Quadratic Vertex D5Aa0F

1. **State the problem:** We are given the quadratic function $$f(x) = -x^2 - 3x + 3$$ and need to express it in vertex form. 2. **Recall the vertex form:** The vertex form of a quadratic function is $$f(x) = a(x-h)^2 + k$$ where $$(h,k)$$ is the vertex. 3. **Complete the square:** Start with $$f(x) = -x^2 - 3x + 3$$. 4. Factor out the coefficient of $$x^2$$ from the first two terms: $$f(x) = -\left(x^2 + 3x\right) + 3$$ 5. To complete the square inside the parentheses, take half of the coefficient of $$x$$, which is $$\frac{3}{2}$$, and square it: $$\left(\frac{3}{2}\right)^2 = \frac{9}{4}$$ 6. Add and subtract $$\frac{9}{4}$$ inside the parentheses: $$f(x) = -\left(x^2 + 3x + \frac{9}{4} - \frac{9}{4}\right) + 3$$ 7. Group the perfect square trinomial and simplify: $$f(x) = -\left(\left(x + \frac{3}{2}\right)^2 - \frac{9}{4}\right) + 3$$ 8. Distribute the negative sign: $$f(x) = -\left(x + \frac{3}{2}\right)^2 + \frac{9}{4} + 3$$ 9. Combine constants: $$3 = \frac{12}{4}$$ so $$f(x) = -\left(x + \frac{3}{2}\right)^2 + \frac{9}{4} + \frac{12}{4} = -\left(x + \frac{3}{2}\right)^2 + \frac{21}{4}$$ 10. **Final vertex form:** $$\boxed{f(x) = -\left(x + \frac{3}{2}\right)^2 + \frac{21}{4}}$$ This shows the vertex at $$\left(-\frac{3}{2}, \frac{21}{4}\right)$$ and the parabola opens downward because $$a = -1 < 0$$.