1. **State the problem:** Find the maximum and minimum points of the quadratic function $f(x) = -2x^2 + 6x - 3$. 2. **Formula and rules:** For a quadratic function $f(x) = ax^2 + bx + c$, the vertex (which gives the max or min point) is at $x = -\frac{b}{2a}$. - If $a > 0$, the parabola opens upwards and the vertex is a minimum point. - If $a < 0$, the parabola opens downwards and the vertex is a maximum point. 3. **Calculate the vertex:** Here, $a = -2$, $b = 6$. $$x = -\frac{6}{2 \times (-2)} = -\frac{6}{-4} = \frac{3}{2}$$ 4. **Find the corresponding $y$ value:** $$f\left(\frac{3}{2}\right) = -2\left(\frac{3}{2}\right)^2 + 6\left(\frac{3}{2}\right) - 3$$ $$= -2 \times \frac{9}{4} + 9 - 3 = -\frac{18}{4} + 6 = -4.5 + 6 = 1.5$$ 5. **Interpretation:** Since $a = -2 < 0$, the parabola opens downward, so the vertex at $\left(\frac{3}{2}, 1.5\right)$ is a maximum point. 6. **Minimum points:** There is no minimum point for this parabola as it opens downward and extends to negative infinity. **Final answer:** The function has a maximum point at $\left(\frac{3}{2}, 1.5\right)$ and no minimum point.