1. **State the problem:** We need to find the maximum or minimum value of the quadratic function $$y = -10x^2 + 20x - 5$$. 2. **Recall the formula:** For a quadratic function $$y = ax^2 + bx + c$$, the vertex (which gives the maximum or minimum) is at $$x = -\frac{b}{2a}$$. 3. **Identify coefficients:** Here, $$a = -10$$, $$b = 20$$, and $$c = -5$$. 4. **Calculate the vertex x-coordinate:** $$x = -\frac{20}{2 \times (-10)} = -\frac{20}{-20} = 1$$. 5. **Calculate the vertex y-coordinate by substituting $$x=1$$ into the function:** $$y = -10(1)^2 + 20(1) - 5 = -10 + 20 - 5 = 5$$. 6. **Determine if it is a maximum or minimum:** Since $$a = -10 < 0$$, the parabola opens downward, so the vertex is a maximum point. 7. **Final answer:** The maximum value of $$y$$ is $$5$$ at $$x = 1$$, so the vertex is $$(1, 5)$$. **Answer:** (1, 5)