Answer
Vertex: $\left(1,-3\right)$
Axis of Symmetry: $x=1$
Domain: all real numbers
Range: $\{y|y\le-3\}$
Graph of $f(x)=-2x^2+4x-5$
Work Step by Step
To find the properties of the given function, $
f(x)=-2x^2+4x-5
,$ convert the equation in the form $y=a(x-h)^2+k$.
Grouping the $x$-variables together and making the coefficient of $x^2$ equal to $1$, the given equation 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 right-side expression by adding $\left(\dfrac{b}{2}\right)^2,$ the equation 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 $a\left(\dfrac{b}{2}\right)^2\Rightarrow
(-2)\left(\dfrac{-2}{2}\right)^2
$ should be subtracted as well to cancel out the term that was added to complete the square.)
Since the vertex of the equation $y=a(x-h)^2+k$ is given by $(h,k)$, then the vertex of the quadratic function above is $
\left(1,-3\right)
$.
The axis of symmetry is given by $x=h$. Hence, the axis of symmetry of the parabola with the given equation is $
x=1
$.
Let $y=f(x).$ Then the given equation is equivalent to $
y=-2x^2+4x-5
$. Substituting values for $x$ and solving for the resulting values of $y$, then
\begin{array}{l|r}
\text{If }x=-1: & \text{If }x=0:
\\\\
y=-2x^2+4x-5 & y=-2x^2+4x-5
\\
y=-2(-1)^2+4(-1)-5 & y=-2(0)^2+4(0)-5
\\
y=-2(1)+4(-1)-5 & y=-2(0)+4(0)-5
\\
y=-2-4-5 & y=0+0-5
\\
y=-11 & y=-5
.\end{array}
Thus the points $
(-1,-11)
$ and $
(0,-5)
$ are points on the parabola. Reflecting these points about the axis of symmetry, then $
(2,-5)
$ and $
(3,-11)
$ are also points on the parabola.
Using the points $\{
(-1,-11),(0,-5),
\left(1,-3\right),
(2,-5),(3,-11)
\}$ the graph of the given equation is derived (see graph above).
Using the graph above, the domain (all $x$-values used in the graph) is the set of all real numbers. The range (all $y$-values used in the graph) is $
\{y|y\le-3\}
$.
