Find the coordinates of the maximum or minimum value of each quadratic equation to the nearest hundredth.

$f\left(x\right)=-4x^2+5x$

Given to determine the coordinates of the maximum or minimum value of the quadratic equation $f\left(x\right)=-4x^2+5x$ 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=-4,b=5

Plugging the values in the equation:

$\begin{array}{l}x=-\frac{b}{2a}\\ x=-\frac{\left(5\right)}{2\left(-4\right)}\\ x=\frac{5}{8}\\ x=0.625\\ x\approx 0.63\end{array}$

Hence the x-coordinate of the vertex is x=0.63.

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

Plugging x=0.625 in the equation:

$\begin{array}{l}f\left(x\right)=-4x^2+5x\\ f\left(0.625\right)=-4{\left(0.625\right)}^2+5\left(0.625\right)\\ f\left(0.625\right)=-1.5625+3.125\\ f\left(0.625\right)=1.5625\\ f\left(0.625\right)\approx 1.56\end{array}$

Hence the coordinates of the vertex is $\left(0.63,1.5625\right)$.

The coordinates of the maximum value of the quadratic equation $f\left(x\right)=-4x^2+5x$ to the nearest hundredth is $\left(0.63,1.5625\right)$.