1. The problem asks to rewrite the quadratic function $f(x) = -2x^2 + 6x - 1$ in the vertex form $a(x+h)^2 + k$, where $a$, $h$, and $k$ are constants. 2. Start with the given function: $$f(x) = -2x^2 + 6x - 1$$ 3. Factor out the coefficient of $x^2$ from the quadratic terms: $$f(x) = -2(x^2 - 3x) - 1$$ 4. Complete the square inside the parentheses. To complete the square, take half of the coefficient of $x$, which is $-3$, so half is $-\frac{3}{2}$. Square it: $$\left(-\frac{3}{2}\right)^2 = \frac{9}{4}$$ 5. Add and subtract $\frac{9}{4}$ inside the parentheses to keep the expression equivalent: $$f(x) = -2 \left(x^2 - 3x + \frac{9}{4} - \frac{9}{4}\right) - 1$$ 6. Group the perfect square trinomial and the constant term: $$f(x) = -2 \left(\left(x - \frac{3}{2}\right)^2 - \frac{9}{4}\right) - 1$$ 7. Distribute $-2$: $$f(x) = -2 \left(x - \frac{3}{2}\right)^2 + 2 \times \frac{9}{4} - 1 = -2 \left(x - \frac{3}{2}\right)^2 + \frac{9}{2} - 1$$ 8. Simplify the constants: $$\frac{9}{2} - 1 = \frac{9}{2} - \frac{2}{2} = \frac{7}{2}$$ 9. The vertex form is therefore: $$f(x) = -2 \left(x - \frac{3}{2}\right)^2 + \frac{7}{2}$$ Final answer: $f(x) = -2(x - \frac{3}{2})^2 + \frac{7}{2}$ where $a = -2$, $h = -\frac{3}{2}$, and $k = \frac{7}{2}$.