Answer
Vertex: $(-4,-6)$
Axis of Symmetry: $x=-4$
Domain: all real numbers
Range: $\{y|y\ge-6\}$
Graph of $f(x)=x^2+8x+10$
Work Step by Step
To find the properties of the given function, $
f(x)=x^2+8x+10
,$ 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+8x)+10
.\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+8x+\left(\dfrac{8}{2}\right)^2\right)+\left[10-\left(\dfrac{8}{2}\right)^2\right]
\\\\
f(x)&=\left(x^2+8x+16\right)+\left[10-16\right]
\\
f(x)&=\left(x+4\right)^2-6
\\
f(x)&=\left(x-(-4)\right)^2-6
.\end{align*}(Note that $a\left(\dfrac{b}{2}\right)^2\Rightarrow
\left(\dfrac{8}{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(-4,-6\right)
$. Thus, the vertex is located on the second quadrant.
The axis of symmetry is given by $x=h$. Hence, the axis of symmetry of the parabola with the given equation is $
x=-4
$.
Let $y=f(x).$ Then the given equation is equivalent to $
y=x^2+8x+10
$. Substituting values for $x$ and solving for the resulting values of $y$, then
\begin{array}{l|r}
\text{If }x=-6: & \text{If }x=-5:
\\\\
y=(-6)^2+8(-6)+10 & y=(-5)^2+8(-5)+10
\\
y=36-48+10 & y=25-40+10
\\
y=-2 & y=-5
.\end{array}
Thus the points $
(-6,-2)
$ and $
(-5,-5)
$ are points on the parabola. Reflecting these points about the axis of symmetry, then $
(-3,-5)
$ and $
(-2,-2)
$ are also points on the parabola.
Using the points $\{
(-6,-2),(-5,-5),
(-4,-6),
(-3,-5),(-2,-2)
\}$ 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-6\}
$.
