Answer
Vertex: $(-1,2)$
Axis of Symmetry: $x=-1$
Domain: all real numbers
Range: $\{y|y\le2\}$
Graph of $f(x)=-\dfrac{1}{2}(x+1)^2+2$
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{1}{2}(x+1)^2+2
,$ has $h=
-1
$ and $k=
2
$.
Since the vertex is given by $(h,k)$, then the vertex of the given quadratic function is $
(-1,2)
$.
With the axis of symmetry given by $x=h$, then the equation of the axis of symmetry is $
x=-1
$.
Let $y=f(x)$. Then $
y=-\dfrac{1}{2}(x+1)^2+2
$. Substituting values of $x$ and solving for $y$ results to
\begin{array}{l|r}
\text{If }x=-5: & \text{If }x=-3:
\\\\
y=-\dfrac{1}{2}(x+1)^2+2 & y=-\dfrac{1}{2}(x+1)^2+2
\\\\
y=-\dfrac{1}{2}(-5+1)^2+2 & y=-\dfrac{1}{2}(-3+1)^2+2
\\\\
y=-\dfrac{1}{2}(-4)^2+2 & y=-\dfrac{1}{2}(-2)^2+2
\\\\
y=-\dfrac{1}{2}(16)+2 & y=-\dfrac{1}{2}(4)+2
\\\\
y=-8+2 & y=-2+2
\\
y=-6 & y=0
.\end{array}
Hence, the points $
(-5,-6)
$ and $
(-3,0)
$ are on the graph of the parabola. Reflecting these points about the axis of symmetry gives the points $
(1,0)
$ and $
(3,-6)
$.
The graph (shown above) is determined using the points $\{
(-5,-6),(-3,0),(-1,2),(1,0),(3,-6)
\}$.
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\le2\}
.$
