Answer
Vertex: $\left(2,4\right)$
Axis of Symmetry: $x=2$
Domain: all real numbers
Range: $\{y|y\le4\}$
Graph of $f(x)=-3x^2+12x-8$
Work Step by Step
To find the properties of the given function, $
f(x)=-3x^2+12x-8
,$ 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)&=(-3x^2+12x)-8
\\
f(x)&=-3(x^2-4x)-8
.\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)&=-3\left(x^2-4x+\left(\dfrac{-4}{2}\right)^2\right)+\left[-8-(-3)\left(\dfrac{-4}{2}\right)^2\right]
\\\\
f(x)&=-3\left(x^2-4x+4\right)+\left[-8+12\right]
\\
f(x)&=-3\left(x-2\right)^2+4
.\end{align*}(Note that $a\left(\dfrac{b}{2}\right)^2\Rightarrow
(-3)\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(2,4\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=-3x^2+12x-8
$. Substituting values for $x$ and solving for the resulting values of $y$, then
\begin{array}{l|r}
\text{If }x=0: & \text{If }x=1:
\\\\
y=-3x^2+12x-8 & y=-3x^2+12x-8
\\
y=-3(0)^2+12(0)-8 & y=-3(1)^2+12(1)-8
\\
y=-3(0)+12(0)-8 & y=-3(1)+12(1)-8
\\
y=0+0-8 & y=-3+12-8
\\
y=-8 & y=1
.\end{array}
Thus the points $
(0,-8)
$ and $
(1,1)
$ are points on the parabola. Reflecting these points about the axis of symmetry, then $
(3,1)
$ and $
(4,-8)
$ are also points on the parabola.
Using the points $\{
(0,-8),(1,1),
\left(2,4\right),
(3,1),(4,-8)
\}$ 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\le4\}
$.
