Answer
Vertex: $(3,-2)$
Axis of Symmetry: $x=3$
Domain: all real numbers
Range: $\{y|y\ge-2\}$
Graph of $f(x)=\dfrac{4}{3}(x-3)^2-2$
Work Step by Step
Using $f(x)=a(x-h)^2+k$ or the standard form of quadratic functions, the given function, $
f(x)=\dfrac{4}{3}(x-3)^2-2
,$ has $h=
3
$ and $k=
-2
$.
Since the vertex is given by $(h,k)$, then the vertex of the given quadratic function is $
(3,-2)
$.
With the axis of symmetry given by $x=h$, then the equation of the axis of symmetry is $
x=3
$.
Let $y=f(x)$. Then $
y=\dfrac{4}{3}(x-3)^2-2
$. Substituting values of $x$ and solving for $y$ results to
\begin{array}{l|r}
\text{If }x=-3: & \text{If }x=0:
\\\\
y=\dfrac{4}{3}(x-3)^2-2 & y=\dfrac{4}{3}(x-3)^2-2
\\\\
y=\dfrac{4}{3}(-3-3)^2-2 & y=\dfrac{4}{3}(0-3)^2-2
\\\\
y=\dfrac{4}{3}(-6)^2-2 & y=\dfrac{4}{3}(-3)^2-2
\\\\
y=\dfrac{4}{3}(36)-2 & y=\dfrac{4}{3}(9)-2
\\\\
y=48-2 & y=12-2
\\
y=46 & y=10
.\end{array}
Hence, the points $
(-3,46)
$ and $
(0,10)
$ are on the graph of the parabola. Reflecting these points about the axis of symmetry gives the points $
(6,10)
$ and $
(9,46)
$.
The graph (shown above) is determined using the points $\{
(-3,46),(0,10),(3,-2),(6,10),(9,46)
\}$.
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-2\}
.$
