substitute $$x=2$$ into the function to find $$y$$: $$y = -(2)^2 + 4(2) - 4 = -4 + 8 - 4 = 0$$. So, the vertex is at $$(2, 0)$$. 4. **Determine the direction of the parabola:** Since $$a = -1 < 0$$, the parabola opens downward. 5. **Find the y-intercept:** Set $$x=0$$: $$y = -0 + 0 - 4 = -4$$. So, the y-intercept is $$(0, -4)$$. 6. **Find the x-intercepts:** Solve $$-x^2 + 4x - 4 = 0$$. Multiply both sides by $$-1$$ to simplify: $$\cancel{-1} \times (-x^2 + 4x - 4) = \cancel{-1} \times 0$$ $$x^2 - 4x + 4 = 0$$. This factors as: $$(x - 2)^2 = 0$$. So, the only root is $$x = 2$$, which matches the vertex x-coordinate. 7. **Summary:** The parabola opens downward, has vertex at $$(2, 0)$$, touches the x-axis at $$(2, 0)$$, and crosses the y-axis at $$(0, -4)$$. This is a downward-opening parabola with a single root at the vertex. **Final answer:** The graph is a downward parabola with vertex at $$(2, 0)$$, y-intercept at $$(0, -4)$$, and x-intercept at $$(2, 0)$$.