Answer
Vertex: $(2,-3)$
Axis of Symmetry: $x=2$
Domain: all real numbers
Range: $\{y|y\ge-3\}$
Graph of $f(x)=2(x-2)^2-3$
Work Step by Step
Using $f(x)=a(x-h)^2+k$ or the standard form of quadratic functions, the given function, $
f(x)=2(x-2)^2-3
,$ has $h=
2
$ and $k=
-3
$.
Since the vertex is given by $(h,k)$, then the vertex of the given quadratic function is $
(2,-3)
$.
With the axis of symmetry given by $x=h$, then the equation of the axis of symmetry is $
x=2
$.
Let $y=f(x)$. Then $
y=2(x-2)^2-3
$. Substituting values of $x$ and solving for $y$ results to
\begin{array}{l|r}
\text{If }x=0: & \text{If }x=1:
\\\\
y=2(x-2)^2-3 & y=2(x-2)^2-3
\\
y=2(0-2)^2-3 & y=2(1-2)^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 $
(0,5)
$ and $
(1,-1)
$ are on the graph of the parabola. Reflecting these points about the axis of symmetry gives the points $
(3,-1)
$ and $
(4,5)
$.
The graph (shown above) is determined using the points $\{
(0,5),(1,-1),(2,-3),(3,-1),(4,5)
\}$.
Using the graph, 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\ge-3\}
.$
