91Ó°ÊÓ

In a quadratic equation $y=a{x}^2+bx+c$, if the coefficient of$x^2$ is positive, then the graph of the parabola is upward open and vertex is at minimum and if the coefficient of$x^2$ is negative, then the graph of the parabola is downward open and vertex is at maximum.

The graph of any quadratic function is a parabola.

Downward open parabola means the coefficient of $x^2$is negative. So, we can ignore the options B and C.

Now the vertex of the parabola is $\left(0,2\right)$.

Since in Option$\text{A}:y=\text{ â¶Ä‰}−\text{ â¶Ä‰}3x^2$ the equationhas the vertex at origin or$\left(0,0\right)$, therefore, this option is not valid.

In option $\text{D}:y=\text{ â¶Ä‰}−\text{ â¶Ä‰}3x^2+2$, the coefficient is negative and it also satisfies the coordinates $\left(0,2\right)$, therefore, option $\text{D}:y=\text{ â¶Ä‰}−\text{ â¶Ä‰}3x^2+2$is the correct choice.