Answer
See below:
Work Step by Step
Consider the quadratic function:
$f\left( x \right)=-3{{x}^{2}}+6x-2$
Now, consider the equation of the vertex for the quadratic formula.
$x=\frac{-b}{2a}$
Substitute the values in the above equation.
$\begin{align}
& x=\frac{-6}{2\left( -3 \right)} \\
& =1
\end{align}$
This shows that the vertex is located, when x is equal to1.
Substitute, x as 1 in the provided equation:
$\begin{align}
& y=-3{{\left( 1 \right)}^{2}}+6\left( 1 \right)-2 \\
& =-3+6-2 \\
& =1
\end{align}$
This shows that the vertex occurs at the coordinate, $\left( 1,1 \right)$.
Now, substitute x as 0 in the above function to calculate the y intercept.
$\begin{align}
& y=-3{{\left( 0 \right)}^{2}}+6\left( 0 \right)-2 \\
& =-2
\end{align}$
This shows that the y intercept occurs at the coordinate, $\left( 0,-2 \right)$.
Now, substitute y as 0 in the above function to calculate the x intercept.
$0=-3{{x}^{2}}+6x-2$
It is further solved using the quadratic function.
$\begin{align}
& x=\frac{-6\pm \sqrt{{{6}^{2}}-\left( 4\times \left( -3 \right)\times \left( -2 \right) \right)}}{2\times \left( -3 \right)} \\
& =\frac{-6\pm 3.464}{-6} \\
& =\frac{-6+3.464}{-6}\ \ \text{or}\ \ \frac{-6-3.464}{-6} \\
& =0.422\ \ \text{or}\ \ 1.577
\end{align}$
This shows that the x intercept occurs at the coordinates, $\left( 1.577,0 \right)$ and $\left( 0.422,0 \right)$.
