Answer
Vertex: $(1,1)$
Opening: up
Shape: narrower
Vertical Axis
Discriminant: -12
Number of $x$-intercepts: 0
Work Step by Step
To find the vertex, the given function, $
f(x)=3x^2-6x+4
,$ should be converted in the form $f(x)=a(x-h)^2+k$.
Grouping the variables together and making the coefficient of $x^2$ equal to $1$, the given function is equivalent to
\begin{align*}
f(x)&=(3x^2-6x)+4
\\
f(x)&=3(x^2-2x)+4
.\end{align*}
Completing the square of the expression with variables by adding $\left(\dfrac{b}{2}\right)^2,$ the expression above is equivalent to
\begin{align*}
f(x)&=3\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)+\left[4-3\left(\dfrac{-2}{2}\right)^2\right]
\\\\
f(x)&=3\left(x^2-2x+1\right)+\left[4-3\right]
\\
f(x)&=3\left(x-1\right)^2+1
.\end{align*}Note that $\left[a\left(\dfrac{b}{2}\right)^2\right]$ should be subtracted as well to negate the addition of that value when completing the square of the expression with variables.
Since the vertex of the quadratic function $f(x)=a(x-h)^2+k$ is given by $(h,k)$, then the vertex of the quadratic function above is $
(1,1)
$.
In the equation $
f(x)=3\left(x-1\right)^2+1
,$ the squared variable is $x.$ Thus, the parabola opens up or down. With $a=3$ (greater than zero), then the parabola opens up.
With $a=3$ (greater than $1$), then the shape of the parabola is narrower than $y=x^2$.
Since $x$ is the squared variable in the equation, then the parabola has a vertical axis. Using $b^2-4ac$ to find the number of $x$-intercepts, with $a=
3
$, $b=
-6
,$ and $c=
4
,$ then
\begin{align*}
b^2-4ac&\Rightarrow
(-6)^2-4(3)(4)
\\&=
36-48
\\&=
-12
.\end{align*}
With a discriminant that is less than zero, then the parabola does not have $x$-intercepts.
