1. The problem is to rewrite the quadratic function $f(x)=-2x^2+6x-1$ in the form $a(x+h)^2+k$ where $a$, $h$, and $k$ are constants. 2. Start by factoring out the coefficient of $x^2$ from the first two terms: $$f(x)=-2(x^2-3x)-1$$ 3. Complete the square inside the parentheses. Take half the coefficient of $x$, which is $-3$, divide by 2 to get $\frac{-3}{2}=-\frac{3}{2}$, then square it: $$\left(-\frac{3}{2}\right)^2=\frac{9}{4}$$ 4. Add and subtract $\frac{9}{4}$ inside the parentheses to keep the expression equivalent: $$f(x)=-2\left(x^2-3x+\frac{9}{4}-\frac{9}{4}\right)-1$$ 5. Group the perfect square trinomial and simplify the constants: $$f(x)=-2\left(\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right)-1$$ 6. Distribute $-2$ and combine constants: $$f(x)=-2\left(x-\frac{3}{2}\right)^2+2\times\frac{9}{4}-1$$ $$f(x)=-2\left(x-\frac{3}{2}\right)^2+\frac{9}{2}-1$$ $$f(x)=-2\left(x-\frac{3}{2}\right)^2+\frac{7}{2}$$ Answer: $$f(x)=-2\left(x-\frac{3}{2}\right)^2+\frac{7}{2}$$ Thus, $a=-2$, $h=-\frac{3}{2}$, and $k=\frac{7}{2}$.