Answer
$\left(\dfrac{2}{3},-\dfrac{2}{3}\right)$
Work Step by Step
To find the vertex of the given equation, $
y=-3x^2+4x-2
,$ 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*}
y&=(-3x^2+4x)-2
\\\\
y&=-3\left(x^2-\dfrac{4}{3}x\right)-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*}
y&=-3\left(x^2-\dfrac{4}{3}x+\left(\dfrac{-4/3}{2}\right)^2\right)+\left[-2-\left(-3\right)\left(\dfrac{-4/3}{2}\right)^2\right]
\\\\
y&=-3\left(x^2-\dfrac{4}{3}x+\dfrac{4}{9}\right)+\left[-2+\dfrac{4}{3}\right]
\\\\
y&=-3\left(x-\dfrac{2}{3}\right)^2+\left[-\dfrac{6}{3}+\dfrac{4}{9}\right]
\\\\
y&=-3\left(x-\dfrac{2}{3}\right)^2-\dfrac{2}{3}
.\end{align*}(Note that $a\left(\dfrac{b}{2}\right)^2\Rightarrow
\left(-3\right)\left(\dfrac{-4/3}{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 equation above is $
\left(\dfrac{2}{3},-\dfrac{2}{3}\right)
$.
