Answer
Vertex: $\left(-1,-3\right)$
Axis of Symmetry: $x=-1$
Domain: all real numbers
Range: $\{y|y\ge-3\}$
Graph of $f(x)=x^2+2x-2$
Work Step by Step
To find the properties of the given function, $
f(x)=x^2+2x-2
,$ convert the equation in the form $y=a(x-h)^2+k$.
Grouping the $x$-variables together and making the coefficient of $x^2$ equal to $1$, the given equation is equivalent to
\begin{align*}
f(x)&=(x^2+2x)-2
.\end{align*}
Completing the square of the right-side expression by adding $\left(\dfrac{b}{2}\right)^2,$ the equation above is equivalent to
\begin{align*}
f(x)&=\left(x^2+2x+\left(\dfrac{2}{2}\right)^2\right)+\left[-2-\left(\dfrac{2}{2}\right)^2\right]
\\\\
f(x)&=\left(x^2+2x+1\right)+\left[-2-1\right]
\\
f(x)&=\left(x+1\right)^2-3
\\
f(x)&=\left(x-(-1)\right)^2-3
.\end{align*}(Note that $a\left(\dfrac{b}{2}\right)^2\Rightarrow
\left(\dfrac{2}{2}\right)^2
$ should be subtracted as well to cancel out the term that was added to complete the square.)
Since the vertex of the equation $y=a(x-h)^2+k$ is given by $(h,k)$, then the vertex of the quadratic function above is $
\left(-1,-3\right)
$.
The axis of symmetry is given by $x=h$. Hence, the axis of symmetry of the parabola with the given equation is $
x=-1
$.
Let $y=f(x).$ Then the given equation is equivalent to $
y=x^2+2x-2
$. Substituting values for $x$ and solving for the resulting values of $y$, then
\begin{array}{l|r}
\text{If }x=-3: & \text{If }x=-2:
\\\\
y=x^2+2x-2 & y=x^2+2x-2
\\
y=(-3)^2+2(-3)-2 & y=(-2)^2+2(-2)-2
\\
y=9-6-2 & y=4-4-2
\\
y=1 & y=-2
.\end{array}
Thus the points $
(-3,1)
$ and $
(-2,-2)
$ are points on the parabola. Reflecting these points about the axis of symmetry, then $
(0,-2)
$ and $
(1,1)
$ are also points on the parabola.
Using the points $\{
(-3,1),(-2,-2),
\left(-1,-3\right),
(0,-2),(1,1)
\}$ the graph of the given equation is derived (see graph above).
Using the graph above, 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-3\}
$.
