Answer
A function
$f(x)=-2(x-1)^2+3$
Domain: set of all real numbers
Range: $\{y|y\le3\}$
Graph of $y=-2(x-1)^2+3$
Work Step by Step
Since the vertex of the equation $y=a(x-h)^2+k$ is given by $(h,k)$, then the vertex of the given quadratic equation, $
y=-2(x-1)^2+3
$, is $
\left(1,3\right)
$.
The axis of symmetry of the equation $y=a(x-h)^2+k$ is given by $x=h$. With $h=
1
$ then the axis of symmetry is $
x=1
$.
To graph the parabola, find points that are on the parabola. This can be done by substituting values of $x$ and solving the corresponding value of $y$. Substituting values of $x$ and solving $y$ results to
\begin{array}{l|r}
\text{If }x=-1: & \text{If }x=0:
\\\\
y=-2(-1-1)^2+3 & y=-2(0-1)^2+3
\\
y=-2(-2)^2+3 & y=-2(-1)^2+3
\\
y=-2(4)+3 & y=-2(1)+3
\\
y=-8+3 & y=-2+3
\\
y=-5 & y=1
.\end{array}
Hence, the points $
(-1,-5)
$ and $
(0,1)
$ are on the parabola. Reflecting these points about the axis of symmetry, the points $
(2,1)
$ and $
(3,-5)
$ are also on the parabola.
Using the points $\{
(-1,-5), (0,1),
\left(1,3\right),
(2,1), (3,-5)
\}$ the graph of the parabola is determined (see graph above).
Using the graph, the domain (values of $x$ used in the graph) is the set of all real numbers. The range (values of $y$ used in the graph) is $
\{y|y\le3\}
$.
Using the graph too, any vertical line drawn on the plane will intersect the graph at only $1$ point. Hence, $y=-2(x-1)^2+3$ is a function and can be written as $f(x)=-2(x-1)^2+3$.
