Answer
Vertex: $(-2,1)$
Axis of Symmetry: $x=-2$
Domain: all real numbers
Range: $\{y|y\le1\}$
Graph of $f(x)=-\dfrac{2}{3}(x+2)^2+1$
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{2}{3}(x+2)^2+1
,$ has $h=
-2
$ and $k=
1
$.
Since the vertex is given by $(h,k)$, then the vertex of the given quadratic function is $
(-2,1)
$.
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=-\dfrac{2}{3}(x+2)^2+1
$. Substituting values of $x$ and solving for $y$ results to
\begin{array}{l|r}
\text{If }x=-8: & \text{If }x=-5:
\\\\
y=-\dfrac{2}{3}(x+2)^2+1 & y=-\dfrac{2}{3}(x+2)^2+1
\\\\
y=-\dfrac{2}{3}(-8+2)^2+1 & y=-\dfrac{2}{3}(-5+2)^2+1
\\\\
y=-\dfrac{2}{3}(-6)^2+1 & y=-\dfrac{2}{3}(-3)^2+1
\\\\
y=-\dfrac{2}{3}(36)+1 & y=-\dfrac{2}{3}(9)+1
\\\\
y=-24+1 & y=-6+1
\\
y=-23 & y=-5
.\end{array}
Hence, the points $
(-8,-23)
$ and $
(-5,-5)
$ are on the graph of the parabola. Reflecting these points about the axis of symmetry gives the points $
(1,-5)
$ and $
(4,-23)
$.
The graph (shown above) is determined using the points $\{
(-8,-23),(-5,-5),(-2,1),(1,-5),(4,-23)
\}$.
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\le1\}
.$
