Answer
Vertex: $(-3,4)$
Axis of Symmetry: $x=-3$
Domain: all real numbers
Range: $\{y|y\le4\}$
Graph of $f(x)=-2(x+3)^2+4$
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+3)^2+4
,$ has $h=
-3
$ and $k=
4
$.
Since the vertex is given by $(h,k)$, then the vertex of the given quadratic function is $
(-3,4)
$.
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=-2(x+3)^2+4
$. Substituting values of $x$ and solving for $y$ results to
\begin{array}{l|r}
\text{If }x=-5: & \text{If }x=-4:
\\\\
y=-2(x+3)^2+4 & y=-2(x+3)^2+4
\\
y=-2(-5+3)^2+4 & y=-2(-4+3)^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 $
(-5,-4)
$ and $
(-4,2)
$ are on the graph of the parabola. Reflecting these points about the axis of symmetry gives the points $
(-2,2)
$ and $
(-1,-4)
$.
The graph (shown above) is determined using the points $\{
(-5,-4),(-4,2),(-3,4),(-2,2),(-1,-4)
\}$.
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\le4\}
.$
