1. **State the problem:** Convert the quadratic function $f(x) = -2x^2 - 4x + 5$ from standard form to vertex form. 2. **Recall the vertex form:** The vertex form of a quadratic is $$f(x) = a(x-h)^2 + k$$ where $(h,k)$ is the vertex. 3. **Start with the given function:** $$f(x) = -2x^2 - 4x + 5$$ 4. **Factor out the coefficient of $x^2$ from the first two terms:** $$f(x) = -2(x^2 + 2x) + 5$$ 5. **Complete the square inside the parentheses:** - Take half of the coefficient of $x$, which is 2, so half is 1. - Square it: $1^2 = 1$. - Add and subtract 1 inside the parentheses to complete the square: $$f(x) = -2(x^2 + 2x + 1 - 1) + 5$$ 6. **Rewrite the trinomial as a perfect square and simplify:** $$f(x) = -2((x + 1)^2 - 1) + 5$$ 7. **Distribute the $-2$ and simplify:** $$f(x) = -2(x + 1)^2 + 2 + 5$$ $$f(x) = -2(x + 1)^2 + 7$$ 8. **Final vertex form:** $$f(x) = -2(x + 1)^2 + 7$$ This shows the vertex is at $(-1, 7)$ and the parabola opens downward because $a = -2 < 0$.