1. **State the problem:** Find the maximum point of the quadratic function $f(x) = -2x^2 + 6x - 3$. 2. **Formula and rules:** For a quadratic function $ax^2 + bx + c$, the vertex (maximum or minimum point) is at $x = -\frac{b}{2a}$. 3. **Calculate the vertex x-coordinate:** Here, $a = -2$ and $b = 6$, so $$x = -\frac{6}{2 \times (-2)} = -\frac{6}{-4} = \frac{3}{2}.$$ 4. **Calculate the y-coordinate of the vertex:** Substitute $x = \frac{3}{2}$ into the function: $$f\left(\frac{3}{2}\right) = -2\left(\frac{3}{2}\right)^2 + 6\left(\frac{3}{2}\right) - 3 = -2 \times \frac{9}{4} + 9 - 3 = -\frac{18}{4} + 6 = -4.5 + 6 = 1.5.$$ 5. **Interpretation:** Since $a = -2 < 0$, the parabola opens downward, so the vertex is a maximum point. **Final answer:** The maximum point is at $\left(\frac{3}{2}, 1.5\right)$.