Answer
Vertex: $\left(-2,-1\right)$
Axis of Symmetry: $x=-2$
Domain: all real numbers
Range: $\{y|y\ge-1\}$
Graph of $f(x)=x^2+4x+3$
Work Step by Step
To find the properties of the given function, $
f(x)=x^2+4x+3
,$ 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+4x)+3
.\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+4x+\left(\dfrac{4}{2}\right)^2\right)+\left[3-\left(\dfrac{4}{2}\right)^2\right]
\\\\
f(x)&=\left(x^2+4x+4\right)+\left[3-4\right]
\\
f(x)&=\left(x+2\right)^2-1
\\
f(x)&=\left(x-(-2)\right)^2-1
.\end{align*}(Note that $a\left(\dfrac{b}{2}\right)^2\Rightarrow
\left(\dfrac{4}{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(-2,-1\right)
$.
The axis of symmetry is given by $x=h$. Hence, the axis of symmetry of the parabola with the given equation is $
x=-2
$.
Let $y=f(x).$ Then the given equation is equivalent to $
y=x^2+4x+3
$. Substituting values for $x$ and solving for the resulting values of $y$, then
\begin{array}{l|r}
\text{If }x=-4: & \text{If }x=-3:
\\\\
y=x^2+4x+3 & y=x^2+4x+3
\\
y=(-4)^2+4(-4)+3 & y=(-3)^2+4(-3)+3
\\
y=16-16+3 & y=9-12+3
\\
y=3 & y=0
.\end{array}
Thus the points $
(-4,3)
$ and $
(-3,0)
$ are points on the parabola. Reflecting these points about the axis of symmetry, then $
(-1,0)
$ and $
(0,3)
$ are also points on the parabola.
Using the points $\{
(-4,3),(-3,0),
\left(-2,-1\right),
(-1,0),(0,3)
\}$ 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-1\}
$.
