How do you write the quadratic function \[y=-2{{x}^{2}}+6x-3\] in vertex form?

Hint: In this given problem, we have to find the vertex form for the given equation. We have to use the formula for vertex form to find the solution. We know that the vertex form of the quadratic equation of the form \[y=a{{x}^{2}}+bx+c\]is \[y=a{{\left( x-h \right)}^{2}}+k\left[ 1 \right]\], from this we have to compare the given equation to the general equation to find the value of a. We have to use some formulas to find the value of k and h, to get the vertex form.

Complete step by step answer:
We know that the given equation is,
\[y=-2{{x}^{2}}+6x-3\] ………. (1)
We also know that the vertex form of the quadratic equation of the form \[y=a{{x}^{2}}+bx+c\] is:
\[y=a{{\left( x-h \right)}^{2}}+k\left[ 1 \right]\] …….. (2)
Now we can compare the general equation and the equation (1), we get
a = -2, b = 6, c = -3.
We know that, to get the vector form, we have to find the value of k and h.
We know that,
\[h=-\dfrac{b}{2a}\] …….. (3)
 \[k=a{{h}^{2}}+bh+c\]……. (4)
Now we can find h from (3) by substituting the value of a and b.
\[\begin{align}
  & \Rightarrow h=\dfrac{-6}{2\left( -2 \right)} \\
 & \Rightarrow h=\dfrac{3}{2} \\
\end{align}\]
Now we can find k from (4) by substituting the value of a, b, c.
\[\begin{align}
  & \Rightarrow k=-2{{\left( \dfrac{3}{2} \right)}^{2}}+\left( 6 \right)\left( \dfrac{3}{2} \right)+\left( -3 \right) \\
 & \Rightarrow k=-\dfrac{9}{2}+\dfrac{18}{2}-3 \\
 & \Rightarrow k=\dfrac{9}{2}-3 \\
 & \Rightarrow k=\dfrac{3}{2} \\
\end{align}\]
Now we can substitute the value a, k, h in equation (2), we get
\[\Rightarrow y=-2{{\left( x-\dfrac{3}{2} \right)}^{2}}+\dfrac{3}{2}\]
Therefore, the vertex form of the equation \[y=-2{{x}^{2}}+6x-3\] is \[y=-2{{\left( x-\dfrac{3}{2} \right)}^{2}}+\dfrac{3}{2}\].

Note:
Students make mistakes in substituting the value of k and h, to find k and h we have formula. Student should remember the vertex form of the general equation \[y=a{{x}^{2}}+bx+c\] is \[y=a{{\left( x-h \right)}^{2}}+k\left[ 1 \right]\], where \[h=-\dfrac{b}{2a}\] and \[k=a{{h}^{2}}+bh+c\]. Students should understand and concentrate on the formula part to find these types of solutions.