Answer
Vertex: $(2,-4)$
Axis of Symmetry: $x=2$
Domain: all real numbers
Range: $\{y|y\ge-4\}$
Graph of $f(x)=2(x-2)^2-4$
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)^2-4
,$ has $h=
2
$ and $k=
-4
$.
Since the vertex is given by $(h,k)$, then the vertex of the given quadratic function is $
(2,-4)
$.
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)^2-4
$. 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)^2-4 & y=2(x-2)^2-4
\\
y=2(0-2)^2-4 & y=2(1-2)^2-4
\\
y=2(-2)^2-4 & y=2(-1)^2-4
\\
y=2(4)-4 & y=2(1)-4
\\
y=8-4 & y=2-4
\\
y=4 & y=-2
.\end{array}
Hence, the points $
(0,4)
$ and $
(1,-2)
$ are on the graph of the parabola. Reflecting these points about the axis of symmetry gives the points $
(3,-2)
$ and $
(4,4)
$.
The graph (shown above) is determined using the points $\{
(0,4),(1,-2),(2,-4),(3,-2),(4,4)
\}$.
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-4\}
.$
