Answer
Vertex: $\left(-\dfrac{1}{2},-3\right)$
Axis of Symmetry: $x=-\dfrac{1}{2}$
Domain: set of all real numbers
Range: $\{y|y\ge-3\}$
Graph of $f(x)=4x^2+4x-2$
Work Step by Step
To find the properties of the given fuction, $
f(x)=4x^2+4x-2
,$ convert to the form $f(x)=a(x-h)^2+k$.
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)&=(4x^2+4x)-2
\\
f(x)&=4(x^2+x)-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)&=4\left(x^2+x+\left(\dfrac{1}{2}\right)^2\right)+\left[-2-4\left(\dfrac{1}{2}\right)^2\right]
\\\\
f(x)&=4\left(x^2+x+\dfrac{1}{4}\right)+\left[-2-1\right]
\\
f(x)&=4\left(x+\dfrac{1}{2}\right)^2-3
.\end{align*}(Note that $a\left(\dfrac{b}{2}\right)^2\Rightarrow
4\left(\dfrac{1}{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 function $f(x)=a(x-h)^2+k$ is given by $(h,k)$, then the vertex of the quadratic function, $
f(x)=4\left(x+\dfrac{1}{2}\right)^2-3
$, is $
\left(-\dfrac{1}{2},-3\right)
$.
The axis of symmetry of the function $f(x)=a(x-h)^2+k$ is given by $x=h$. With $h=
-\dfrac{1}{2}
$ then the axis of symmetry is $
x=-\dfrac{1}{2}
$.
To graph the parabola, find points that are on the parabola. This can be done by substituting values of $x$ and solving the corresponding value of $y$. Let $y=f(x).$ Then $
y=4x^2+4x-2
$. Substituting values of $x$ and solving $y$ results to
\begin{array}{l|r}
\text{If }x=-2: & \text{If }x=-1:
\\\\
y=4x^2+4x-2 & y=4x^2+4x-2
\\
y=4(-2)^2+4(-2)-2 & y=4(-1)^2+4(-1)-2
\\
y=4(4)+4(-2)-2 & y=4(1)+4(-1)-2
\\
y=16-8-2 & y=4-4-2
\\
y=6 & y=-2
.\end{array}
Hence, the points $
(-2,6)
$ and $
(-1,-2)
$ are on the parabola. Reflecting these points about the axis of symmetry, the points $
(0,-2)
$ and $
(1,6)
$ are also on the parabola.
Using the points $\{
(-2,6), (-1,-2),
\left(-\dfrac{1}{2},-3\right),
(0,-2), (1,6)
\}$ the graph of the parabola is determined (see graph above).
Using the graph, the domain (values of $x$ used in the graph) is the set of all real numbers. The range (values of $y$ used in the graph) is $
\{y|y\ge-3\}
$.
