91Ó°ÊÓ

Given to determine the coordinates of the maximum or minimum value of the quadratic equation $f\left(x\right)=-6x^2+9x$ to the nearest hundredth.

The maximum or minimum value of a quadratic function lies at the vertex of the graph.

For an equation of the form $f\left(x\right)=a{x}^2+bx+c$:

If $a>0$, it is an upwards opening parabola and so has a minimum value

If $a<0$, it is a downwards opening parabola and so has a maximum value.

So, the given quadratic equation has a maximum value.

For an equation of the form $f\left(x\right)=a{x}^2+bx+c$, the x-coordinate of the vertex is given by $x=-\frac{b}{2a}$

Here for the given equation, $a=-6,b=9$

Plugging the values in the equation:

$\begin{array}{l}x=−\frac{b}{2a}\\ x=−\frac{\left(9\right)}{2\left(−6\right)}\\ x=\frac{9}{12}\\ x=0.75\end{array}$

Hence the x-coordinate of the vertex is $x=0.75.$

So, the y-coordinate of the vertex can be obtained by plugging the value in the equation.

Plugging $x=0.75$ in the equation:

$\begin{array}{l}f\left(x\right)=−6x^2+9x\\ f\left(0.75\right)=−6{\left(0.75\right)}^2+9\left(0.75\right)\\ f\left(0.75\right)=−3.375+6.75\\ f\left(0.75\right)=3.375\end{array}$

Hence the coordinates of the vertex is $\left(0.75,3.375\right)$.