Answer
$(1,-3)$
Work Step by Step
To find the vertex, the given function, $
f(x)=-2x^2+4x-5
,$ should be converted in the form $f(x)=a(x-h)^2+k$.
Grouping the variables together and making the coefficient of $x^2$ equal to $1$, the given function is equivalent to
\begin{align*}
f(x)&=(-2x^2+4x)-5
\\
f(x)&=-2(x^2-2x)-5
.\end{align*}
Completing the square of the expression with variables by adding $\left(\dfrac{b}{2}\right)^2,$ the expression above is equivalent to
\begin{align*}
f(x)&=-2\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)+\left[-5-(-2)\left(\dfrac{-2}{2}\right)^2\right]
\\\\
f(x)&=-2\left(x^2-2x+1\right)+\left[-5+2\right]
\\\\
f(x)&=-2\left(x-1\right)^2-3
.\end{align*}Note that $\left[a\left(\dfrac{b}{2}\right)^2\right]$ should be subtracted as well to negate the addition of that value when completing the square of the expression with variables.
Since the vertex of the quadratic function $f(x)=a(x-h)^2+k$ is given by $(h,k)$, then the vertex of the quadratic function above is $
(1,-3)
$.
